Chapter 09 Homework
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Transcript of Chapter 09 Homework
4/24/2014 Chapter 9 Homework
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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
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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))
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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
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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
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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
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Part C
Calculate the instantaneous value of the angular velocity at time .
Express your answer in radians per second.
ANSWER:
Correct
Part D
Calculate the average angular velocity for the time interval to seconds.
Express your answer in radians per second.
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
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Hint 1. How to approach the problem
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
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ANSWER:
Correct
Part B
How long does it take for the salad spinner to come to rest?
Express your answer numerically in seconds.
Hint 1. How to approach the problem
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
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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
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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
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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)
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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
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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.
Express your answer in terms of and .
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.
Express your answer in terms of and .
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
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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.
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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?
Hint 1. Distinguishing tangential, centripetal, and 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!).
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
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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
Part E
Who moves with greater magnitude of angular acceleration?
Hint 1. Distinguishing tangential, centripetal, and 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
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 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
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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
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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
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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
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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
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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
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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
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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
4/24/2014 Chapter 9 Homework
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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
4/24/2014 Chapter 9 Homework
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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
4/24/2014 Chapter 9 Homework
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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
4/24/2014 Chapter 9 Homework
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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