chapter 13 skill building

MasteringPhysics: Assignment Print View

PSS 13.1: Let's Go for a Spin

Learning Goal: To practice Problem-Solving Strategy 13.1 for problems involving rotational dynamics.

A uniform board of mass and length is pivoted on one end and is supported in the horizontal

position by a rope attached to the other end. Another rope, attached to the board a distance

from the pivot point, is being pulled straight down with a constant force of magnitude .

Suddenly, the rope attached to the end of the board breaks. What is the instantaneous angular acceleration of the board?

MODEL: Model the object as a simple shape.

VISUALIZE: Draw a pictorial representation to clarify the situation, define cordinates and symbols, and list known information.

■ Identify the axis about which the object rotates. ■ Identify forces and determine their distance from the axis. For most problems it will be

useful to draw a free-body diagram. ■ Identify any torques caused by the forces and the signs of the torques.

SOLVE: The mathematical representation is based on Newton's 2nd law for rotational motion:

.

■ Find the moment of inertia in Table 13.13 or, if needed, calculate it as an integral or by using the Parallel-Axis Theorem.

■ Use rotational kinematics to find angles and angular velocities.

ASSESS: Check that your result has the correct units, is reasonable, and answers the question.

Identify the object for which you will write the equation of Newton's 2nd law. Make reasonable simplifying assumptions.

Part A

http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1043419 (1 of 26)4/19/2006 10:59:53 PM


The board should be modeled as

ANSWER: a thin rod in rotational equilibrium.

a thin rod in total equilibrium.

a thin rod rotating about its center.

a thin rod rotating about one end.

Draw the pictorial representation and list all known information. Mark all forces. Identify the axis of rotation and show the distances from that axis to each force applied. Use your sketch to answer the following questions.

Part B

How far is the axis of rotation from the center of mass of the board?

Express your answer in terms of quantities given in the problem introduction.

ANSWER:

Part C

How far is the axis of rotation from the point of application of force ?

Express your answer in terms of quantities given in the problem introduction.

ANSWER:

Part D

How many forces are acting on the board after the rope attached to the end of the board breaks?

ANSWER: one two three four

Part E



When you write Newton's 2nd law for the rotational motion of the board, how many nonzero torques will it contain?

ANSWER: one two three four

Although the reaction force acting on the board at the pivot point is not zero, its torque about that point is zero because that force is applied at the axis of rotation, and the moment arm is zero.

The figure shows what a good pictorial diagram for this problem might look like.

Now use the information and the insights that you have accumulated to construct the necessary mathematical expressions and to derive the solution.

Part F

Find the angular acceleration of the board immediately after the rope breaks.

Part F.1 Find the net torque acting on the board

Part not displayed

Part F.2 Find the moment of inertia of the board



Find the moment of inertia of the board with respect to the axis of rotation.

Express your answer in terms of the given quantities and appropriate constants. You may or may not use all given quantities.

ANSWER:

=

Express your answer in terms of the given quantities and appropriate constants. You may or may not use all given quantities.

ANSWER: =

When you work on a problem on your own, without the computer-provided feedback, only you can assess whether your answer seems right. The following questions will help you practice the skills necessary for such an assessment.

Part G

Which of these mathematical expressions have units of angular acceleration?

A.

B.

C.

D.

E.

Type alphabetically the letters corresponding to the correct answers. Do not use commas. For instance, if if options A, C, and D have the correct units, enter ACD.

ANSWER: BD

Part H



Intuitively, what can you say about the initial angular acceleration if the magnitude of the force

acting on the board were increased and all other conditions remained the same?

ANSWER: would increase.

would decrease.

would stay the same.

The increase of the magnitude of increases the net torque acting on the board, thus increasing

the magnitude of . Checking the mathematical expression for , you can see that it agrees with intuition.

Part I

If the length of the board were increased and all other conditions remained the same, how would this affect the angular acceleration of the board?

Choose the answer that not only indicates the correct change in the angular acceleration but also gives the correct explanation.

ANSWER: would increase because the torque acting on the board increases.

would increase because the torque due to the weight of the board is larger.

would decrease because the moment of inertia of the board increases with length.

would decrease because the moment of inertia increases more than the torque on the board.

would stay the same because both the moment of inertia and the torque on the board increase.

The increase of the length of the board by a certain factor increases the net torque acting on the board by that same factor; however, it also increases the moment of inertia of the board by the square of that same factor. The net result is that the magnitude of decreases.

Center of Mass and External Forces



Learning Goal: Understand that, for many purposes, a system can be treated as a point-like particle with its mass concentrated at the center of mass.A complex system of objects, both point-like and extended ones, can often be treated as a point particle, located at the system's center of mass. Such an approach can greatly simplify problem solving.

Before you use the center of mass approach, you should first understand the following terms: System: Any collection of objects that are of interest to you in a particular situation. In many problems, you have a certain freedom in choosing your system. Making a wise choice for the system is often the first step in solving the problem efficiently. Center of mass: The point that represents the "average" position of the entire mass of a system. To find the coordinate of the center of mass of a system in a given frame of reference, the following definition can be used:

In this definition, each is the mass of one of the objects within the system and is the corresponding -coordinate of that object. Of course, and can be found in a similar way. Internal force: Any force that results from an interaction between the objects inside your system. As we will show, the internal forces do not affect the motion of the system's center of mass. External force: Any force acting on an object inside your system that results from an interaction with an object outside your system. Consider a system of two blocks that have masses and :



Assume that the blocks are point-like particles and have coordinates and as shown. In this problem, the blocks' motion is restricted along the x-axis.

Part A

Find the x-coordinate of the center of mass of the system.

Express your answer in terms of , , and .

ANSWER: =

Part B

If , then the center of mass is located:

ANSWER: to the left of at a distance much greater than

to the left of at a distance much less than

to the right of at a distance much less than

to the right of at a distance much greater than

to the right of at a distance much less than

to the left of at a distance much less than

Part C

If , then the center of mass is located:

ANSWER: at

at

half-way between and

the answer depends on and

Part D



Let us assume that the blocks are in motion, and the -components of their velocities at a certain moment are and , respectively. Find the -component of the velocity of the center of mass at that moment. Remember that, in general,


ANSWER: =

Of course, and can be positive or negative or equal to zero.

Part E

Suppose that and have equal magnitudes. Also, is directed to the right and is directed

to the left. The velocity of the center of mass is then:

ANSWER: directed to the left

directed to the right

zero

the answer depends on the ratio

Part F

Let us assume that the blocks are in motion, and the x-components of their momenta at a certain moment are and , respectively. Find the x-component of the velocity of the center of mass at that moment.


ANSWER: =

Part G



Suppose that . Therefore, the following must be true:

ANSWER:

none of the above

Part H

Let us assume that the blocks are accelerating, and the -xcomponents of their accelerations at a certain moment are and , respectively. Find the x-component of the acceleration of the center of mass at that moment. Remember that, in general,


ANSWER: =

Of course, and can be positive or negative or equal to zero.

We will now consider the effect of external and internal forces on the acceleration of the center of mass.

Part I

Consider the same system of two blocks. An external force is now acting on block . No

forces are applied to block .



Find the acceleration of the center of mass of the system.

Hint I.1

Hint not displayed

Express your answer in terms of the x-component of the force, and .

ANSWER: =

Part J

Consider the same system of two blocks. Now, there are two forces involved. An external force

is acting on block and another external force is acting on block .


Express your answer in terms of the x-components (not magnitudes!) and

of the forces, and .

ANSWER:

=

Note that, in both cases, the acceleration of mass can be found as

where is the net external force applied to the system, and is the total mass of the

system. Even though each force is only applied to one object, it affects the acceleration of the center of mass of the entire system.



This result is especially useful since it can be applied to a general case, involving any number of objects moving in all directions and being acted upon by any number of external forces.

Part K

Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that and represent the components, not the magnitudes, of the corresponding

forces.

ANSWER:

cannot possibly be zero if two external forces are involved

Part L

Consider the same system of two blocks. Now, there are two internal forces involved. An internal

force is applied to block by block and another internal force is applied to block

by block .


Express your answer in terms of the x-components (not magnitudes!) and

of the forces, and .

ANSWER:

=

Newton's 3rd law tells you that . From your answers above,

you can conclude that .

The internal forces do not change the velocity of the center of mass of the system.



In the absence of any external forces, and is constant.

You just demonstrated this to be the case for the two-body situation; however, it is true in more general cases as well.

Circular Motion Tutorial

Learning Goal: Understand how to find the equation of motion of a particle undergoing uniform circular motion.Consider a particle--the small red block in the figure--that is constrained to move in a circle of

radius . We can specify its position solely by , the angle that the vector from the origin to the

block makes with our chosen reference axis at time . Following the standard conventions we

measure in the counterclockwise direction from the positive x axis.

Part A

What is the position vector as a function of

angle . For later remember that is itself a

function of time.

Part A.1

Part not displayed

Part A.2

Part not displayed

Give your answer in terms of , , and unit

vectors and corresponding to the coordinate

system in the figure.

ANSWER:

=

Uniform Circular Motion

A frequently encountered kind of circular motion is uniform circular motion, where changes at

a constant rate . In other words,



.

Usually, .

Part B

For uniform circular motion, find at an arbitrary time .

Give your answer in terms of and .

ANSWER: =

Part C

What does become now?

Express your answer in terms of , , , and unit vectors and .

ANSWER: =

Part D

Find , a position vector at time .

Hint D.1 Finding

Hint not displayed

Give your answer in terms of and unit vectors and/or .

ANSWER: =

Part E



Determine an expression for the position vector of a particle that starts on the positive y axis at

(i.e., at , ) and subsequently moves with constant .

Hint E.1

Hint not displayed

Part E.2

Part not displayed

Express your answer in terms of , , , and unit vectors and .

ANSWER: =

From this excersice you have learned that even though the motion takes place in the plane there is only one degree of freedom, angle , and that changing the initial coordinates introduces a phase

angle in the equation.

Introduction to Moments of Inertia

Learning Goal: To understand the definition and the meaning of moment of inertia; to be able to calculate the moments of inertia for a group of particles and for a continuous mass distribution with a high degree of symmetry.

By now, you may be familiar with a set of equations describing rotational kinematics. One thing that you may have noticed was the similarity between translational and rotational formulas. Such similarity also exists in dynamics and in the work-energy domain.

For a particle of mass moving at a constant speed , the kinetic energy is given by the formula

. If we consider instead a rigid object of mass rotating at a constant angular speed ,

the kinetic energy of such an object cannot be found by using the formula directly:

different parts of the object have different linear speeds. However, they all have the same angular speed. It would be desirable to obtain a formula for kinetic energy of rotational motion that is

similar to the one for translational motion; such a formula would include the term instead of .

Such a formula can, indeed, be written: for rotational motion of a system of small particles or for a rigid object with continuous mass distribution, the kinetic energy can be written as

.



Here, is called the moment of inertia of the object (or of the system of particles). It is the quantity

representing the inertia with respect to rotational motion.

It can be shown that for a discrete system, say of particles, the moment of inertia (also known as rotational inertia) is given by

.

In this formula, is the mass of the ith particle and is the distance of that particle from the axis of rotation.

For a rigid object, consisting of infinitely many particles, the analogue of such summation is integration over the entire object:

.

In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, and its applicability. It is recommended that you read the corresponding sections in your textbook before attempting these questions.

Part A

On which of the following does the moment of inertia of an object depend?

A. linear speed B. linear acceleration C. angular speed D. angular acceleration E. total mass F. shape and density of the object G. location of the axis of rotation

Type the letters corresponding to the correct answers. Do not use commas. For instance, if you think that only assumptions C and D are correct, type CD.

ANSWER: EFG

Unlike mass, the moment of inertia depends not only on the amount of matter in an object but also on the distribution of mass in space. The moment of inertia is also dependent on the axis of rotation. The same object, rotating with the same angular speed, may have different kinetic energy



depending on the axis of rotation.

Consider the system of two particles, a and b, shown in the figure . Particle a has mass , and particle b has mass .

Part B

What is the moment of inertia of

particle a? ANSWER:

undefined: an axis of rotation has not been specified.

Part C

Find the moment of inertia of particle a with respect to the x axis (that is, if the x axis is the axis

of rotation), the moment of inertia of particle a with respect to the y axis, and the moment of

inertia of particle a with respect to the z axis (the axis that passes through the origin

perpendicular to both the x and y axes).

Express your answers in terms of and separated by commas.

ANSWER: , , =

Part D

Find the total moment of inertia of the system of two particles shown in the diagram with

respect to the y axis.

Express your answer in terms of and .

ANSWER: =

For parts G to J, suppose that both particles rotate with the same angular speed about the y axis while maintaining their distances from the y axis.

Part E



Using the total moment of inertia of the system found in Part F, find the total kinetic energy

of the system.


ANSWER: =

It is useful to see how the formula for rotational kinetic energy agrees with the formula

for the kinetic energy of an object that is not rotating. To see the connection, let us find the kinetic energy of each particle.

Part F

Using the formula for kinetic energy of a moving particle , find the kinetic energy

of particle a and the kinetic energy of particle b.

Part F.1

Part not displayed

Express your answers in terms of , , and separated by a comma.

ANSWER: , =

Part G

Using the results for the kinetic energy of each particle, find the total kinetic energy of the

system of particles.


ANSWER: =

Not surprisingly, the formulas and give the same result. They should, of

course, since the rotational kinetic energy of a system of particles is simply the sum of the kinetic energies of the individual particles making up the system. A system of two (or more) discrete particles is relatively easy to deal with; however, when we analyze an extended object with mass continuously distributed in space, things get trickier. Such an object can be viewed as a system of a very large number of particles, each of a very small mass. To calculate the moment of inertia, one must add a very large number (in the limit, an infinite number) of terms. In the next part, you are asked to perform such a calculation.



Parallel Axis Theorem

The parallel axis theorem relates , the moment of inertia of an object about an axis passing

through its 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 . 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.

Part A.1

Part 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 of length . The moment of inertia of the cube

about an axis through its center of mass and perpendicular to one of its faces is given by

. Find , the moment of inertia about an axis p through one of the edges of the cube



Part B.1

Part not displayed

Express in terms of and . Use

fractions rather than decimal numbers in your answer.

ANSWER: =

PSS 13.2: Walking the Plank

Learning Goal: To practice Problem-Solving Strategy 13.2 for problems involving equilbrium of rigid bodies.A horizontal uniform bar of mass and length is hung horizontally on two vertical strings.

String 1 is attached to the end of the bar and string 2 is attached a distance from the other end.

A monkey of mass walks from one end of the bar to the other. Find the tension in string 1

at the moment that the monkey is halfway between the ends of the bar.

MODEL: Model the object as a simple shape.

VISUALIZE: Draw a pictorial representation that shows all forces and distances. List known information.

■ Pick any point you wish as a pivot point. The net torque about this point is zero. ■ Determine the moment arms of all forces about this pivot point. ■ Determine the sign of each torque about this pivot point.

SOLVE: The mathematical representation is based on the fact that an object in total equilibrium experiences neither a net force nor a net torque. Thus,

.

■ Write equations for , , and .



■ Solve the three simultaneous equations.

ASSESS: Check that your answer is reasonable and answers the question.

We start by making simplifying assumptions appropriate for the situation.

Part A

The bar should be modeled as

ANSWER: a particle moving with constant acceleration.

a flexible rod moving with constant acceleration.

a rigid rod moving with constant acceleration.

a particle in total equilibrium.

a flexible rod in total equilibrium.

a rigid rod in total equilibrium.

Now draw a sketch, labeling all forces and distances. Use your sketch to answer the following questions.

Part B

Which of the following diagrams correctly represents the forces acting at the moment described in the problem introduction? (Note that the forces are not necessarily drawn to scale.)

ANSWER: A B C D



Notice that the forces are not labeled on these drawings; on your own sketch, they should be.

Part C

What is the distance between the monkey and string 2?

Express your answer in terms of .

ANSWER:

Part D

Which of the following choices of pivot point would lead to a torque equation in which the only unknown quantity is ?

ANSWER: the center of mass of the bar

the point of attachment of string 1

the point of attachment of string 2

the end of the bar closest to string 2

If you choose the point of attachment of string 2 as the pivot point, the torque equation will not depend on the (unknown) tension in string 2. Eliminating terms, particularly those involving unknown quantities, from the torque equation generally makes it easier to solve. Other choices of pivot would also work, but they might lead to more complicated mathematics. In particular, choosing the end of the rod as a pivot leads to the most complicated torque equation involving all the forces in the problem.

Now use the information and the insights that you have accumulated to construct the necessary mathematical expressions and to derive the solution.

Part E

Find , the magnitude of the force of tension in string 1.

Hint E.1 How to approach the problem

Hint not displayed

Hint E.2 Force exerted by the monkey on the bar

Hint not displayed

Express your answer in terms of appropriate constants and any of the given variables.

ANSWER: =



When you work on a problem on your own, without the computer-provided feedback, only you can assess whether your answer seems right. The following questions will help you practice the skills necessary for such an assessment.

Part F

Look at the pictorial diagram that you drew for this problem. Consider the pivot point to be at the center of mass of the bar. Intuitively, what can you say about the forces of tension?

ANSWER:

The torques due to the forces of tension must balance out each other. Because the moment arm for is greater than that for , it follows that .

Part G

For the bar to experience no net force, what must the tension in string 2 be?

Express your answer in terms of appropriate constants and any of the given variables.

ANSWER: =

Note that this answer confirms the previous result that .

Axis of Rotation and Moment of Inertia Ranking Task

Two identical uniform solid spheres are attached by a solid uniform thin rod, as shown in the figure. The rod lies on a line connecting the centers of mass of the two spheres. The axes A, B, C, and D are in the plane of the page (which also contains the centers of mass of the spheres and the rod), while axes E and F (represented by black dots) are perpendicular to the page.



Part A

Rank the moments of inertia of this object about the axes indicated.

Hint A.1

Hint not displayed

Hint A.2

Hint not displayed

Rank from largest to smallest.

ANSWER:

View

Kinetic Energy and Moment of Inertia

Consider a particle of mass that is revolving with angular speed around an axis. The perpendicular distance from the particle to the axis is .

Part A

Which of the following are legitimate units for expressing rotational velocity, commonly denoted by ? (The rotational velocity appears in many rotational motion equations such as and

.)

a. radians per second b. degrees per second c. meters per second d. arc seconds e. revolutions per second

Enter "t" for each unit that can be used to express rotational velocity, and "f" for each that cannot be used. For example, if


C=F > B > A=E > D


only the last unit can be used, you would enter "f,f,f,f,t".

ANSWER: t,f,f,f,f

The rate of rotation can be expressed in radians per second, degrees per second, or revolutions per second. However, the angular speed is most often defined as

, with measured in radians. In this

problem, we will be using radians per second to express angular speed.

Part B

Find the kinetic energy of the rotating particle.

Part B.1

Part not displayed


ANSWER: =

Part C

The kinetic energy of a rotating body is generally written as , where is the moment of

inertia. Find the moment of inertia of the particle described in the problem introduction with respect to the axis about which it is rotating.

Hint C.1

Hint not displayed

Give your answer in terms of and , not .

ANSWER: =

Consider a system of several point masses all rotating about the same axis. The total kinetic energy of the system is the sum of the kinetic energies of all of the masses. For point masses,



.

Comparing this to the expression for the kinetic energy given in Part C, we see that the moment of inertia is

.

In other words, the total moment of inertia is the sum of the moments of inertia of each mass.

Part D

Find the moment of inertia of a hoop of radius and mass with respect to an axis

perpendicular to the hoop and passing through its center. Part D.1

Part not displayed

Express your answer in terms of and .

ANSWER: =

The End of the Song



As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration.

Part A

If the CD rotates at 500 while the last song is playing, and then spins down to zero angular speed in 2.60 with constant angular acceleration, what is , the magnitude of the angular acceleration of the CD, as it spins to a stop?

Hint A.1 Angular acceleration

Hint not displayed

Express your answer in radians per second squared.

ANSWER: = 20.1

Part B

How many complete revolutions does the CD make as it spins to a stop?

Part B.1

Part not displayed

Your answer should be an integer.

ANSWER: 10.0 revolutions


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