Physics Homework #10 (SOLUTIONS)

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4/10/09 12:01 AM

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Course PHY121(TTH) HW #10 Due at 11:59pm on Sunday, April 12, 2009 View Grading Details

Parallel Axis Theorem The parallel axis theorem theorem relates

, the moment of of inertia of an object about an axis passing through its center center of mass, to

,

the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and

is the mass of the object.

Part A Suppose a uniform slender rod has length

and mass mass

. The moment of inertia of the rod about about an axis that is

perpendicular to the rod and that passes 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

in terms of

and

. Use fractions rather than decimal numbers in your answer.

ANSWER: =

Part B Now consider a cube of mass

with edges edges of length

. The 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 center p , 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 terms of

and

. Use fractions rather than decimal numbers in your answer.

=

Problem 9.70 Engineers are designing a system by which a falling mass 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 allowed to fall through through 6.00

, it gives 300.0

of

kinetic energy to the drum.

http://session.masteringphysics.com/myct

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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

xy plane, and lies in the

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, the angle

, in terms of

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

) components as shown. Find the

magnitude of the radial and tangential components, You may assume that

Hint A.1

) and

and

.

is between zero and 90 degrees.

Magnitude of Hint not displayed

Enter your answer as an ordered pair. Express ANSWER:

and

in terms 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 terms of

and

or in terms of


,

, and

.

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MasteringPhysics 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 terms 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 terms of ANSWER:

and

or in terms 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 thetangential force.

Part A What is

, the magnitude of the tangential force that acts on the pole due to the tension in the rope?


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 terms 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.


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 terms 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 terms 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.


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.


,

, 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 passing 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 terms of the cylinder's radius


and the magnitude of the acceleration due to gravity

.

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ANSWER: =

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,

, in terms of

,

,

, and

(but not

).

ANSWER: =

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 terms of the displacement

the moment of inertia of the wheel

, the magnitude

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 terms 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 terms 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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Physics Homework #10 (SOLUTIONS)

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