Physics Homework #10 (solutions) 2l722z
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MasteringPhysics
4/10/09 12:01 AM
[ Assignment View ]
Course PHY121(TTH) HW #10 Due at 11:59pm on Sunday, April 12, 2009 View Grading Details
Parallel Axis Theorem The parallel axis theorem relates
, the moment of inertia of an object about an axis ing through its center of mass, to
,
the moment of inertia of the same object about a parallel axis ing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that es through point p, and
is the mass of the object.
Part A Suppose a uniform slender rod has length
and mass
. The moment of inertia of the rod about about an axis that is
perpendicular to the rod and that es through its center of mass is given by
. Find
, the moment of inertia
of the rod with respect to a parallel axis through one end of the rod. Hint A.1
Find the distance from the axis to the center of mass Hint not displayed
Express ANSWER:
in of
and
. Use fractions rather than decimal numbers in your answer.
=
Part B Now consider a cube of mass
with edges of length . The moment of inertia
of mass and perpendicular to one of its faces is given by
. Find
of the cube about an axis through its center , the moment of inertia about an axis p
through one of the edges of the cube
Hint B.1
Find the distance from the
axis to the
axis
Hint not displayed Express ANSWER:
in of
and . Use fractions rather than decimal numbers in your answer.
=
Problem 9.70 Engineers are deg a system by which a falling mass
imparts kinetic energy to a rotating uniform drum to which it is
attached by thin, very light wire wrapped around the rim of the drum (the figure ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 . In the earth tests, when is set to 19.0
and allowed to fall through 6.00
, it gives 300.0
of
kinetic energy to the drum.
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Part A If the system is operated on Mars, through what distance would the 19.0-
mass have to fall to give the same amount of
kinetic energy to the drum? ANSWER:
= 15.8
Part B How fast would the 19.0ANSWER:
mass be moving on Mars just as the drum gained 300.0
of kinetic energy?
= 9.27
Problem 9.89 Two metal disks, one with radius = 1.57
= 2.37
and mass
= 0.850
and the other with radius
= 5.00
and mass
, are welded together and mounted on a frictionless axis through their common center. .
Part A What is the total moment of inertia of the two disks? ANSWER:
= 2.20×10 −3
Part B A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.03 above the floor, what is its speed just before it strikes the floor? ANSWER:
= 3.32
Part C Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk. ANSWER:
= 5.01
Torque about the z Axis Learning Goal: To understand two different techniques for computing the torque on an object due to an applied force. Imagine an object with a pivot point p at the origin of the coordinate system shown . The force vector this force of magnitude plane. The vector p to the point where
lies in the xy plane, and
acts on the object at a point in the xy
is the position vector relative to the pivot point is applied.
The torque on the object due to the force
is equal to the cross
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product
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. When, as in this problem, the force vector and
lever arm both lie in the xy plane of the paper or computer screen, only the z component of torque is nonzero. When the torque vector is parallel to the z axis (
), it is
easiest to find the magnitude and sign of the torque, , in of the angle
between the position and force vectors using one of two
simple methods: the Tangential Component of the Force method or the Moment Arm of the Force method. Note that in this problem, the positive z direction is perpendicular to the computer screen and points toward you (given by the right-hand rule ), so a positive torque would cause counterclockwise rotation about the z axis. Tangential component of the force
Part A Decompose the force vector tangential (perpendicular to
into radial (i.e., parallel to
magnitude of the radial and tangential components, You may assume that
Hint A.1
) and
) components as shown. Find the and
.
is between zero and 90 degrees.
Magnitude of Hint not displayed
Enter your answer as an ordered pair. Express ANSWER:
and
in of
and .
=
Part B Is the following statement true or false? The torque about point p is proportional to the length ANSWER:
of the position vector
.
true false
Part C Is the following statement true or false? Both the radial and tangential components of ANSWER:
generate torque about point p.
true false
Part D Is the following statement true or false? In this problem, the tangential force vector would tend to turn an object clockwise around pivot point p. ANSWER:
true false
Part E Find the torque
about the pivot point p due to force
. Your answer should correctly express both the magnitude and sign of
. Express your answer in of
and
or in of
, , and .
ANSWER:
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=
Moment arm of the force In the figure, the dashed line extending from the force vector is called the line of action of
. The perpendicular distance
from the pivot point p to the line of action is called the moment arm of the force. Part F What is the length,
, of the moment arm of the force
about
point p? Express your answer in of
ANSWER:
and .
=
Part G Find the torque
about p due to
. Your answer should correctly express both the magnitude and sign of .
Express your answer in of ANSWER:
and
or in of , , and
.
=
Three equivalent expressions for expressing torque about the z axis have been discussed in this problem: 1. Torque is defined as the cross product between the position and force vectors. When both
and
lie in the xy plane,
only the z component of torque is nonzero, and the cross product simplifies to: . Note that a positive value for
indicates a counterclockwise direction about the z axis.
2. Torque is generated by the component of
that is tangential to the position vector
(the tangential component of
force): . 3. The magnitude of torque is the product of the force and the perpendicular distance between the z axis and the line of action of a force, , called the moment arm of the force: .
Calculating Torques Using Two Standard Methods Learning Goal: To understand the two most common procedures for finding torques when the forces and displacements are all in one plane: the moment arm method and the tangential force method. The purpose of this problem is to give you further practice finding torques in two-dimensional situations. In this case it is overkill to use the full cross product definition of the torque because the only nonzero component of the torque is the component perpendicular to the plane containing the problem. There are two common methods for finding torque in a two-dimensional problem: the tangential force method and the moment arm method. Both of these methods will be illustrated in this problem. Throughout the problem, torques that would cause counterclockwise rotation are considered to be positive. Consider a uniform pole of length so that the pole makes an angle
, attached at its base (via a pivot) to a wall. The other end of the pole is attached to a cable, with respect to the wall, and the cable is horizontal. The tension in the cable is
. The pole is
attached to the wall at point A.
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Tangential force method The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force.
Part A What is
, the magnitude of the tangential force that acts on the pole due to the tension in the rope?
Express your answer in of ANSWER:
and .
=
When using the tangential force method, you calculate the torque using the equation , where
is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by
checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque). Part B What is the magnitude of the torque Express your answer in of ANSWER:
on the pole, about point A, due to the tension in the rope? ,
, and .
=
Moment arm method The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm. For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above the point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.
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Part C Find
, the length of the moment arm of the force.
Express your answer in of ANSWER:
and .
=
To calculate the torque using the moment arm method, use the equation , where
is the moment arm perpendicular to the applied force.
Part D Find the magnitude of the torque Express your answer in of ANSWER:
on the pole, about point A, due to the tension in the rope. ,
, and .
=
For this problem, the two methods of finding torque involved nearly the same of amount of algebra, and either method could be used. Of course, both methods lead to the same final result. Now consider a woman standing on the ball of her foot as shown . A normal force of magnitude acts upward on the ball of her foot. The Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a force . This small bone has a length , and the angle between this bone and the Achilles' tendon is
. The horizontal displacement
between the ball of the foot and the point P is
.
Part E Suppose you were asked to find the torque about point P due to the normal force
in of given quantities. Which
method of finding the torque would be the easiest to use? ANSWER:
tangential force method moment arm method
Part F Find
, the torque about point P due to the normal force.
Express your answer in of ANSWER:
and any of the other quantities given in the figure.
=
Part G Suppose you were asked to find the torque about point P due to the force of magnitude
in the Achilles' tendon. Which of the
following statements is correct? ANSWER:
The tangential force method must be used. The moment arm method must be used. Either method may be used. Neither method can be used.
Part H Find
, the torque about point P due to the force applied by the Achilles' tendon.
Express your answer in of ANSWER:
,
, and
.
=
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Exercise 10.4 Three forces are applied to a wheel of radius 0.350 tangent to it, and the other one makes a 40.0
, as shown in the figure . One force is perpendicular to the rim, one is
angle with the
radius.
Part A What is the magnitude of the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and ing through its center? ANSWER:
= 0.310
Part B What is the direction of the net torque in part (A). ANSWER:
into the page out of the page.
Exercise 10.13 A grindstone in the shape of a solid disk with diameter 0.500 You press an ax against the rim with a normal force of
and a mass of
= 50.0
is rotating at
= 810
.
= 210
, and the grindstone comes to rest in 7.30 .
Part A Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings. ANSWER:
0.346
Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord's weight can be ignored.
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Part A Which of the following statements correctly describes the system shown in the figure? Hint A.1
Conditions for equilibrium Hint not displayed
Hint A.2
Rotational analogue of Newton's second law Hint not displayed
Hint A.3
Relation between linear and angular acceleration Hint not displayed
Check all that apply. ANSWER:
The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero.
Part B What happens when block B moves downward? Hint B.1
How to approach the problem Hint not displayed
Hint B.2
Find the net torque on the pulley Hint not displayed
ANSWER:
The right cord pulls on the pulley with greater force than the left cord.
Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration.
Acceleration of a Pulley A string is wrapped around a uniform solid cylinder of radius , as shown in the figure . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass . Note that the positive y direction is downward and counterclockwise torques are positive.
Part A Find the magnitude Hint A.1
of the angular acceleration of the cylinder as the block descends.
How to approach the problem Hint not displayed
Hint A.2
Find the net force on the block Hint not displayed
Hint A.3
Find the net torque on the pulley Hint not displayed
Hint A.4
Relate linear and angular acceleration Hint not displayed
Hint A.5
Putting it together Hint not displayed
Express your answer in of the cylinder's radius
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and the magnitude of the acceleration due to gravity .
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ANSWER:
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=
Note that the magnitude of the linear acceleration of the block is
, which does not depend on the value of .
Pulling a String to Accelerate a Wheel A bicycle wheel is mounted on a fixed, frictionless axle, as shown . A massless string is wound around the wheel's rim, and a constant horizontal force of magnitude starts pulling the string from the top of the wheel starting at time
when the wheel is not
rotating. Suppose that at some later time
the string has been
pulled through a distance , where
. The wheel has moment of inertia
is a dimensionless number less than 1,
the wheel's mass, and
is
is its radius. Assume that the string does
not slip on the wheel.
Part A Find
, the angular acceleration of the wheel, which results from
pulling the string to the left. Use the standard convention
that counterclockwise angular accelerations are positive. Hint A.1
Relate torque about the axle to force applied to the wheel Hint not displayed
Hint A.2
Relate torque on wheel to angular acceleration Hint not displayed
Express the angular acceleration, ANSWER:
, in of
, ,
, and
(but not
).
=
Part B The force
pulling the string is constant; therefore the magnitude of the angular acceleration
of the wheel is constant for
this configuration. Find the magnitude of the angular velocity
of the wheel when the string has been pulled a distance
Note that there are two ways to find an expression for Hint B.1
.
; these expressions look very different but are equivalent.
What the no-slip case means Hint not displayed
Hint B.2
Review of translational motion with constant acceleration Hint not displayed
Hint B.3
When has the string been pulled a distance
?
Hint not displayed Hint B.4
Relating translational acceleration and angular acceleration Hint not displayed
Express the angular velocity
of the wheel in of the displacement , the magnitude
the moment of inertia of the wheel
of the applied force, and
, if you've found such a solution. Otherwise, following the hints for this part
should lead you to express the angular velocity
of the wheel in of the displacement , the wheel's radius , and
. ANSWER: =
This solution can be obtained from the equations of rotational motion and the equations of motion with constant acceleration. An alternate approach is to calculate the work done over the displacement by the force and equate this
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work to the increase in rotational kinetic energy of rotation of the wheel
Part C Find , the speed of the string after it has been pulled by Hint C.1
Relating the speed of the string
over a distance
.
to the angular velocity Hint not displayed
Express the speed of the string in of
,
, , and
; do not include
,
, or
in your answer.
ANSWER: =
Note that this is the speed that an object of mass with constant magnitude
(which is less than
) would attain if pulled a distance
by a force
.
Exercise 10.20 A string is wrapped several times around the rim of a small hoop with radius 8.00
and mass 0.180
. The free end of the
string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 55.0 , calculate
Part A the angular speed of the rotating hoop and ANSWER:
= 29.0
Part B the speed of its center. ANSWER:
Summary
= 2.32
11 of 11 items complete (100.62% avg. score) 12.15 of 12 points
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[ Assignment View ]
Course PHY121(TTH) HW #10 Due at 11:59pm on Sunday, April 12, 2009 View Grading Details
Parallel Axis Theorem The parallel axis theorem relates
, the moment of inertia of an object about an axis ing through its center of mass, to
,
the moment of inertia of the same object about a parallel axis ing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that es through point p, and
is the mass of the object.
Part A Suppose a uniform slender rod has length
and mass
. The moment of inertia of the rod about about an axis that is
perpendicular to the rod and that es through its center of mass is given by
. Find
, the moment of inertia
of the rod with respect to a parallel axis through one end of the rod. Hint A.1
Find the distance from the axis to the center of mass Hint not displayed
Express ANSWER:
in of
and
. Use fractions rather than decimal numbers in your answer.
=
Part B Now consider a cube of mass
with edges of length . The moment of inertia
of mass and perpendicular to one of its faces is given by
. Find
of the cube about an axis through its center , the moment of inertia about an axis p
through one of the edges of the cube
Hint B.1
Find the distance from the
axis to the
axis
Hint not displayed Express ANSWER:
in of
and . Use fractions rather than decimal numbers in your answer.
=
Problem 9.70 Engineers are deg a system by which a falling mass
imparts kinetic energy to a rotating uniform drum to which it is
attached by thin, very light wire wrapped around the rim of the drum (the figure ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 . In the earth tests, when is set to 19.0
and allowed to fall through 6.00
, it gives 300.0
of
kinetic energy to the drum.
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Part A If the system is operated on Mars, through what distance would the 19.0-
mass have to fall to give the same amount of
kinetic energy to the drum? ANSWER:
= 15.8
Part B How fast would the 19.0ANSWER:
mass be moving on Mars just as the drum gained 300.0
of kinetic energy?
= 9.27
Problem 9.89 Two metal disks, one with radius = 1.57
= 2.37
and mass
= 0.850
and the other with radius
= 5.00
and mass
, are welded together and mounted on a frictionless axis through their common center. .
Part A What is the total moment of inertia of the two disks? ANSWER:
= 2.20×10 −3
Part B A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.03 above the floor, what is its speed just before it strikes the floor? ANSWER:
= 3.32
Part C Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk. ANSWER:
= 5.01
Torque about the z Axis Learning Goal: To understand two different techniques for computing the torque on an object due to an applied force. Imagine an object with a pivot point p at the origin of the coordinate system shown . The force vector this force of magnitude plane. The vector p to the point where
lies in the xy plane, and
acts on the object at a point in the xy
is the position vector relative to the pivot point is applied.
The torque on the object due to the force
is equal to the cross
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product
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. When, as in this problem, the force vector and
lever arm both lie in the xy plane of the paper or computer screen, only the z component of torque is nonzero. When the torque vector is parallel to the z axis (
), it is
easiest to find the magnitude and sign of the torque, , in of the angle
between the position and force vectors using one of two
simple methods: the Tangential Component of the Force method or the Moment Arm of the Force method. Note that in this problem, the positive z direction is perpendicular to the computer screen and points toward you (given by the right-hand rule ), so a positive torque would cause counterclockwise rotation about the z axis. Tangential component of the force
Part A Decompose the force vector tangential (perpendicular to
into radial (i.e., parallel to
magnitude of the radial and tangential components, You may assume that
Hint A.1
) and
) components as shown. Find the and
.
is between zero and 90 degrees.
Magnitude of Hint not displayed
Enter your answer as an ordered pair. Express ANSWER:
and
in of
and .
=
Part B Is the following statement true or false? The torque about point p is proportional to the length ANSWER:
of the position vector
.
true false
Part C Is the following statement true or false? Both the radial and tangential components of ANSWER:
generate torque about point p.
true false
Part D Is the following statement true or false? In this problem, the tangential force vector would tend to turn an object clockwise around pivot point p. ANSWER:
true false
Part E Find the torque
about the pivot point p due to force
. Your answer should correctly express both the magnitude and sign of
. Express your answer in of
and
or in of
, , and .
ANSWER:
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=
Moment arm of the force In the figure, the dashed line extending from the force vector is called the line of action of
. The perpendicular distance
from the pivot point p to the line of action is called the moment arm of the force. Part F What is the length,
, of the moment arm of the force
about
point p? Express your answer in of
ANSWER:
and .
=
Part G Find the torque
about p due to
. Your answer should correctly express both the magnitude and sign of .
Express your answer in of ANSWER:
and
or in of , , and
.
=
Three equivalent expressions for expressing torque about the z axis have been discussed in this problem: 1. Torque is defined as the cross product between the position and force vectors. When both
and
lie in the xy plane,
only the z component of torque is nonzero, and the cross product simplifies to: . Note that a positive value for
indicates a counterclockwise direction about the z axis.
2. Torque is generated by the component of
that is tangential to the position vector
(the tangential component of
force): . 3. The magnitude of torque is the product of the force and the perpendicular distance between the z axis and the line of action of a force, , called the moment arm of the force: .
Calculating Torques Using Two Standard Methods Learning Goal: To understand the two most common procedures for finding torques when the forces and displacements are all in one plane: the moment arm method and the tangential force method. The purpose of this problem is to give you further practice finding torques in two-dimensional situations. In this case it is overkill to use the full cross product definition of the torque because the only nonzero component of the torque is the component perpendicular to the plane containing the problem. There are two common methods for finding torque in a two-dimensional problem: the tangential force method and the moment arm method. Both of these methods will be illustrated in this problem. Throughout the problem, torques that would cause counterclockwise rotation are considered to be positive. Consider a uniform pole of length so that the pole makes an angle
, attached at its base (via a pivot) to a wall. The other end of the pole is attached to a cable, with respect to the wall, and the cable is horizontal. The tension in the cable is
. The pole is
attached to the wall at point A.
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Tangential force method The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force.
Part A What is
, the magnitude of the tangential force that acts on the pole due to the tension in the rope?
Express your answer in of ANSWER:
and .
=
When using the tangential force method, you calculate the torque using the equation , where
is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by
checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque). Part B What is the magnitude of the torque Express your answer in of ANSWER:
on the pole, about point A, due to the tension in the rope? ,
, and .
=
Moment arm method The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm. For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above the point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.
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Part C Find
, the length of the moment arm of the force.
Express your answer in of ANSWER:
and .
=
To calculate the torque using the moment arm method, use the equation , where
is the moment arm perpendicular to the applied force.
Part D Find the magnitude of the torque Express your answer in of ANSWER:
on the pole, about point A, due to the tension in the rope. ,
, and .
=
For this problem, the two methods of finding torque involved nearly the same of amount of algebra, and either method could be used. Of course, both methods lead to the same final result. Now consider a woman standing on the ball of her foot as shown . A normal force of magnitude acts upward on the ball of her foot. The Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a force . This small bone has a length , and the angle between this bone and the Achilles' tendon is
. The horizontal displacement
between the ball of the foot and the point P is
.
Part E Suppose you were asked to find the torque about point P due to the normal force
in of given quantities. Which
method of finding the torque would be the easiest to use? ANSWER:
tangential force method moment arm method
Part F Find
, the torque about point P due to the normal force.
Express your answer in of ANSWER:
and any of the other quantities given in the figure.
=
Part G Suppose you were asked to find the torque about point P due to the force of magnitude
in the Achilles' tendon. Which of the
following statements is correct? ANSWER:
The tangential force method must be used. The moment arm method must be used. Either method may be used. Neither method can be used.
Part H Find
, the torque about point P due to the force applied by the Achilles' tendon.
Express your answer in of ANSWER:
,
, and
.
=
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Exercise 10.4 Three forces are applied to a wheel of radius 0.350 tangent to it, and the other one makes a 40.0
, as shown in the figure . One force is perpendicular to the rim, one is
angle with the
radius.
Part A What is the magnitude of the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and ing through its center? ANSWER:
= 0.310
Part B What is the direction of the net torque in part (A). ANSWER:
into the page out of the page.
Exercise 10.13 A grindstone in the shape of a solid disk with diameter 0.500 You press an ax against the rim with a normal force of
and a mass of
= 50.0
is rotating at
= 810
.
= 210
, and the grindstone comes to rest in 7.30 .
Part A Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings. ANSWER:
0.346
Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord's weight can be ignored.
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Part A Which of the following statements correctly describes the system shown in the figure? Hint A.1
Conditions for equilibrium Hint not displayed
Hint A.2
Rotational analogue of Newton's second law Hint not displayed
Hint A.3
Relation between linear and angular acceleration Hint not displayed
Check all that apply. ANSWER:
The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero.
Part B What happens when block B moves downward? Hint B.1
How to approach the problem Hint not displayed
Hint B.2
Find the net torque on the pulley Hint not displayed
ANSWER:
The right cord pulls on the pulley with greater force than the left cord.
Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration.
Acceleration of a Pulley A string is wrapped around a uniform solid cylinder of radius , as shown in the figure . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass . Note that the positive y direction is downward and counterclockwise torques are positive.
Part A Find the magnitude Hint A.1
of the angular acceleration of the cylinder as the block descends.
How to approach the problem Hint not displayed
Hint A.2
Find the net force on the block Hint not displayed
Hint A.3
Find the net torque on the pulley Hint not displayed
Hint A.4
Relate linear and angular acceleration Hint not displayed
Hint A.5
Putting it together Hint not displayed
Express your answer in of the cylinder's radius
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and the magnitude of the acceleration due to gravity .
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MasteringPhysics
ANSWER:
4/10/09 12:01 AM
=
Note that the magnitude of the linear acceleration of the block is
, which does not depend on the value of .
Pulling a String to Accelerate a Wheel A bicycle wheel is mounted on a fixed, frictionless axle, as shown . A massless string is wound around the wheel's rim, and a constant horizontal force of magnitude starts pulling the string from the top of the wheel starting at time
when the wheel is not
rotating. Suppose that at some later time
the string has been
pulled through a distance , where
. The wheel has moment of inertia
is a dimensionless number less than 1,
the wheel's mass, and
is
is its radius. Assume that the string does
not slip on the wheel.
Part A Find
, the angular acceleration of the wheel, which results from
pulling the string to the left. Use the standard convention
that counterclockwise angular accelerations are positive. Hint A.1
Relate torque about the axle to force applied to the wheel Hint not displayed
Hint A.2
Relate torque on wheel to angular acceleration Hint not displayed
Express the angular acceleration, ANSWER:
, in of
, ,
, and
(but not
).
=
Part B The force
pulling the string is constant; therefore the magnitude of the angular acceleration
of the wheel is constant for
this configuration. Find the magnitude of the angular velocity
of the wheel when the string has been pulled a distance
Note that there are two ways to find an expression for Hint B.1
.
; these expressions look very different but are equivalent.
What the no-slip case means Hint not displayed
Hint B.2
Review of translational motion with constant acceleration Hint not displayed
Hint B.3
When has the string been pulled a distance
?
Hint not displayed Hint B.4
Relating translational acceleration and angular acceleration Hint not displayed
Express the angular velocity
of the wheel in of the displacement , the magnitude
the moment of inertia of the wheel
of the applied force, and
, if you've found such a solution. Otherwise, following the hints for this part
should lead you to express the angular velocity
of the wheel in of the displacement , the wheel's radius , and
. ANSWER: =
This solution can be obtained from the equations of rotational motion and the equations of motion with constant acceleration. An alternate approach is to calculate the work done over the displacement by the force and equate this
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work to the increase in rotational kinetic energy of rotation of the wheel
Part C Find , the speed of the string after it has been pulled by Hint C.1
Relating the speed of the string
over a distance
.
to the angular velocity Hint not displayed
Express the speed of the string in of
,
, , and
; do not include
,
, or
in your answer.
ANSWER: =
Note that this is the speed that an object of mass with constant magnitude
(which is less than
) would attain if pulled a distance
by a force
.
Exercise 10.20 A string is wrapped several times around the rim of a small hoop with radius 8.00
and mass 0.180
. The free end of the
string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 55.0 , calculate
Part A the angular speed of the rotating hoop and ANSWER:
= 29.0
Part B the speed of its center. ANSWER:
Summary
= 2.32
11 of 11 items complete (100.62% avg. score) 12.15 of 12 points
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