Projectile Motion Experiment
By Monday, August 19, perform the following experiments online at the Projectile Motion – Galileo website. (http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html)
Fill in the tables, and post your answers to the questions.
Question 1
Procedure: Keep the initial velocity fixed at 50 m/s. Perform the experiment for each of the following angles:
Record in table 1:
Range
) traveled by the projectile at various angles
Table 1: Experiment 1 – Range & Time of flight
|
S. No. |
Initial Velocity |
Angle of Projection |
Time of Flight |
Maximum Height Attained |
||
Observe and Analyze:
Question 2
Procedure: Set the angle to 45 degrees. Perform the experiment with the following initial velocities.
Record in Table 2: The horizontal and vertical components of the velocity and the ranges for each of the velocities.
Table 2: Experiment 2 –
Range at 45°
|
Horizontal Velocity |
Vertical Velocity |
Observe and Analyze:
Attached are assignment related online lectures and textbook chapters
Projectile Motion Experiment
By Monday, August 19, perform the following
experiments online
at the Projectile Motion – Galileo website. (
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
)
Fill in the tables, and post your answers to the questions.
Question 1
Procedure: Keep the initial velocity fixed at 50 m/s. Perform the experiment for each of the following angles:
· 15 degrees
· 30 degrees
· 45 degrees
· 60 degrees
· 75 degrees
Record in table 1:
· The maximum horizontal distance (Range) traveled by the projectile at various angles
· The total time of flight at various angles
· The maximum height attained
Table 1: Experiment 1 – Range & Time of flight
|
S. No. |
Initial Velocity |
Angle of Projection |
Range |
Time of Flight
|
Maximum Height Attained |
|||||||||||||||||||||||||||||||||||||
|
1. |
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|
2. |
||||||||||||||||||||||||||||||||||||||||||
|
3. |
||||||||||||||||||||||||||||||||||||||||||
|
4. |
||||||||||||||||||||||||||||||||||||||||||
|
5. |
Observe and Analyze:
· What angle produces the maximum range? Why?
· What angle produces the maximum height? Why?
· Are there angles which produce the same range? If so, how would you explain this?
Question 2
Procedure: Set the angle to 45 degrees. Perform the experiment with the following initial velocities.
· 30 m/s
· 40 m/s
· 50 m/s
· 60 m/s
Record in Table 2: The horizontal and vertical components of the velocity and the ranges for each of the velocities.
Table 2: Experiment 2 – Range at 45°
|
Horizontal Velocity |
Vertical Velocity |
Range at 45° |
Observe and Analyze:
· As the velocity is increased what happens to the Range?
· Using the initial horizontal and vertical velocities from your table, verify mathematically that the range is correct for the initial speed of 60 m/s.
Attached are assignment related online lectures and textbook chapters
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7 ROTATIONAL MOTION AND THE LAW OF GRAVITY
Astronauts fall around the Earth at thousands of meters per second, held by the centripetal force provided by gravity.
Rotational motion is an important part of everyday life. The rotation of the Earth creates the cycle of day and night, the rotation of wheels enables easy vehicular motion, and modern technology depends on circular motion in a variety of contexts, from the tiny gears in a Swiss watch to the operation of lathes and other machinery. The concepts of angular speed, angular acceleration, and centripetal acceleration are central to understanding the motions of a diverse range of phenomena, from a car moving around a circular racetrack to clusters of galaxies orbiting a common center.
Rotational motion, when combined with Newton’s law of universal gravitation and his laws of motion, can also explain certain facts about space travel and satellite motion, such as where to place a satellite so it will remain fixed in position over the same spot on the Earth. The generalization of gravitational potential energy and energy conservation offers an easy route to such results as planetary escape speed. Finally, we present Kepler’s three laws of planetary motion, which formed the foundation of Newton’s approach to gravity.
In the study of linear motion, the important concepts are displacement Ax, velocity v, and acceleration a. Each of these concepts has its analog in rotational motion: angular displacement Δθ, angular velocity ω, and angular acceleration a.
The radian, a unit of angular measure, is essential to the understanding of these concepts. Recall that the distance s around a circle is given by s = 2πr, where r is the radius of the circle. Dividing both sides by r results in s/r = 2π. This quantity is dimensionless because both s and r have dimensions of length, but the value 2π corresponds to a displacement around a circle. A half circle would give an answer of p, a quarter circle an answer of π/2. The numbers 2π, π, and π/2 correspond to angles of 360°, 180°, and 90°, respectively, so a new unit of angular measure, the radian, can be introduced, with 180° = π rad relating degrees to radians.
The angle θ subtended by an arc length s along a circle of radius r, measured in radians counterclockwise from the positive x-axis, is
The angle θ in
Equation 7.1
is actually an angular displacement from the positive x-axis, and s the corresponding displacement along the circular arc, again measured from the positive x-axis.
Figure 7.1
illustrates the size of 1 radian, which is approximately 57°. Converting from degrees to radians requires multiplying by the ratio (π rad/180°). For example, 45° (π rad/180°) = (π/4) rad.
Equation 7.1
uses angles measured in radians. Angles expressed in terms of degrees must first be converted to radians. Also, be sure to check whether your calculator is in degree or radian mode when solving problems involving rotation.
For very short time intervals, the average angular speed approaches the instantaneous angular speed, just as in the linear case.
The instantaneous angular speed ω of a rotating rigid object is the limit of the average speed Δθ/Δt as the time interval Δt approaches zero:
SI unit: radian per second (rad/s)
We take ω to be positive when θ is increasing (counterclockwise motion) and negative when θ is decreasing (clockwise motion). When the angular speed is constant, the instantaneous angular speed is equal to the average angular speed.
Goal Convert an angular speed in revolutions per minute to radians per second.
Problem The rotor on a helicopter turns at an angular speed of 3.20 × 102 revolutions per minute. (In this book, we sometimes use the abbreviation rpm, but in most cases we use rev/min.) (a) Express this angular speed in radians per second. (b) If the rotor has a radius of 2.00 m, what arclength does the tip of the blade trace out in 3.00 × 102 s?
Strategy During one revolution, the rotor turns through an angle of 2π radians. Use this relationship as a conversion factor.
Solution
· (a) Express this angular speed in radians per second.Apply the conversion factors 1 rev = 2π rad and 60 s = 1 min:
· (b) Multiply the angular speed by the time to obtain the angular displacement:Multiply the angular displacement by the radius to get the arc length:
Remarks In general, it’s best to express angular speeds in radians per second. Consistent use of radian measure minimizes errors.
When a rigid object rotates about a fixed axis, as does the bicycle wheel, every portion of the object has the same angular speed and the same angular acceleration. This fact is what makes these variables so useful for describing rotational motion. In contrast, the tangential (linear) speed and acceleration of the object take different values that depend on the distance from a given point to the axis of rotation.
A number of parallels exist between the equations for rotational motion and those for linear motion. For example, compare the defining equation for the average angular speed,
with that of the average linear speed,
In these equations, v takes the place of v and θ takes the place of x, so the equations differ only in the names of the variables. In the same way, every linear quantity we have encountered so far has a corresponding “twin” in rotational motion.
The procedure used in
Section 2.5
to develop the kinematic equations for linear motion under constant acceleration can be used to derive a similar set of equations for rotational motion under constant angular acceleration. The resulting equations of rotational kinematics, along with the corresponding equations for linear motion, are as follows:
|
Linear Motion with a Constant (Variables: x and v) |
Rotational Motion About a Fixed Axis with a Constant (Variables: θ and ω) |
EXAMPLE 7.2 A Rotating Wheel
Goal Apply the rotational kinematic equations.
Problem A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s?
Strategy The angular acceleration is constant, so this problem just requires substituting given values into Equations 7.7 and 7.8.
Solution
· (a) Find the angular displacement after 2.00 s, in both radians and revolutions.Use Equation 7.8, setting ωi = 2.00 rad/s, α = 3.5 rad/s2, and t = 2.00 s:Convert radians to revolutions.
· (b) What is the angular speed of the wheel at t = 2.00 s?Substitute the same values into Equation 7.7:
Remarks The result of part (b) could also be obtained from Equation 7.9 and the results of part (a).
Goal Apply the time-independent rotational kinematic equation.
Problem An airplane propeller slows from an initial angular speed of 12.5 rev/s to a final angular speed of 5.00 rev/s. During this process, the propeller rotates through 21.0 revolutions. Find the angular acceleration of the propeller in radians per second squared, assuming it’s constant.
Strategy The given quantities are the angular speeds and the displacement, which suggests applying Equation 7.9, the time-independent rotational kinematic equation, to find a.
Solution
First, convert the angular displacement to radians and the angular speeds to rad/s:
Substitute these values into Equation 7.9 to find the angular acceleration a:
Solve for α:
Remark Waiting until the end to convert revolutions to radians is also possible and requires only one conversion instead of three.
The tangential acceleration of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular acceleration. Again, radian measure must be used for the angular acceleration term in this equation.
One last equation that relates linear quantities to angular quantities will be derived in the next section.
Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea’s angular speed is (a) twice Chuck’s (b) the same as Chuck’s (c) half of Chuck’s (d) impossible to determine.
When the merry-go-round of
Quick Quiz 7.4
is rotating at a constant angular speed, Andrea’s tangential speed is (a) twice Chuck’s (b) the same as Chuck’s (c) half of Chuck’s (d) impossible to determine.
Why is the launch area for the European Space Agency in South America and not in Europe?
Explanation Satellites are boosted into orbit on top of rockets, which provide the large tangential speed necessary to achieve orbit. Due to its rotation, the surface of Earth is already traveling toward the east at a tangential speed of nearly 1 700 m/s at the equator. This tangential speed is steadily reduced farther north because the distance to the axis of rotation is decreasing. It finally goes to zero at the North Pole. Launching eastward from the equator gives the satellite a starting initial tangential speed of 1 700 m/s, whereas a European launch provides roughly half that speed (depending on the exact latitude).
EXAMPLE 7.4 Compact Discs
Goal Apply the rotational kinematics equations in tandem with tangential acceleration and speed.
Problem A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, find the tangential speed of a microbe riding on the rim of the disc when t = 0.892 s. (d) What is the magnitude of the tangential acceleration of the microbe at the given time?
Strategy We can solve parts (a) and (b) by applying the kinematic equations for angular speed and angular displacement (Eqs. 7.7 and 7.8). Multiplying the radius by the angular acceleration yields the tangential acceleration at the rim, whereas multiplying the radius by the angular speed gives the tangential speed at that point.
Solution
· (a) Find the angular acceleration of the disc.Apply the angular velocity equation ω = ω i+ αt, taking ωi = 0 at t = 0:
· (b) Through what angle does the disc turn?Use Equation 7.8 for angular displacement, with t = 0.892 s and ωi = 0:
· (c) Find the final tangential speed of a microbe at r = 4.45 cm.Substitute into
Equation 7.10
:
· (d) Find the tangential acceleration of the microbe at r = 4.45 cm.Substitute into
Equation 7.11
:
Remarks Because 2π rad = 1 rev, the angular displacement in part (b) corresponds to 2.23 rev. In general, dividing the number of radians by 6 gives a good approximation to the number of revolutions, because 2π ~ 6.
Phonograph Records and Compact Discs
Before compact discs became the medium of choice for recorded music, phonographs were popular. There are similarities and differences between the rotational motion of phonograph records and that of compact discs. A phonograph record rotates at a constant angular speed. Popular angular speeds were 33⅓ rev/min for long-playing albums (hence the nickname “LP”), 45 rev/min for “singles,” and 78 rev/min used in very early recordings. At the outer edge of the record, the pickup needle (stylus) moves over the vinyl material at a faster tangential speed than when the needle is close to the center of the record. As a result, the sound information is compressed into a smaller length of track near the center of the record than near the outer edge.
CDs, on the other hand, are designed so that the disc moves under the laser pickup at a constant tangential speed. Because the pickup moves radially as it follows the tracks of information, the angular speed of the compact disc must vary according to the radial position of the laser. Because the tangential speed is fixed, the information density (per length of track) anywhere on the disc is the same.
Example 7.5
demonstrates numerical calculations for both compact discs and phonograph records.
Goal Relate angular to linear variables.
Problem In a compact disc player, as the read head moves out from the center of the disc, the angular speed of the disc changes so that the linear speed at the position of the head remains at a constant value of about 1.3 m/s. (a) Find the angular speed of the compact disc when the read head is at r = 2.0 cm and again at r = 5.6 cm. (b) An old-fashioned record player rotates at a constant angular speed, so the linear speed of the record groove moving under the detector (the stylus) changes. Find the linear speed of a 45.0-rpm record at points 2.0 and 5.6 cm from the center. (c) In both the CD and phonograph record, information is recorded in a continuous spiral track. Calculate the total length of the track for a CD designed to play for 1.0 h.
Strategy This problem is just a matter of substituting numbers into the appropriate equations. Part (a) requires relating angular and linear speed with
Equation 7.10
, vt = rω, solving for ω and substituting given values. In part (b), convert from rev/min to rad/s and substitute straight into
Equation 7.10
to obtain the linear speeds. In part (c), linear speed multiplied by time gives the total distance.
Solution
Figure 7.6a
shows a car moving in a circular path with constant linear speed v. Even though the car moves at a constant speed, it still has an acceleration. To understand this, consider the defining equation for average acceleration:
The numerator represents the difference between the velocity vectors
f and
i. These vectors may have the same magnitude, corresponding to the same speed, but if they have different directions, their difference can’t equal zero. The direction of the car’s velocity as it moves in the circular path is continually changing, as shown in
Figure 7.6b
. For circular motion at constant speed, the acceleration vector always points toward the center of the circle. Such an acceleration is called a centripetal (center-seeking) acceleration. Its magnitude is given by
To derive
Equation 7.13
, consider
Figure 7.7a
. An object is first at point with velocity
i at time ti and then at point with velocity
f at a later time tf. We assume
i and
f differ only in direction; their magnitudes are the same (vi = vf = v). To calculate the acceleration, we begin with
Equation 7.12
,
where Δ
v
=
f − v
i is the change in velocity. When Δt is very small, Δs and Δθ are also very small. In
Figure 7.7b
f is almost parallel to
i, and the vector Δ
v
is approximately perpendicular to them, pointing toward the center of the circle. In the limiting case when Δt becomes vanishingly small, Δ
v
points exactly toward the center of the circle, and the average acceleration av becomes the instantaneous acceleration . From
Equation 7.14
, and Δ
point in the same direction (in this limit), so the instantaneous acceleration points to the center of the circle.
An object moves in a circular path with constant speed v. Which of the following statements is true concerning the object? (a) Its velocity is constant, but its acceleration is changing. (b) Its acceleration is constant, but its velocity is changing. (c) Both its velocity and acceleration are changing. (d) Its velocity and acceleration remain constant.
Goal Apply the concepts of centripetal acceleration and tangential speed.
Problem A race car accelerates uniformly from a speed of 40.0 m/s to a speed of 60.0 m/s in 5.00 s while traveling counterclockwise around a circular track of radius 4.00 × 102 m. When the car reaches a speed of 50.0 m/s, find (a) the magnitude of the car’s centripetal acceleration, (b) the angular speed, (c) the magnitude of the tangential acceleration, and (d) the magnitude of the total acceleration.
Strategy Substitute values into the definitions of centripetal acceleration (
Eq. 7.13
), tangential speed (
Eq. 7.10
), and total acceleration (
Eq. 7.18
). Dividing the change in linear speed by the time yields the tangential acceleration.
Solution
FIGURE 7.9 (a) The right-hand rule for determining the direction of the angular velocity vector . (b) The direction of is in the direction of advance of a right-handed screw.
When we discussed linear motion in
Chapter 2
, we emphasized that displacement, velocity, and acceleration are all vector quantities. In describing rotational motion, angular displacement, angular velocity, and angular acceleration are also vector quantities.
The direction of the angular velocity vector can be found with the right-hand rule, as illustrated in
Figure 7.9a
. Grasp the axis of rotation with your right hand so that your fingers wrap in the direction of rotation. Your extended thumb then points in the direction of .
Figure 7.9b
shows that is also in the direction of advance of a rotating right-handed screw.
We can apply this rule to a rotating disk viewed along the axis of rotation, as in
Figure 7.10
. When the disk rotates clockwise (
Fig. 7.10a
), the right-hand rule shows that the direction of is into the page. When the disk rotates counterclockwise (
Fig. 7.10b
), the direction of is out of the page.
Finally, the directions of the angular acceleration and the angular velocity are the same if the angular speed v (the magnitude of ) is increasing with time, and are opposite each other if the angular speed is decreasing with time.
An object can have a centripetal acceleration only if some external force acts on it. For a ball whirling in a circle at the end of a string, that force is the tension in the string. In the case of a car moving on a flat circular track, the force is friction between the car and track. A satellite in circular orbit around Earth has a centripetal acceleration due to the gravitational force between the satellite and Earth.
Some books use the term “centripetal force,” which can give the mistaken impression that it is a new force of nature. This is not the case: The adjective “centripetal” in “centripetal force” simply means that the force in question acts toward a center. The gravitational force and the force of tension in the string of a yo-yo whirling in a circle are examples of centripetal forces, as is the force of gravity on a satellite circling the Earth.
Consider a ball of mass m that is tied to a string of length r and is being whirled at constant speed in a horizontal circular path, as illustrated in
Figure 7.11
. Its weight is supported by a frictionless table. Why does the ball move in a circle? Because of its inertia, the tendency of the ball is to move in a straight line; however, the string prevents motion along a straight line by exerting a radial force on the ball—a tension force—that makes it follow the circular path. The tension is directed along the string toward the center of the circle, as shown in the figure.
In general, applying Newton’s second law along the radial direction yields the equation relating the net centripetal force Fc —the sum of the radial components of all forces acting on a given object—with the centripetal acceleration:
“Centripetal force” is a classification that includes forces acting toward a central point, like string tension on a tetherball or gravity on a satellite. A centripetal force must be supplied by some actual, physical force.
A net force causing a centripetal acceleration acts toward the center of the circular path and effects a change in the direction of the velocity vector. If that force should vanish, the object would immediately leave its circular path and move along a straight line tangent to the circle at the point where the force vanished.
Astronauts spending lengthy periods of time in space experience a number of negative effects due to weightlessness, such as weakening of muscle tissue and loss of calcium in bones. These effects may make it very difficult for them to return to their usual environment on Earth. How could artificial gravity be generated in space to overcome such complications?
Solution A rotating cylindrical space station creates an environment of artificial gravity. The normal force of the rigid walls provides the centripetal force, which keeps the astronauts moving in a circle (
Fig. 7.12
). To an astronaut, the normal force can’t be easily distinguished from a gravitational force as long as the radius of the station is large compared with the astronaut’s height. (Otherwise there are unpleasant inner ear effects.) This same principle is used in certain amusement park rides in which passengers are pressed against the inside of a rotating cylinder as it tilts in various directions. The visionary physicist Gerard O’Neill proposed creating a giant space colony a kilometer in radius that rotates slowly, creating Earth-normal artificial gravity for the inhabitants in its interior. These inside-out artificial worlds could enable safe transport on a several- thousand-year journey to another star system.
PROBLEM-SOLVING STRATEGY
FORCES THAT CAUSE CENTRIPETAL ACCELERATION
Use the following steps in dealing with centripetal accelerations and the forces that produce them:
· 1. Draw a free-body diagram of the object under consideration, labeling all forces that act on it.
· 2. Choose a coordinate system that has one axis perpendicular to the circular path followed by the object (the radial direction) and one axis tangent to the circular path (the tangential, or angular, direction). The normal direction, perpendicular to the plane of motion, is also often needed.
· 3. Find the net force Fc toward the center of the circular path, Fc = ∑Fr, where ∑Fc is the sum of the radial components of the forces. This net radial force causes the centripetal acceleration.
· 4. Use Newton’s second law for the radial, tangential, and normal directions, as required, writing . Remember that the magnitude of the centripetal acceleration for uniform circular motion can always be written
· 5. Solve for the unknown quantities.
EXAMPLE 7.7 Buckle Up for Safety
Goal Calculate the frictional force that causes an object to have a centripetal acceleration.
Problem A car travels at a constant speed of 30.0 mi/h (13.4 m/s) on a level circular turn of radius 50.0 m, as shown in the bird’s-eye view in
Figure 7.13a
. What minimum coefficient of static friction, ms, between the tires and roadway will allow the car to make the circular turn without sliding?
Strategy In the car’s free-body diagram (
Fig. 7.13b
) the normal direction is vertical and the tangential direction is into the page (Step 2). Use Newton’s second law. The net force acting on the car in the radial direction is the force of static friction toward the center of the circular path, which causes the car to have a centripetal acceleration. Calculating the maximum static friction force requires the normal force, obtained from the normal component of the second law.
EXAMPLE 7.8 Daytona International Speedway
Goal Solve a centripetal force problem involving two dimensions.
Problem The Daytona International Speedway in Daytona Beach, Florida, is famous for its races, especially the Daytona 500, held every spring. Both of its courses feature four-story, 31.0° banked curves, with maximum radius of 316 m. If a car negotiates the curve too slowly, it tends to slip down the incline of the turn, whereas if it’s going too fast, it may begin to slide up the incline. (a) Find the necessary centripetal acceleration on this banked curve so the car won’t slip down or slide up the incline. (Neglect friction.) (b) Calculate the speed of the race car.
Strategy Two forces act on the race car: the force of gravity and the normal force . (See
Fig. 7.14
.) Use Newton’s second law in the upward and radial directions to find the centripetal acceleration ac. Solving ac = v2/r for v then gives the race car’s speed.
FIGURE 7.14 (
Example 7.8
) Front view of a car rounding a banked roadway. Vector components are shown to the right.
Solution
· (a) Find the centripetal acceleration.Write Newton’s second law for the car:Use the y-component of Newton’s second law to solve for the normal force n:Obtain an expression for the horizontal component of , which is the centripetal force Fc in this example:Substitute this expression for Fc into the radial component of Newton’s second law and divide by m to get the centripetal acceleration:
· (b) Find the speed of the race car.Apply
Equation 7.13
:
APPLICATION
Banked Roadways
Remarks Both banking and friction assist in keeping the race car on the track.
QUESTION 7.8
What three physical quantities determine a minimum safe speed on a banked racetrack?
EXERCISE 7.8
A racetrack is to have a banked curve with radius of 245 m. What should be the angle of the bank if the normal force alone is to allow safe travel around the curve at 58.0 m/s?
Answer 54.5°
EXAMPLE 7.9 Riding the Tracks
Goal Combine centripetal force with conservation of energy.
Problem
Figure 7.15a
shows a roller-coaster car moving around a circular loop of radius R. (a) What speed must the car have so that it will just make it over the top without any assistance from the track? (b) What speed will the car subsequently have at the bottom of the loop? (c) What will be the normal force on a passenger at the bottom of the loop if the loop has a radius of 10.0 m?
Strategy This problem requires Newton’s second law and centripetal acceleration to find an expression for the car’s speed at the top of the loop, followed by conservation of energy to find its speed at the bottom. If the car just makes it over the top, the force must become zero there, so the only force exerted on the car at that point is the force of gravity, . At the bottom of the loop, the normal force acts up toward the center and the gravity force acts down, away from the center. The difference of these two is the centripetal force. The normal force can then be calculated from Newton’s second law.
Fictitious Forces
Anyone who has ridden a merry-go-round as a child (or as a fun-loving grown-up) has experienced what feels like a “center-fleeing” force. Holding onto the railing and moving toward the center feels like a walk up a steep hill.
Actually, this so-called centrifugal force is fictitious. In reality, the rider is exerting a centripetal force on his body with his hand and arm muscles. In addition, a smaller centripetal force is exerted by the static friction between his feet and the platform. If the rider’s grip slipped, he wouldn’t be flung radially away; rather, he would go off on a straight line, tangent to the point in space where he let go of the railing. The rider lands at a point that is farther away from the center, but not by “fleeing the center” along a radial line. Instead, he travels perpendicular to a radial line, traversing an angular displacement while increasing his radial displacement. (See
Fig. 7.16
.)
Tip 7.3 Centrifugal Force
A so-called centrifugal force is most often just the absence of an adequate centripetal force, arising from measuring phenomena from a noninertial (accelerating) frame of reference such as a merry-go-round.
7.5 NEWTONIAN GRAVITATION
Prior to 1686, a great deal of data had been collected on the motions of the Moon and planets, but no one had a clear understanding of the forces affecting them. In that year, Isaac Newton provided the key that unlocked the secrets of the heavens. He knew from the first law that a net force had to be acting on the Moon. If it were not, the Moon would move in a straight-line path rather than in its almost circular orbit around Earth. Newton reasoned that this force arose as a result of an attractive force between the Moon and the Earth, called the force of gravity, and that it was the same kind of force that attracted objects—such as apples—close to the surface of the Earth.
In 1687 Newton published his work on the law of universal gravitation:
Law of universal gravitation
8 ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS
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8 ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS
Rotational motion is key in the harnessing of energy for power and propulsion, as illustrated by a steamboat. The rotating paddle-wheel, driven by a steam engine, pushes water backwards, and the reaction force of the water thrusts the boat forward.
In the study of linear motion, objects were treated as point particles without structure. It didn’t matter where a force was applied, only whether it was applied or not.
The reality is that the point of application of a force does matter. In football, for example, if the ball carrier is tackled near his midriff, he might carry the tackler several yards before falling. If tackled well below the waistline, however, his center of mass rotates toward the ground, and he can be brought down immediately. Tennis provides another good example. If a tennis ball is struck with a strong horizontal force acting through its center of mass, it may travel a long distance before hitting the ground, far out of bounds. Instead, the same force applied in an upward, glancing stroke will impart topspin to the ball, which can cause it to land in the opponent’s court.
The concepts of rotational equilibrium and rotational dynamics are also important in other disciplines. For example, students of architecture benefit from understanding the forces that act on buildings and biology students should understand the forces at work in muscles and on bones and joints. These forces create torques, which tell us how the forces affect an object’s equilibrium and rate of rotation.
We will find that an object remains in a state of uniform rotational motion unless acted on by a net torque. This principle is the equivalent of Newton’s first law. Further, the angular acceleration of an object is proportional to the net torque acting on it, which is the analog of Newton’s second law. A net torque acting on an object causes a change in its rotational energy.
Finally, torques applied to an object through a given time interval can change the object’s angular momentum. In the absence of external torques, angular momentum is conserved, a property that explains some of the mysterious and formidable properties of pulsars—remnants of supernova explosions that rotate at equatorial speeds approaching that of light.
8.1 TORQUE
Forces cause accelerations; torques cause angular accelerations. There is a definite relationship, however, between the two concepts.
Figure 8.1
depicts an overhead view of a door hinged at point O. From this viewpoint, the door is free to rotate around an axis perpendicular to the page and passing through O. If a force is applied to the door, there are three factors that determine the effectiveness of the force in opening the door: the magnitude of the force, the position of application of the force, and the angle at which it is applied.
The vectors and lie in a plane. As discussed in detail shortly in conjunction with
Figure 8.4
, the torque is then perpendicular to this plane. The point O is usually chosen to coincide with the axis the object is rotating around, such as the hinge of a door or hub of a merry-go-round. (Other choices are possible as well.) In addition, we consider only forces acting in the plane perpendicular to the axis of rotation. This criterion excludes, for example, a force with upward component on a merry-go-round railing, which cannot affect the merry-go-round’s rotation.
Under these conditions, an object can rotate around the chosen axis in one of two directions. By convention, counterclockwise is taken to be the positive direction, clockwise the negative direction. When an applied force causes an object to rotate counterclockwise, the torque on the object is positive. When the force causes the object to rotate clockwise, the torque on the object is negative. When two or more torques act on an object at rest, the torques are added. If the net torque isn’t zero, the object starts rotating at an ever-increasing rate. If the net torque is zero, the object’s rate of rotation doesn’t change. These considerations lead to the rotational analog of the first law: the rate of rotation of an object doesn’t change, unless the object is acted on by a net torque.
EXAMPLE 8.1 Battle of the Revolving Door
Goal Apply the basic definition of torque.
Problem Two disgruntled businesspeople are trying to use a revolving door, as in
Figure 8.2
. The woman on the left exerts a force of 625 N perpendicular to the door and 1.20 m from the hub’s center, while the man on the right exerts a force of 8.50 × 102 N perpendicular to the door and 0.800 m from the hub’s center. Find the net torque on the revolving door.
Strategy Calculate the individual torques on the door using the definition of torque,
Equation 8.1
, and then sum to find the net torque on the door. The woman exerts a negative torque, the man a positive torque. Their positions of application also differ.
FIGURE 8.2 (Example 8.1)
Solution
Calculate the torque exerted by the woman. A negative sign must be supplied because 1, if unopposed, would cause a clockwise rotation:
Calculate the torque exerted by the man. The torque is positive because 2, if unopposed, would cause a counterclockwise rotation:
Sum the torques to find the net torque on the door:
Remark The negative result here means that the net torque will produce a clockwise rotation.
QUESTION 8.1
What happens if the woman suddenly slides closer to the hub by 0.400 m?
EXERCISE 8.1
A businessman enters the same revolving door on the right, pushing with 576 N of force directed perpendicular to the door and 0.700 m from the hub, while a boy exerts a force of 365 N perpendicular to the door, 1.25 m to the left of the hub. Find (a) the torques exerted by each person and (b) the net torque on the door.
Answers
EXAMPLE 8.2 The Swinging Door
Goal Apply the more general definition of torque.
Problem (a) A man applies a force of F = 3.00 × 102 N at an angle of 60.0° to the door of
Figure 8.5a
, 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?
Strategy Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn’t open, so the sum of these torques must be zero, a condition that can be used to find the wedge force.
FIGURE 8.5 (Example 8.2a) (a) Top view of a door being pushed by a 300-N force. (b) The components of the 300-N force.
Solution
(a) Compute the torque due to the applied force exerted at 60.0°.
Substitute into the general torque equation:
(b) Calculate the force exerted by the wedge on the other side of the door.
Set the sum of the torques equal to zero:
The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of −90.0°, opposite Fy.
Remark Notice that the angle from the position vector to the wedge force is −90°. This is because, starting at the position vector, it’s necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition.
Figure 8.5b
illustrates the fact that the component of the force perpendicular to the lever arm causes the torque.
QUESTION 8.2
To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob?
EXERCISE 8.2
A man ties one end of a strong rope 8.00 m long to the bumper of his truck, 0.500 m from the ground, and the other end to a vertical tree trunk at a height of 3.00 m. He uses the truck to create a tension of 8.00 × 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.
Answer 2.28 × 103 N · m
8.2 TORQUE AND THE TWO CONDITIONS FOR EQUILIBRIUM
An object in mechanical equilibrium must satisfy the following two conditions:
· 1. The net external force must be zero: Σ
· 2. The net external torque must be zero:
The first condition is a statement of translational equilibrium: The sum of all forces acting on the object must be zero, so the object has no translational acceleration, 5 0. The second condition is a statement of rotational equilibrium: The sum of all torques on the object must be zero, so the object has no angular acceleration, 5 0. For an object to be in equilibrium, it must both translate and rotate at a constant rate.
Because we can choose any location for calculating torques, it’s usually best to select an axis that will make at least one torque equal to zero, just to simplify the net torque equation.
This large balanced rock at the Garden of the Gods in Colorado Springs, Colorado, is in mechanical equilibrium.
EXAMPLE 8.3 Balancing Act
Goal Apply the conditions of equilibrium and illustrate the use of different axes for calculating the net torque on an object.
Problem A woman of mass m = 55.0 kg sits on the left end of a seesaw—a plank of length L = 4.00 m, pivoted in the middle as in
Figure 8.6
. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 12.0 kg. (c) Repeat part (b), but this time compute the torques about an axis through the left end of the plank.
Tip 8.1 Specify Your Axis
Choose the axis of rotation and use that axis exclusively throughout a given problem. The axis need not correspond to a physical axle or pivot point. Any convenient point will do.
8.3 THE CENTER OF GRAVITY
In the example of the seesaw in the previous section, we guessed that the torque due to the force of gravity on the plank was the same as if all the plank’s weight were concentrated at its center. This is a general procedure: To compute the torque on a rigid body due to the force of gravity, the body’s entire weight can be thought of as concentrated at a single point. The problem then reduces to finding the location of that point. If the body is homogeneous (its mass is distributed evenly) and symmetric, it’s usually possible to guess the location of that point, as in
Example 8.3
. Otherwise, it’s necessary to calculate the point’s location, as explained in this section.
Consider an object of arbitrary shape lying in the xy-plane, as in
Figure 8.7
. The object is divided into a large number of very small particles of weight m1g, m2g, m3g, … having coordinates (x1, y1), (x2, y2), (x3, y3), …. If the object is free to rotate around the origin, each particle contributes a torque about the origin that is equal to its weight multiplied by its lever arm. For example, the torque due to the weight m1g is m1gx1, and so forth.
We wish to locate the point of application of the single force of magnitude w = F = Mg (the total weight of the object), where the effect on the rotation of the object is the same as that of the individual particles. This point is called the object’s center of gravity. Equating the torque exerted by w at the center of gravity to the sum of the torques acting on the individual particles gives
We assume that g is the same everywhere in the object (which is true for all objects we will encounter). Then the g factors in the preceding equation cancel, resulting in
where xcg is the x-coordinate of the center of gravity. Similarly, the y-coordinate and z-coordinate of the center of gravity of the system can be found from
and
These three equations are identical to the equations for a similar concept called center of mass. The center of mass and center of gravity of an object are exactly the same when g doesn’t vary significantly over the object.
It’s often possible to guess the location of the center of gravity. The center of gravity of a homogeneous, symmetric body must lie on the axis of symmetry. For example, the center of gravity of a homogeneous rod lies midway between the ends of the rod, and the center of gravity of a homogeneous sphere or a homogeneous cube lies at the geometric center of the object. The center of gravity of an irregularly shaped object, such as a wrench, can be determined experimentally by suspending the wrench from two different arbitrary points (
Fig. 8.8
). The wrench is first hung from point A, and a vertical line AB (which can be established with a plumb bob) is drawn when the wrench is in equilibrium. The wrench is then hung from point C, and a second vertical line CD is drawn. The center of gravity coincides with the intersection of these two lines. In fact, if the wrench is hung freely from any point, the center of gravity always lies straight below the point of support, so the vertical line through that point must pass through the center of gravity.
Several examples in
Section 8.4
involve homogeneous, symmetric objects where the centers of gravity coincide with their geometric centers. A rigid object in a uniform gravitational field can be balanced by a single force equal in magnitude to the weight of the object, as long as the force is directed upward through the object’s center of gravity.
FIGURE 8.7 The net gravitational torque on an object is zero if computed around the center of gravity. The object will balance if supported at that point (or at any point along a vertical line above or below that point).
FIGURE 8.8 An experimental technique for determining the center of gravity of a wrench. The wrench is hung freely from two different pivots, A and C. The intersection of the two vertical lines, AB and CD, locates the center of gravity.
EXAMPLE 8.4 Where Is the Center of Gravity?
Goal Find the center of gravity of a system of particles.
Problem Three particles are located in a coordinate system as shown in
Figure 8.9
. Find the center of gravity.
Strategy The y-coordinate and z-coordinate of the center of gravity are both zero because all the particles are on the x-axis. We can find the x-coordinate of the center of gravity using
Equation 8.3a
.
FIGURE 8.9 (Example 8.4) Locating the center of gravity of a system of three particles.
Solution
Apply
Equation 8.3a
to the system of three particles:
Compute the numerator of Equation (1):
Substitute the denominator, Σmi = 11.0 kg, and the numerator into Equation (1).
QUESTION 8.4
If 1.00 kg is added to the masses on the left and right, does the center of mass (a) move to the left, (b) move to the right, or (c) remain in the same position.
EXERCISE 8.4
If a fourth particle of mass 2.00 kg is placed at x = 0, y = 0.250 m, find the x- and y-coordinates of the center of gravity for this system of four particles.
Answer xcg = 0.115 m; ycg = 0.038 5 m
EXAMPLE 8.5 Locating Your Lab Partner’s Center of Gravity
Goal Use torque to find a center of gravity.
Problem In this example we show how to find the location of a person’s center of gravity. Suppose your lab partner has a height L of 173 cm (5 ft, 8 in) and a weight w of 715 N (160 lb). You can determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in
Figure 8.10
. If the board’s weight wb is 49 N and the scale reading F is 3.50 × 102 N, find the distance of your lab partner’s center of gravity from the left end of the board.
Strategy To find the position x of the center of gravity, compute the torques using an axis through O. Set the sum of the torques equal to zero and solve for xcg.
FIGURE 8.10 (Example 8.5) Determining your lab partner’s center of gravity.
Solution
Apply the second condition of equilibrium. There is no torque due to the normal force because its moment arm is zero about an axis through O.
Solve for xcg and substitute known values:
Remarks The given information is sufficient only to determine the x-coordinate of the center of gravity. The other two coordinates can be estimated, based on the body’s symmetry.
QUESTION 8.5
What would happen if a support is placed exactly at x = 79 cm followed by the removal of the supports at the subject’s head and feet?
EXERCISE 8.5
Suppose a 416-kg alligator of length 3.5 m is stretched out on a board of the same length weighing 65 N. If the board is supported on the ends as in
Figure 8.10
, and the scale reads 1 880 N, find the x-component of the alligator’s center of gravity.
Answer 1.59 m
Tip 8.2 Rotary Motion under Zero Torque
If a net torque of zero is exerted on an object, it will continue to rotate at a constant angular speed—which need not be zero. However, zero torque does imply that the angular acceleration is zero.
8.4 EXAMPLES OF OBJECTS IN EQUILIBRIUM
Recall from
Chapter 4
that when an object is treated as a geometric point, equilibrium requires only that the net force on the object is zero. In this chapter we have shown that for extended objects a second condition for equilibrium must also be satisfied: The net torque on the object must be zero. The following general procedure is recommended for solving problems that involve objects in equilibrium.
PROBLEM-SOLVING STRATEGY
OBJECTS IN EQUILIBRIUM
· 1. Diagram the system. Include coordinates and choose a convenient rotation axis for computing the net torque on the object.
· 2. Draw a free-body diagram of the object of interest, showing all external forces acting on it. For systems with more than one object, draw a separate diagram for each object. (Most problems will have a single object of interest.)
· 3. Apply Στi = 0, the second condition of equilibrium. This condition yields a single equation for each object of interest. If the axis of rotation has been carefully chosen, the equation often has only one unknown and can be solved immediately.
· 4. Apply Σ
Fx
= 0 and Σ
Fy
= 0, the first condition of equilibrium. This yields two more equations per object of interest.
· 5. Solve the system of equations. For each object, the two conditions of equilibrium yield three equations, usually with three unknowns. Solve by substitution.
EXAMPLE 8.6 A Weighted Forearm
Goal Apply the equilibrium conditions to the human body.
Problem A 50.0-N (11-lb) bowling ball is held in a person’s hand with the forearm horizontal, as in
Figure 8.11a
. The biceps muscle is attached 0.030 0 m from the joint, and the ball is 0.350 m from the joint. Find the upward force exerted by the biceps on the forearm (the ulna) and the downward force exerted by the humerus on the forearm, acting at the joint. Neglect the weight of the forearm and slight deviation from the vertical of the biceps.
Strategy The forces acting on the forearm are equivalent to those acting on a bar of length 0.350 m, as shown in
Figure 8.11b
. Choose the usual x- and y-coordinates as shown and the axis at O on the left end. (This completes Steps 1 and 2.) Use the conditions of equilibrium to generate equations for the unknowns, and solve.
FIGURE 8.11 (Example 8.6) (a) A weight held with the forearm horizontal. (b) The mechanical model for the system.
Solution
Apply the second condition for equilibrium (Step 3) and solve for the upward force F:
Apply the first condition for equilibrium (Step 4) and solve for the downward force R:
Remarks The magnitude of the force supplied by the biceps must be about ten times as large as the bowling ball it is supporting!
QUESTION 8.6
Suppose the biceps were surgically reattached three centimeters farther toward the person’s hand. If the same bowling ball were again held in the person’s hand, how would the force required of the biceps be affected? Explain.
EXERCISE 8.6
Suppose you wanted to limit the force acting on your joint to a maximum value of 8.00 × 102 N. (a) Under these circumstances, what maximum weight would you attempt to lift? (b) What force would your biceps apply while lifting this weight?
Answers (a) 75.0 N (b) 875 N
EXAMPLE 8.7 Walking a Horizontal Beam
Goal Solve an equilibrium problem with nonperpendicular torques.
Problem A uniform horizontal beam 5.00 m long and weighing 3.00 × 102 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (
Fig. 8.12a
, page 238). If a person weighing 6.00 × 102 N stands 1.50 m from the wall, find the magnitude of the tension in the cable and the force exerted by the wall on the beam.
Strategy See
Figures 8.12b
and
8.12c
. The second condition of equilibrium, Στi = 0, with torques computed around the pin, can be solved for the tension T in the cable. The first condition of equilibrium, Σ
i = 0, gives two equations and two unknowns for the two components of the force exerted by the wall, Rx and Ry.
FIGURE 8.12 (
Example 8.7
) (a) A uniform beam attached to a wall and supported by a cable. (b) A free-body diagram for the beam. (c) The component form of the free-body diagram. (d) (Exercise 8.7)
Solution
From
Figure 8.12
, the forces causing torques are the wall force , the gravity forces on the beam and the man, wB and wM, and the tension force . Apply the condition of rotational equilibrium:
Compute torques around the pin at O, so τR = 0 (zero moment arm). The torque due to the beam’s weight acts at the beam’s center of gravity.
Substitute L = 5.00 m and the weights, solving for T:
Now apply the first condition of equilibrium to the beam:
· (1)
· (2)
Substituting the value of T found in the previous step and the weights, obtain the components of :
Remarks Even if we selected some other axis for the torque equation, the solution would be the same. For example, if the axis were to pass through the center of gravity of the beam, the torque equation would involve both T and Ry. Together with Equations (1) and (2), however, the unknowns could still be found—a good exercise.
QUESTION 8.7
What happens to the tension in the cable if the man in
Figure 8.12a
moves farther away from the wall?
EXERCISE 8.7
A person with mass 55.0 kg stands 2.00 m away from the wall on a 6.00-m beam, as shown in
Figure 8.12d
. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in the wire.
Answers T = 751 N, Rx = −6.50 × 102 N, Ry = 556 N
EXAMPLE 8.8 Don’t Climb the Ladder
Goal Apply the two conditions of equilibrium.
Problem A uniform ladder 10.0 m long and weighing 50.0 N rests against a smooth vertical wall as in
Figure 8.13a
. If the ladder is just on the verge of slipping when it makes a 50.0° angle with the ground, find the coefficient of static friction between the ladder and ground.
Strategy
Figure 8.13b
is the free-body diagram for the ladder. The first condition of equilibrium, Σ
i = 0, gives two equations for three unknowns: the magnitudes of the static friction force f and the normal force n, both acting on the base of the ladder, and the magnitude of the force of the wall, P, acting on the top of the ladder. The second condition of equilibrium, Στi = 0, gives a third equation (for P), so all three quantities can be found. The definition of static friction then allows computation of the coefficient of static friction.
FIGURE 8.13 (Interactive Example 8.8) (a) A ladder leaning against a frictionless wall. (b) A free-body diagram of the ladder. (c) Lever arms for the force of gravity and .
Solution
Apply the first condition of equilibrium to the ladder:
Apply the second condition of equilibrium, computing torques around the base of the ladder, with τ;grav standing for the torque due to the ladder’s 50.0-N weight:
The torques due to friction and the normal force are zero about O because their moment arms are zero. (Moment arms can be found from
Fig. 8.13c
.)
From Equation (1), we now have f = P = 21.0 N. The ladder is on the verge of slipping, so write an expression for the maximum force of static friction and solve for μs:
Remarks Note that torques were computed around an axis through the bottom of the ladder so that only and the force of gravity contributed nonzero torques. This choice of axis reduces the complexity of the torque equation, often resulting in an equation with only one unknown.
QUESTION 8.8
If a 50.0 N monkey hangs from the middle rung, would the coefficient of static friction be (a) doubled, (b) halved, or (c) unchanged?
EXERCISE 8.8
If the coefficient of static friction is 0.360, and the same ladder makes a 60.0° angle with respect to the horizontal, how far along the length of the ladder can a 70.0-kg painter climb before the ladder begins to slip?
Answer 6.33 m
8.5 RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION
When a rigid object is subject to a net torque, it undergoes an angular acceleration that is directly proportional to the net torque. This result, which is analogous to Newton’s second law, is derived as follows.
FIGURE 8.14 An object of mass m attached to a light rod of length r moves in a circular path on a friction-less horizontal surface while a tangential force t acts on it.
The system shown in
Figure 8.14
consists of an object of mass m connected to a very light rod of length r. The rod is pivoted at the point O, and its movement is confined to rotation on a frictionless horizontal table. Assume that a force Ft acts perpendicular to the rod and hence is tangent to the circular path of the object. Because there is no force to oppose this tangential force, the object undergoes a tangential acceleration at in accordance with Newton’s second law:
Multiply both sides of this equation by r:
Substituting the equation at = ra relating tangential and angular acceleration into the above expression gives
The left side of
Equation 8.4
is the torque acting on the object about its axis of rotation, so we can rewrite it as
Equation 8.5
shows that the torque on the object is proportional to the angular acceleration of the object, where the constant of proportionality mr2 is called the moment of inertia of the object of mass m. (Because the rod is very light, its moment of inertia can be neglected.)
QUICK QUIZ 8.1
Using a screwdriver, you try to remove a screw from a piece of furniture, but can’t get it to turn. To increase the chances of success, you should use a screwdriver that (a) is longer, (b) is shorter, (c) has a narrower handle, or (d) has a wider handle.
Torque on a Rotating Object
Consider a solid disk rotating about its axis as in
Figure 8.15a
. The disk consists of many particles at various distances from the axis of rotation. (See
Fig. 8.15b
.) The torque on each one of these particles is given by
Equation 8.5
. The net torque on the disk is given by the sum of the individual torques on all the particles:
Because the disk is rigid, all of its particles have the same angular acceleration, so a is not involved in the sum. If the masses and distances of the particles are labeled with subscripts as in
Figure 8.15b
, then
This quantity is the moment of inertia, I, of the whole body:
Moment of inertia
FIGURE 8.15 (a) A solid disk rotating about its axis. (b) The disk consists of many particles, all with the same angular acceleration.
The moment of inertia has the SI units kg · m2. Using this result in
Equation 8.6
, we see that the net torque on a rigid body rotating about a fixed axis is given by
Rotational analog of Newton’s second law
FIGURE 8.16 (Quick Quiz 8.3)
APPLICATION
Bicycle Gears
The gear system on a bicycle provides an easily visible example of the relationship between torque and angular acceleration. Consider first a five-speed gear system in which the drive chain can be adjusted to wrap around any of five gears attached to the back wheel (
Fig. 8.17
). The gears, with different radii, are concentric with the wheel hub. When the cyclist begins pedaling from rest, the chain is attached to the largest gear. Because it has the largest radius, this gear provides the largest torque to the drive wheel. A large torque is required initially, because the bicycle starts from rest. As the bicycle rolls faster, the tangential speed of the chain increases, eventually becoming too fast for the cyclist to maintain by pushing the pedals. The chain is then moved to a gear with a smaller radius, so the chain has a smaller tangential speed that the cyclist can more easily maintain. This gear doesn’t provide as much torque as the first, but the cyclist needs to accelerate only to a somewhat higher speed. This process continues as the bicycle moves faster and faster and the cyclist shifts through all five gears. The fifth gear supplies the lowest torque, but now the main function of that torque is to counter the frictional torque from the rolling tires, which tends to reduce the speed of the bicycle. The small radius of the fifth gear allows the cyclist to keep up with the chain’s movement by pushing the pedals.
A 15-speed bicycle has the same gear structure on the drive wheel, but has three gears on the sprocket connected to the pedals. By combining different positions of the chain on the rear gears and the sprocket gears, 15 different torques are available.
More on the Moment of Inertia
As we have seen, a small object (or a particle) has a moment of inertia equal to mr2 about some axis. The moment of inertia of a composite object about some axis is just the sum of the moments of inertia of the object’s components. For example, suppose a majorette twirls a baton as in
Figure 8.18
. Assume that the baton can be modeled as a very light rod of length 2l with a heavy object at each end. (The rod of a real baton has a significant mass relative to its ends.) Because we are neglecting the mass of the rod, the moment of inertia of the baton about an axis through its center and perpendicular to its length is given by
Equation 8.7
:
FIGURE 8.17 The drive wheel and gears of a bicycle.
FIGURE 8.18 A baton of length 2l and mass 2m. (The mass of the connecting rod is neglected.) The moment of inertia about the axis through the baton’s center and perpendicular to its length is 2ml;2.
Because this system consists of two objects with equal masses equidistant from the axis of rotation, r = l for each object, and the sum is
If the mass of the rod were not neglected, we would have to include its moment of inertia to find the total moment of inertia of the baton.
We pointed out earlier that I is the rotational counterpart of m. However, there are some important distinctions between the two. For example, mass is an intrinsic property of an object that doesn’t change, whereas the moment of inertia of a system depends on how the mass is distributed and on the location of the axis of rotation.
Example 8.9
illustrates this point.
EXAMPLE 8.9 The Baton Twirler
Goal Calculate a moment of inertia.
Problem In an effort to be the star of the half-time show, a majorette twirls an unusual baton made up of four spheres fastened to the ends of very light rods (
Fig. 8.19
). Each rod is 1.0 m long. (a) Find the moment of inertia of the baton about an axis perpendicular to the page and passing through the point where the rods cross. (b) The majorette tries spinning her strange baton about the axis OO’, as shown in
Figure 8.20
. Calculate the moment of inertia of the baton about this axis.
Strategy In
Figure 8.19
, all four balls contribute to the moment of inertia, whereas in
Figure 8.20
, with the new axis, only the two balls on the left and right contribute. Technically, the balls on the top and bottom still make a small contribution because they’re not really point particles. However, their moment of inertia can be neglected because the radius of the sphere is much smaller than the radius formed by the rods.
FIGURE 8.19 (Example 8.9a) Four objects connected to light rods rotating in the plane of the page.
Solution
(a) Calculate the moment of inertia of the baton when oriented as in
Figure 8.19
.
Apply
Equation 8.7
, neglecting the mass of the connecting rods:
(b) Calculate the moment of inertia of the baton when oriented as in
Figure 8.20
(page 243).
Apply
Equation 8.7
again, neglecting the radii of the 0.20-kg spheres.
Remarks The moment of inertia is smaller in part (b) because in this configuration the 0.20-kg spheres are essentially located on the axis of rotation.
QUESTION 8.9
If one of the rods is lengthened, which one would cause the larger change in the moment of inertia, the rod connecting bars one and three or the rod connecting balls two and four?
EXERCISE 8.9
Yet another bizarre baton is created by taking four identical balls, each with mass 0.300 kg, and fixing them as before, except that one of the rods has a length of 1.00 m and the other has a length of 1.50 m. Calculate the moment of inertia of this baton (a) when oriented as in
Figure 8.19
; (b) when oriented as in
Figure 8.20
, with the shorter rod vertical; and (c) when oriented as in
Figure 8.20
, but with longer rod vertical.
Answers (a) 0.488 kg · m2 (b) 0.338 kg · m2 (c) 0.150 kg middot; m2
FIGURE 8.20 (Example 8.9b) A double baton rotating about the axis OO’.
Calculation of Moments of Inertia for Extended Objects
The method used for calculating moments of inertia in
Example 8.9
is simple when only a few small objects rotate about an axis. When the object is an extended one, such as a sphere, a cylinder, or a cone, techniques of calculus are often required, unless some simplifying symmetry is present. One such extended object amenable to a simple solution is a hoop rotating about an axis perpendicular to its plane and passing through its center, as shown in
Figure 8.21
. (A bicycle tire, for example, would approximately fit into this category.)
To evaluate the moment of inertia of the hoop, we can still use the equation I = Σmr2 and imagine that the mass of the hoop M is divided into n small segments having masses m1, m2, m3, …, mn, as in
Figure 8.21
, with M = m1 + m2 + m3 + … + mn. This approach is just an extension of the baton problem described in the preceding examples, except that now we have a large number of small masses in rotation instead of only four.
We can express the sum for I as
All of the segments around the hoop are at the same distance R from the axis of rotation, so we can drop the subscripts on the distances and factor out R2 to obtain
This expression can be used for the moment of inertia of any ring-shaped object rotating about an axis through its center and perpendicular to its plane. Note that the result is strictly valid only if the thickness of the ring is small relative to its inner radius.
The hoop we selected as an example is unique in that we were able to find an expression for its moment of inertia by using only simple algebra. Unfortunately, for most extended objects the calculation is much more difficult because the mass elements are not all located at the same distance from the axis, so the methods of integral calculus are required. The moments of inertia for some other common shapes are given without proof in
Table 8.1
(page 244). You can use this table as needed to determine the moment of inertia of a body having any one of the listed shapes.
If mass elements in an object are redistributed parallel to the axis of rotation, the moment of inertia of the object doesn’t change. Consequently, the expression I = MR2 can be used equally well to find the axial moment of inertia of an embroidery hoop or of a long sewer pipe. Likewise, a door turning on its hinges is described by the same moment-of-inertia expression as that tabulated for a long thin rod rotating about an axis through its end.
FIGURE 8.21 A uniform hoop can be divided into a large number of small segments that are equidistant from the center of the hoop.
Tip 8.3 No Single Moment of Inertia
Moment of inertia is analogous to mass, but there are major differences. Mass is an inherent property of an object. The moment of inertia of an object depends on the shape of the object, its mass, and the choice of rotation axis.
TABLE 8.1 Moments of Inertia for Various Rigid Objects of Uniform Composition
EXAMPLE 8.10 Warming Up
Goal Find a moment of inertia and apply the rotational analog of Newton’s second law.
Problem A baseball player loosening up his arm before a game tosses a 0.150kg baseball, using only the rotation of his forearm to accelerate the ball (
Fig. 8.22
). The forearm has a mass of 1.50 kg and a length of 0.350 m. The ball starts at rest and is released with a speed of 30.0 m/s in 0.300 s. (a) Find the constant angular acceleration of the arm and ball. (b) Calculate the moment of inertia of the system consisting of the forearm and ball. (c) Find the torque exerted on the system that results in the angular acceleration found in part (a).
Strategy The angular acceleration can be found with rotational kinematic equations, while the moment of inertia of the system can be obtained by summing the separate moments of inertia of the ball and forearm. Multiplying these two results together gives the torque.
FIGURE 8.22 (Example 8.10) A ball being tossed by a pitcher. The forearm is used to accelerate the ball.
Solution
(a) Find the angular acceleration of the ball.
The angular acceleration is constant, so use the angular velocity kinematic equation with ωi = 0:
The ball accelerates along a circular arc with radius given by the length of the forearm. Solve v = rω for ω and substitute:
(b) Find the moment of inertia of the system (forearm plus ball).
Find the moment of inertia of the ball about an axis that passes through the elbow, perpendicular to the arm:
Obtain the moment of inertia of the forearm, modeled as a rod, by consulting
Table 8.1
:
Sum the individual moments of inertia to obtain the moment of inertia of the system (ball plus forearm):
(c) Find the torque exerted on the system.
Apply
Equation 8.8
, using the results of parts (a) and (b):
Remarks Notice that having a long forearm can greatly increase the torque and hence the acceleration of the ball. This is one reason it’s advantageous for a pitcher to be tall—the pitching arm is proportionately longer. A similar advantage holds in tennis, where taller players can usually deliver faster serves.
QUESTION 8.10
Why do pitchers step forward when delivering the pitch? Why is the timing important?
EXERCISE 8.10
A catapult with a radial arm 4.00 m long accelerates a ball of mass 20.0 kg through a quarter circle. The ball leaves the apparatus at 45.0 m/s. If the mass of the arm is 25.0 kg and the acceleration is constant, find (a) the angular acceleration, (b) the moment of inertia of the arm and ball, and (c) the net torque exerted on the ball and arm.
Hint: Use the time-independent rotational kinematics equation to find the angular acceleration, rather than the angular velocity equation.
Answers (a) 40.3 rad/s2 (b) 453 kg · m2 (c) 1.83 × 104 N · m
EXAMPLE 8.11 The Falling Bucket
Goal Combine Newton’s second law with its rotational analog.
Problem A solid, frictionless cylindrical reel of mass M = 3.00 kg and radius R = 0.400 m is used to draw water from a well (
Fig. 8.23a
). A bucket of mass m = 2.00 kg is attached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and acceleration a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall?
Strategy This problem involves three equations and three unknowns. The three equations are Newton’s second law applied to the bucket, ma = ΣFi; the rotational version of the second law applied to the cylinder, Iα = Στi; and the relationship between linear and angular acceleration, a = rα, which connects the dynamics of the bucket and cylinder. The three unknowns are the acceleration a of the bucket, the angular acceleration a of the cylinder, and the tension T in the rope. Assemble the terms of the three equations and solve for the three unknowns by substitution. Part (b) is a review of kinematics.
FIGURE 8.23 (Example 8.11) (a) A water bucket attached to a rope passing over a frictionless reel. (b) A free-body diagram for the bucket. (c) The tension produces a torque on the cylinder about its axis of rotation. (d) A falling cylinder (Exercise 8.11).
Solution
(a) Find the tension in the cord and the acceleration of the bucket.
Apply Newton’s second law to the bucket in
Figure 8.23b
. There are two forces: the tension acting upwards and gravity acting downwards.
Apply τ; = Iα to the cylinder in
Figure 8.23c
:
Notice the angular acceleration is clockwise, so the torque is negative. The normal and gravity forces have zero moment arm and don’t contribute any torque.
Solve for T and substitute α = a/R (notice that both α and a are negative):
Substitute the expression for T in Equation (3) into Equation (1), and solve for the acceleration:
Substitute the values for m, M, and g, getting a, then substitute a into Equation (3) to get T:
(b) Find the distance the bucket falls in 3.00 s.
Apply the displacement kinematic equation for constant acceleration, with t = 3.00 s and v0 = 0:
Remarks Proper handling of signs is very important in these problems. All such signs should be chosen initially and checked mathematically and physically. In this problem, for example, both the angular acceleration a and the acceleration a are negative, so a = a/R applies. If the rope had been wound the other way on the cylinder, causing counterclockwise rotation, the torque would have been positive, and the relationship would have been α = –a/R, with the double negative making the right-hand side positive, just like the left-hand side.
QUESTION 8.11
How would the acceleration and tension change if most of the reel’s mass were at its rim?
EXERCISE 8.11
A hollow cylinder of mass 0.100 kg and radius 4.00 cm has a string wrapped several times around it, as in
Figure 8.23d
. If the string is attached to a rigid support and the cylinder allowed to drop from rest, find (a) the acceleration of the cylinder and (b) the speed of the cylinder when a meter of string has unwound off of it.
Answers (a) −4.90 m/s2 (b) 3.13 m/s
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Week 3 Overview |
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Last week, we covered multiple forces acting on an object. This week we will cover motion in two dimensions, inclined planes, circular motion, and rotation. |
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Forces in Two Dimensions (1 of 2) |
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So far you have dealt with single forces acting on a body or more than two forces that act parallel to each other. But in real life situations more than one force may act on a body. How are Newton’s laws applied to such cases? We will restrict the forces to two dimensions. Since force and acceleration are vectors, Newton’s law can be applied independently to the X and Y-axes of a coordinate system. For a given problem you can choose a suitable coordinate system. But once a coordinate system is chosen, we have to stick with it for that problem. The example that follows shows how to find the acceleration of a body when two forces act on it at right angles to each other. |
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Forces in Two Dimensions (2 of 2) |
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To find the resultant acceleration we draw an arrow OA of length 3 units along the X-axis and then an arrow AB of length 4 units along the Y-axis. The resultant acceleration is the arrow OB with the length of 5 units. Therefore, the acceleration is 5 m/s2 in the direction of OB. Also when you measure the angle AOB with a protractor, we find it to be 53°. The acceleration caused by the two forces is 5 m/s2 at an angle of 53°. |
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Uniform Circular Motion |
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When an object travels in a circular path at a constant speed, its motion is referred to as uniform circular motion, and the object is accelerated towards the center of the circle. If the radius of the circular path is r, the magnitude of this acceleration is ac = v2 / r, where v is its speed and ac is called the centripetal acceleration. A centripetal force is responsible for the centripetal acceleration, which constantly pulls the object towards the center of the circular path. There cannot be any circular motion without a centripetal force. |
Banking
When there is a sharp turn in the road or when a turn has to be taken at a high speed as in a racetrack, the outer part of the road or the track is raised from the inner part of the track. This is called banking. It provides additional centripetal force to a turning vehicle so that it doesn’t skid.
The angle of banking is kept just right so that it provides all the centripetal force required and a motorist does not have to depend on the friction force at all.
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Inclined Planes |
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Forces on an Inclined Plane The inclined plane is a device that reduces the force needed to lift objects. Consider the forces acting on a block on an inclined surface. The inclined surface exerts a normal force FN on the block that is perpendicular to the incline. The force of gravity, FG, points downward. If there is no friction, the net force, Fnet, acting on the block is the resultant of FN and FG. By Newton’s second law the net force must point down the incline because the block moves only along the incline and not perpendicular to it. The vector triangle shows that the magnitude of the net force is always less than the weight, FG. Example: A 2 kg block is on an incline which is only able to support 16 N of its weight. Find the acceleration of the block along the incline. Solution: First, from FG = mg, a 2 kg block weighs 20 N. Also, the normal force is given as 16 N. We now draw a free body diagram showing all the forces acting on the block: We can find the net force on the block using Pythagorean’s Theorem: |
Now, we can put Fnet into F=ma to find the acceleration of the block:
The block will accelerate down the incline at 6 m/s2.
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Projectile Motion (1 of 2) |
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When a ball is thrown or a shell is fired from a gun at an angle to the horizontal, the ball or the shell follows a curved path known as a parabola. The motion is in two dimensions and is called projectile motion. The moving object is called a projectile. The vertical motion and horizontal motion can be analyzed separately for a projectile. The horizontal direction can be the x-axis and the vertical direction the y-axis. |
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Vertical Motion In the vertical direction only the force of gravity acts on the projectile. The vertical motion of the projectile is the same as the motion of an object thrown vertically upward with the same initial vertical velocity. If vy is the initial vertical component of velocity, the projectile will reach a maximum height of h = vy2 / (2 g), which is the maximum height reached by an object thrown vertically upward with velocity vy. The time for which the projectile remains in flight is again determined by the initial vertical component of velocity and is given by t = 2 vy / g, which is also the total time for which an object thrown vertically upward with velocity vy stays in the air. Using the navigation on the left, please proceed to the next page. |
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Projectile Motion (2 of 2) |
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Horizontal Motion In the horizontal direction there is no force acting on the projectile. So by Newton’s law there is no acceleration in the horizontal direction. If the projectile is given an initial velocity of vx it remains the same throughout the duration of the flight. The horizontal distance, x, that the projectile travels is given by: x = vxt The total time of flight has already been stated to be 2 vy / g. Therefore, the range is: x = vx (2 vy / g) The range is determined by both the horizontal and vertical components of velocity. |
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Rotational Motion (1 of 2) |
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The CD drive in your computer makes the CD spin at a high rate to either read or write data. This is an example of rotation about a fixed axis. Until now, we have dealt with objects that translate or move along a straight or curved path. Just as the motion along a path is described in terms of distance, speed, and acceleration, rotational motion is measured in terms of the angular position, speed, and acceleration of a body. Angular Displacement, Velocity, and Acceleration Angular displacement is the angle through which a body rotates and is measured in radians. When a body makes one full rotation its angular displacement, θ, is 360° or 2π radians or 6.28 radians. A radian is approximately 57.3°. You should always use radians when working with any circular motion problem. To convert between units, one revolution = 360 degrees = 2π radians. To convert an angular displacement to a distance, multiply by the radius of the circle: A rotating rigid body has an angular speed. For translation, speed is the rate at which distance is covered. Angular velocity, ω, is the rate at which the angular position of the body changes. To convert from an angular velocity to a tangential velocity, multiply by the radius of the circle: If the angular velocity of a body changes, the body has an angular acceleration, α. Angular acceleration is the rate at which angular velocity changes. To convert from an angular acceleration to a tangential acceleration, multiply by the radius of the circle: |
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Rotational Motion (2 of 2) |
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Torque When the doorknob is pushed or pulled, the force rotates the door about its hinge. The turning action depends on the product of the force component perpendicular to the lever arm and the lever arm. This quantity is called torque, τ, and its units are N m. Torque = Force perpendicular to lever arm x length of lever arm τ=Fperpr |
Newton’s Law for Rotation
Newton’s second law, F = ma, is for translation. For rotation the Newton’s law is τ= Iα.
In this equation
τ is the torque,
I is the rotational inertia of the rigid body, and
α its angular acceleration.
A pair of dumbbells that has its two masses farther away from each other, is much more difficult to twist than one in which the masses are close together. Rotational inertia, I, depends not only on the mass of the object, but also on how this mass is distributed about the axis of rotation. The farther the mass is from the axis of rotation, the greater the rotational inertia.
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Rotational Equilibrium |
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An object is in rotational equilibrium when the net torque acting on the object is zero. This in turn means that the angular acceleration is zero. Rotational equilibrium defines when an object is in balance, and can also be used to define when a lever will be able to raise a mass. |
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Week 3 Summary |
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This week covered forces in two dimensions, circular motion, inclined planes, projectiles, and rotation. You learned how to apply Newton’s laws in situations with different types of motion. Relate the concepts covered in this week to more real life situations. |
Tutor, my answers and notes to you are in red
Projectile Motion Experiment
By Monday, August 19, perform the following
experiments online
at the Projectile Motion – Galileo website. (
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
)
Fill in the tables, and post your answers to the questions.
Question 1
Procedure: Keep the initial velocity fixed at 50 m/s. Perform the experiment for each of the following angles:
· 15 degrees
· 30 degrees
· 45 degrees
· 60 degrees
· 75 degrees
Record in table 1:
· The maximum horizontal distance (Range) traveled by the projectile at various angles
· The total time of flight at various angles
· The maximum height attained
Table 1: Experiment 1 – Range & Time of flight
|
S. No. |
Initial Velocity |
Angle of Projection |
Range |
Time of Flight
|
Maximum Height Attained |
||||||
|
1. |
50 m/s |
15° |
127.42 m |
2. 64 s |
8.54 m |
||||||
| 2. |
30° |
220.7 m |
5. 1 s |
31.85 m |
|||||||
|
3. |
45° |
25 4. 84 m |
7.21 s |
63.71 m |
|||||||
| 4. |
60° |
8.83 s |
95.56 m |
||||||||
| 5. |
75° |
127.42 m |
9.85 s |
118.88 m |
Observe and Analyze:
· What angle produces the maximum range? Why?
· What angle produces the maximum height? Why?
· Are there angles which produce the same range? If so, how would you explain this?
Question 2
Procedure: Set the angle to 45 degrees. Perform the experiment with the following initial velocities.
· 30 m/s
· 40 m/s
· 50 m/s
· 60 m/s
Record in Table 2: The horizontal and vertical components of the velocity and the ranges for each of the velocities.
Table 2: Experiment 2 – Range at 45°
|
Horizontal Velocity |
Vertical Velocity |
Range at 45° |
||||||||||
|
30 m/s |
||||||||||||
|
40 m/s |
||||||||||||
|
60 m/s |
Tutor, the experiment online only gives me the values of Max distance, Max height, end velocity, and total time. It does not give horizontal or vertical velocities. So, I’m just going to write down my findings from the experiments and you can try to figure out what the horizontal and vertical velocities are because I have no clue what they’re talking about in Table 2 listed above.
30 m/s – Max distance: 91.74 m, Max height: 22.93 m, End velocity: 30 m/s, Total time: 4.32 s
40 m/s – Max distance: 163.1 m, Max height: 40.77 m, End velocity: 40 m/s, Total time: 5.77 s
50 m/s – Max distance: 254.84 m, Max height: 63.71 m, End velocity: 50 m/s, Total time: 7.21 s
60 m/s – Max distance: 366.97 m, Max height: 91.74 m, End velocity: 60 m/s, Total time: 8.65 s
Observe and Analyze:
· As the velocity is increased what happens to the Range?
· Using the initial horizontal and vertical velocities from your table, verify mathematically that the range is correct for the initial speed of 60 m/s.
Attached are assignment related online lectures and textbook chapters
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