Chapter 09 Homework

4/24/2014 Chapter 9 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 1/27

Chapter 9 HomeworkDue: 10:00pm on Wednesday, April 9, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Exercise 9.1

Part A

What angle in radians is subtended by an arc of 1.56 in length on the circumference of a circle of radius 2.56 ?

ANSWER:

Correct

Part B

What is this angle in degrees?

ANSWER:

Correct

Part C

An arc of length 14.2 on the circumference of a circle subtends an angle of 124 . What is the radius of thecircle?

ANSWER:

Correct

Part D

The angle between two radii of a circle with radius 1.47 is 0.660 . What length of arc is intercepted on thecircumference of the circle by the two radii?

ANSWER:

m m

= 0.609 θ rad

= 34.9 θ ∘

cm ∘

= 6.56 r cm

m rad



Correct

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 specifyits position solely by , the angle that the vector from the origin to the block makes with our chosen reference axis attime . 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.

Give your answer in terms of , , and unit vectors and corresponding to the coordinate system in

the figure.

Hint 1. x coordinate

What is the x coordinate of the particle?

Your answer should be in terms of and .

ANSWER:

Hint 2. y coordinate

= 0.970 L m

Rθ(t)

t θ(t)

(t)r θ(t) θ(t)

R θ(t) i j

R θ(t)

= x Rcos(θ(t))



What is the y coordinate of the particle?

Your answer should be in terms of and .

ANSWER:

ANSWER:

Correct

Uniform Circular Motion A frequently encountered kind of circular motion is uniform circular motion, where changes at a constant rate . Inother words,

.

Usually, .

Part B

For uniform circular motion, find at an arbitrary time .

Give your answer in terms of and .

ANSWER:

Correct

Part C

What does become now?

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

ANSWER:

R θ(t)

= y Rsin(θ(t))

= (t)r Rcos(θ(t)) + Rsin(θ(t))i j

θ(t) ω

ω = dθ(t)dt

θ(t = 0) = 0

θ(t) t

ω t

= θ(t) ωt

(t)r

R ω t i j

= (t)r Rcos(ωt) + Rsin(ωt)i j



Correct

Part D

Find , a position vector at time .

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

Hint 1. Finding

Simply plug into your expression for the components of .

ANSWER:

Correct

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 .

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

Hint 1. Adding a phase

You can think of changing the initial position as adding a phase angle to the equation for . That is,

.

Hint 2. Finding a phase

From previous parts you found that and . What should the angle be

for and to be equal to 0 and respectively?

Express your answer as a fraction of the number , for example or .

ANSWER:

ANSWER:

r t = 0

R i j

r

t = 0 (t)r

= r Ri

t = 0 t = 0( , ) = (0,R)x0 y0 ω

R ω t i j

ϕ θ(t)θ(t) = ωt + ϕ

x = R cos(θ(t)) y = R sin(θ(t)) θ

x y R

π (3/4)π (1/4)π

= 1.57θ



Correct

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

Pushing a Merry-Go-Round

A child is pushing a playground merry-go-round. The angle through which the merry-go-round has turned varies with timeaccording to , where and .

Part A

Calculate the angular velocity of the merry-go-round as a function of time.

Express your answer in radians per second in terms of , , and .

Hint 1. Position versus velocity

Recall that the angular velocity of an object is just the time derivative of its angular position.

ANSWER:

Correct

Part B

What is the initial value of the angular velocity?

Express your answer in radians per second.

Hint 1. Position versus velocity

Recall that the angular velocity of an object is just the time derivative of its angular position. The initial valueis just the value at .

ANSWER:

= (t)ryaxis Rcos(ωt + 1.57) + Rsin(ωt + 1.57)i j

θ

θ(t) = γt + βt3 γ = 0.400 rad/s β = 0.0120 rad/s3

γ β t

= ω(t) γ + 3βt2 rad/sec

ω0

t = 0 s

= 0.4 ω0 rad/s



Correct

Part C

Calculate the instantaneous value of the angular velocity at time .


ANSWER:

Correct

Part D

Calculate the average angular velocity for the time interval to seconds.


Hint 1. How to approach the problem

In order to find the average angular velocity, just take the total angular displacement and divide by the totaltime. You can find the total angular displacement from the formula in the introduction for angulardisplacement .

ANSWER:

Correct

Constant Angular Acceleration in the Kitchen

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 secondsand then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinnerslows down with constant angular acceleration.

Part A

What is the magnitude of the angular acceleration of the salad spinner as it slows down?

Express your answer numerically in radians per second per second.

ω(t) t = 5.00 s

= 1.3 ω(5.00) rad/s

ωav t = 0 t = 5.00

θ(t)

= 0.7 ωav rad/s




Recall from your study of kinematics the three equations of motion derived for systems undergoing constantlinear acceleration. You are now studying systems undergoing constant angular acceleration and will needto work with the three analogous equations of motion. Collect your known quantities and then determinewhich of the angular kinematic equations is appropriate to find the angular acceleration .

Hint 2. Find the angular velocity of the salad spinner while Dario is spinning it

What is the angular velocity of the salad spinner as Dario is spinning it?

Express your answer numerically in radians per second.

Hint 1. Converting rotations to radians

When the salad spinner spins through one revolution, it turns through radians.

ANSWER:

Hint 3. Find the angular distance the salad spinner travels as it comes to rest

Through how many radians does the salad spinner rotate as it comes to rest?

Express your answer numerically in radians.

Hint 1. Converting rotations to radians

One revolution is equivalent to radians.

ANSWER:

Hint 4. Determine which equation to use

You know the initial and final velocities of the system and the angular distance through which the spinnerrotates as it comes to a stop. Which equation should be used to solve for the unknown constant angularacceleration ?

ANSWER:

α

2π

= 25.1 ω0 radians/s

Δθ = θ − θ0

2π

= 37.7 Δθ radians

α

θ = + t+ αθ0 ω012

t2

ω = + αtω0

= + 2α(θ − )ω2 ω20 θ0



ANSWER:

Correct

Part B

How long does it take for the salad spinner to come to rest?

Express your answer numerically in seconds.


Again, you will need the equations of rotational kinematics that apply to situations of constant angularacceleration. Collect your known quantities and then determine which of the angular kinematic equations isappropriate to find .

Hint 2. Determine which equation to use

You have the initial and final velocities of the system and the angular acceleration, which you found in theprevious part. Which is the best equation to use to solve for the unknown time ?

ANSWER:

ANSWER:

Correct

Marching Band

A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in anevent, such as a parade, and must round a curve in the road, the musician on the outside of the curve must walk aroundthe curve in the same amount of time as the musician on the inside of the curve. This motion can be approximated by adisk rotating at a constant rate about an axis perpendicular to its plane. In this case, the axis of rotation is at the insideof the curve.

= 8.38 α radians/s2

t

t

θ = + t+ αθ0 ω012 t2

ω = + αtω0

= + 2α(θ − )ω2 ω20 θ0

= 3.00 t s



Consider two musicians, Alf and Beth. Beth is four times the distance from the inside of the curve as Alf.

Part A

If Beth travels a distance during time , how far does Alf travel during the same amount of time?

Hint 1. Find the angle through which Alf rotates

If Beth rotates through an angle of during time , through what angle does Alf rotate during the sameamount of time?

Hint 1. Angular velocity

At any given instant, every part of a rigid body has the same angular velocity , where is given bythe relationship

.

ANSWER:

Hint 2. Arc length

If an angle (measured in radians) is subtended by an arc of length on a circle of radius , as shown inthe figure, then

.

s Δt

θ Δt

ω ω

ω = ΔθΔt

4θ

2θ

θ

θ12

θ14

θ s r

s = rθ



Use this formula to compare the lengths of the arcs that Alf and Beth trace out during equal time intervals.

ANSWER:

Correct

The musician on the outside of the curve must travel farther than the musician on the inside of the curve inorder to maintain the marching band's straight, even rows.

Part B

If Alf moves with speed , what is Beth's speed? Speed in this case means the magnitude of the linear velocity, notthe magnitude of the angular velocity.

ANSWER:

Correct

The musician on the outside of the curve must travel faster than the musician on the inside of the curve. This iswhy most of the musicians on the outside of a curve appear to be jogging while their colleagues on the insideof the curve march in place.

Constrained Rotation and Translation

Learning Goal:

To understand that contact between rolling objects and what they roll against imposes constraints on the change inposition(velocity) and angle (angular velocity).

The way in which a body makes contact with the world often imposes a constraint relationship between its possiblerotation and translational motion. A ball rolling on a road, a yo-yo unwinding as it falls, and a baseball leaving thepitcher's hand are all examples of constrained rotation and translation. In a similar manner, the rotation of one body andthe translation of another may be constrained, as happens when a fireman unrolls a hose from its storage drum.

Situations like these can be modeled by constraint equations, relating the coupled angular and linear motions. Although

4s

2s

s12

s14s

v

4vv

v14



these equations fundamentally involve position (the angle of the wheel at a particular distance down the road), it isusually the relationship of velocities and accelerations that are relevant in solving a problem involving such constraints.The velocities are needed in the conservation equations for momentum and angular momentum, and the accelerationsare needed for the dynamical equations.

It is important to use the standard sign conventions: positive for counterclockwise rotation and positive for motion towardthe right. Otherwise, your dynamical equations will have to be modified. Unfortunately, a frequent result will be theappearance of negative signs in the constraint equations.

Consider a measuring tape unwinding from a drum of radius .The center of the drum is not moving; the tape unwinds as itsfree end is pulled away from the drum. Neglect the thicknessof the tape, so that the radius of the drum can be assumednot to change as the tape unwinds. In this case, the standardconventions for the angular velocity and for the(translational) velocity of the end of the tape result in aconstraint equation with a positive sign (e.g., if , that is,the tape is unwinding, then also).

Part A

Assume that the function represents the length of tape that has unwound as a function of time. Find , the

angle through which the drum will have rotated, as a function of time.

Express your answer (in radians) in terms of and any other given quantities.

Hint 1. Find the amount of tape that unrolls in one complete revolution of the drum

If the measuring tape unwinds one complete revolution ( ), how much tape, , will have unwound?

ANSWER:

ANSWER:

Correct

Part B

r

ωv

v > 0ω > 0

x(t) θ(t)

x(t)

θ = 2π x2π

= x2π 2πr

= radians θ(t) x(t)r

ω(t)



The tape is now wound back into the drum at angular rate . With what velocity will the end of the tape move?

(Note that our drawing specifies that a positive derivative of implies motion away from the drum. Be careful with

your signs! The fact that the tape is being wound back into the drum implies that , and for the end of the

tape to move closer to the drum, it must be the case that .

Answer in terms of and other given quantities from the problem introduction.

Hint 1. How to approach the probelm

The function is given by the derivative of with respect to time. Compute this derivative using the

expression for found in Part A and the fact that .

Express your answer in terms of and .

ANSWER:

ANSWER:

Correct

Part C

Since is a positive quanitity, the answer you just obtained implies that will always have the same sign as

. If the tape is unwinding, both quanitites will be positive. If the tape is being wound back up, both quantities

will be negative. Now find , the linear acceleration of the end of the tape.

Express your answer in terms of , the angular acceleration of the drum: .

ANSWER:

Correct

Part D

Perhaps the trickiest aspect of working with constraint equations for rotational motion is determining the correctsign for the kinematic quantities. Consider a tire of radius rolling to the right, without slipping, with constant x

ω(t)x(t)

ω(t) < 0v(t) < 0

ω(t)

ω(t) θ(t)

θ(t) = v(t)dx(t)dt

v(t) r

= ω(t) v(t)r

= v(t) rω(t)

r v(t)ω(t)

a(t)

α(t) α(t) = dω(t)dt

= a(t) rα(t)

r



velocity . Find , the (constant) angular velocity of the

tire. Be careful of the signs in your answer; recall thatpositive angular velocity corresponds to rotation in thecounterclockwise direction.


ANSWER:

Correct

This is an example of the appearance of negative signs in constraint equations--a tire rolling in the positivedirection translationally exhibits negative angular velocity, since rotation is clockwise.

Part E

Assume now that the angular velocity of the tire, which continues to roll without slipping, is not constant, but ratherthat the tire accelerates with constant angular acceleration . Find , the linear acceleration of the tire.


ANSWER:

Correct

Linear and Rotational Quantities Conceptual Question

A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at pointA and Bobby is riding at point B.

vx ω

vx r

= ω−vx

r

α ax

α r

= ax −rα



Part A

Which child moves with greater magnitude of velocity?

Hint 1. Distinguishing between velocity and angular velocity

Ana’s (or Bobby’s) velocity is determined by the actual distance traveled (typically in meters) in a given timeinterval. The angular velocity is determined by the angle through which he rotates (typically in radians) in agiven time interval.

ANSWER:

Correct

Part B

Who moves with greater magnitude of angular velocity?

Hint 1. Distinguishing between velocity and angular velocity

Ana’s (or Bobby’s) velocity is determined by the actual distance he travels (typically in meters) in a giventime interval. His angular velocity is determined by the angle through which he rotates (typically in radians) ina given time interval.

ANSWER:

Ana has the greater magnitude of velocity.

Bobby has the greater magnitude of velocity.

Both Ana and Bobby have the same magnitude of velocity.



Correct

Part C

Who moves with greater magnitude of tangential acceleration?

Hint 1. Distinguishing tangential, centripetal, and angular acceleration

Ana’s tangential and centripetal acceleration are components of his acceleration vector. During circularmotion, if Ana’s speed is changing (meaning the merry-go-round is speeding up or slowing down) he willhave a nonzero tangential acceleration. However, even if the merry-go-round is turning at constant angularspeed, he will experience a centripetal acceleration, because the direction of his velocity vector is changing(you can’t move along a circular path unless your direction of travel is changing!).Both tangential and centripetal accelerations have units of , since they are the two-dimensional

components of linear acceleration. Angular acceleration, on the other hand, is a measure of the change inAna’s angular velocity. If his rate of rotation is changing, he will have a nonzero angular acceleration. Thus,angular acceleration has units of .

ANSWER:

Correct

Both Ana and Bobby are maintaining a constant speed, so they both have a tangential acceleration of zero(thus they are equal)!

Part D

Who has the greater magnitude of centripetal acceleration?


Ana’s tangential and centripetal acceleration are components of his acceleration vector. For circular motion,if Ana’s speed is changing (meaning the merry-go-round is speeding up or slowing down) he will have anonzero tangential acceleration. However, even if the merry-go-round is turning at constant angular speed, hewill experience a centripetal acceleration, because the direction of his velocity vector is changing (you can’tmove along a circular path unless your direction of travel is changing!).

Ana has the greater magnitude of angular velocity.

Bobby has the greater magnitude of angular velocity.

Both Ana and Bobby have the same magnitude of angular velocity.

m/s2

rad/s2

Ana has the greater magnitude of tangential acceleration.

Bobby has the greater magnitude of tangential acceleration.

Both Ana and Bobby have the same magnitude of tangential acceleration.

m/ 2



Both tangential and centripetal accelerations have units of , since they are the two-dimensional


ANSWER:

Correct

Part E

Who moves with greater magnitude of angular acceleration?


Ana’s tangential and centripetal acceleration are components of his acceleration vector. For circular motion,if Ana’s speed is changing (meaning the merry-go-round is speeding up or slowing down) he will have anonzero tangential acceleration. However, even if the merry-go-round is turning at constant angular speed, hewill experience a centripetal acceleration, because the direction of his velocity vector is changing (you can’tmove along a circular path unless your direction of travel is changing!).Both tangential and centripetal accelerations have units of , since they are the two-dimensional


ANSWER:

Correct

Both Ana and Bobby are maintaining a constant angular velocity, so they both have an angular acceleration ofzero (thus they are equal)!

Exercise 9.20

m/s2

rad/s2

Ana has the greater magnitude of centripetal acceleration.

Bobby has the greater magnitude of centripetal acceleration.

Both Ana and Bobby have the same magnitude of centripetal acceleration.

m/s2

rad/s2

Ana has the greater magnitude of angular acceleration.

Bobby has the greater magnitude of angular acceleration.

Both Ana and Bobby have the same magnitude of angular acceleration.

m−7



A compact disc (CD) stores music in a coded pattern of tiny pits deep. The pits are arranged in a track thatspirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 and 58.0 ,respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 .

Part A

What is the angular speed of the CD when scanning the innermost part of the track?

ANSWER:

Correct

Part B

What is the angular speed of the CD when scanning the outermost part of the track?

ANSWER:

Correct

Part C

The maximum playing time of a CD is 74.0 . What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?

ANSWER:

Correct

Part D

What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take thedirection of rotation of the disc to be positive.

ANSWER:

Correct

m10−7

mm mmm/s

= 50.0 ω rad/s

= 21.6 ω rad/s

min

= 5.55 L km

= −6.41×10−3 αav rad/s2



Exercise 9.26

Part A

Derive an equation for the radial acceleration that includes and , but not .

ANSWER:

Correct

Part B

You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 when

the tangential velocity of that point has magnitude 2.00 . What angular velocity is required to achieve these

values?

ANSWER:

Correct

Weight and Wheel

Consider a bicycle wheel that initially is not rotating. A block of mass is attached to the wheel and is allowed to fall adistance . Assume that the wheel has a moment of inertia about its rotation axis.

Part A

Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius

. Find , the angular speed of the wheel after the block has fallen a distance , for this case.

Express in terms of , , , , and .

v ω r

= arad vω

m/s2

m/s

= 0.250 ω rad/s

mh I

rA

ωA h

ωA m g h rA I



Hint 1. How to approach this problem

The most straighforward way to solve this problem is to use conservation of mechanical energy. The totalinitial energy of the system is equal to the total final energy of the system (where the system consists of thewheel and the block). In other words,

.

Where is the initial energy of the system, is the final energy of the block and is the final energy

of the wheel.

Hint 2. Initial energy of the system

Initially, the wheel is not rotating. The initial energy of the system consists of the gravitational potentialenergy stored in the block, since it is not moving either. Supposing that the gravitiational potential energy ofthe block is zero at "ground level," find the initial energy of the system.

ANSWER:

Hint 3. Final energy of block

Find the final energy of the block.

Express the final energy of the block in terms of given quantities (excluding ) and the unknownfinal angular velocity of the wheel, .

Hint 1. Final velocity of the block

Find , the magnitude of the final velocity of the block.

Express the velocity in terms of and the final angular velocity of the wheel, .

ANSWER:

= +E i Ebf Ewf

Ei Ebf Ewf

= E i mgh

h

ωA

vf

rA ωA

= vf rAωA



ANSWER:

Hint 4. Final energy of wheel

Find the final kinetic energy of the wheel.

Express your answer in terms of (the wheel's moment of inertia) and .

ANSWER:

ANSWER:

Correct

Part B

Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel ofradius . Find , the angular speed of the wheel after

the block has fallen a distance , for this case.

Express in terms of , , , , and .

Hint 1. Similarity to previous part

The derivation of is exactly the same as the derivation for , using instead of .

= Ebf m12 ( )rAωA

2

I ωA

= Ewf I12

ωA2

= ωA

2mgh

m +IrA2

− −−−−−−√

rB ωB

h

ωB m g h rB I

ωB ωA rB rA



ANSWER:

Correct

Part C

Which of the following describes the relationship between and ?

Hint 1. How to approach this question

To figure out which angular velocity is greater ( or ), you only need to consider the radius dependence

of the expression for . Ignoring all of the other parameters, you should have found that goes as 1/radius(where "radius" refers to where the string is attached, which is not necessarily the outer radius of the wheel).The problem then reduces to figuring out which is greater, or .

ANSWER:

Correct

This is related to why gears are found on the inside rather than the outside of a wheel.

Exercise 9.30

Four small spheres, each of which you can regard as a point of mass 0.200 , are arranged in a square 0.400 on aside and connected by light rods .

= ωB

2mgh

m +IrB2

− −−−−−−√

ωA ωB

ωA ωB

ω ω

1/rA 1/rB

>ωA ωB

>ωB ωA

=ωA ωB

kg m



Part A

Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane(an axis through point O in the figure).

ANSWER:

Correct

Part B

Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis alongthe line AB in the figure).

ANSWER:

Correct

Part C

Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lowerright spheres and through point O.

ANSWER:

= 6.40×10−2 I kg ⋅ m2

= 3.20×10−2 I kg ⋅ m2

= 3.20×10−2 I kg ⋅ m2



Correct

Exercise 9.31

Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult TableMoments of Inertia of Various Bodies in the Textbook as needed.

Part A

A thin 2.50- rod of length 90.0 , about an axis perpendicular to it and passing through one end.

ANSWER:

Correct

Part B

A thin 2.50- rod of length 90.0 , about an axis perpendicular to it and passing through its center.

ANSWER:

Correct

Part C

A thin 2.50- rod of length 90.0 , about an axis parallel to the rod and passing through it.

ANSWER:

Correct

Part D

A 4.50- sphere 30.0 in diameter, about an axis through its center, if the sphere is solid.

ANSWER:

kg cm

= 0.675 I kg ⋅ m2

kg cm

= 0.169 I kg ⋅ m2

kg cm

= 0 I kg ⋅ m2

kg cm



Correct

Part E

A 4.50- sphere 30.0 in diameter, about an axis through its center, if the sphere is a thin-walled hollow shell.

ANSWER:

Correct

Part F

An 7.50- cylinder, of length 14.0 and diameter 10.0 , about the central axis of the cylinder, if the cylinder

is thin-walled and hollow.

ANSWER:

Correct

Part G

An 7.50- cylinder, of length 14.0 and diameter 10.0 , about the central axis of the cylinder, if the cylinder

is solid.

ANSWER:

Correct

Exercise 9.35

A wagon wheel is constructed as shown in the figure . The radius of the wheel is 0.300 , and the rim has mass 1.45 . Each of the eight spokes, that lie along a diameter and are 0.300 long, has mass 0.200 .

= 4.05×10−2 I kg ⋅ m2

kg cm

= 6.75×10−2 I kg ⋅ m2

kg cm cm

= 1.88×10−2 I kg ⋅ m2

kg cm cm

= 9.38×10−3 I kg ⋅ m2

mkg m kg



Part A

What is the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of thewheel?

ANSWER:

Correct

Exercise 9.44

A light, flexible rope is wrapped several times around a hollow cylinder with a weight of 40.0 and a radius of 0.25 ,that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligiblemoment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force P for a distanceof 5.00 , at which point the end of the rope is moving at 6.00 .

Part A

If the rope does not slip on the cylinder, what is the value of P?

ANSWER:

Correct

Parallel Axis Theorem

= 0.179 I kg ⋅ m2

N m

m m/s

= 14.7 P N

Icm



The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center ofmass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematicalstatement of the theorem is , where is the perpendicular distance from the center of mass to theaxis 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.

Express in terms of and . Use fractions rather than decimal numbers in your answer.

Hint 1. Find the distance from the axis to the center of mass

Find the distance appropriate to this problem. That is, find the perpendicular distance from the center ofmass of the rod to the axis passing through one end of the rod.

ANSWER:

ANSWER:

Correct

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

Express in terms of and . Use fractions rather than decimal numbers in your answer.

Icm

Ip

= + MIp Icm d2 dM

L m

= mIcm112

L2

Iend

Iend m L

d

= d L2

= IendmL2

3

m a Icm

= mIcm16

a2 Iedge

Iedge m a



Hint 1. Find the distance from the axis to the axis

Find the perpendicular distance from the center of mass axis to the new edge axis (axis labeled p in thefigure).

ANSWER:

ANSWER:

Correct

Score Summary:Your score on this assignment is 99.8%.You received 14.97 out of a possible total of 15 points.

o p

d

= da

2√

= Iedge2ma2

3

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Chapter 09 Homework

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