EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
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Chapter 13, Problem 73PQ

A uniform disk of mass m = 10.0 kg and radius r = 34.0 cm mounted on a frictionless axle through its center, and initially at rest, is acted upon by two tangential forces of equal magnitude F, acting on opposite sides of its rim until a point on the rim experiences a centripetal acceleration of 4.00 m/s2 (Fig. P13.73). a. What is the angular momentum of the disk at this time? b. If F = 2.00 N, how long do the forces have to be applied to the disk to achieve this centripetal acceleration?

Chapter 13, Problem 73PQ, A uniform disk of mass m = 10.0 kg and radius r = 34.0 cm mounted on a frictionlessaxle through its

FIGURE P13.73

(a)

Expert Solution
Check Mark
To determine

Angular momentum of the disk.

Answer to Problem 73PQ

Angular momentum of the disk is 1.98kgm2/s_.

Explanation of Solution

The disk in the question rotates about an axis passing through the center of the disk. Also the axis is perpendicular to the surface of the disk.

Write the equation to find the moment of inertia of the disk about an axis passing through its center.

  I=12Mr2                                                                                                          (I)

Here, I is the moment of inertia of the disk about the axis passing through the center, M is the mass of the disk , and r is the radius of the disk.

Write the equation to find the centripetal acceleration felt by the disk.

  a=v2r                                                                                                                (II)

Here, a is the centripetal acceleration, v is the linear velocity, and r is the radius of the disk.

Write the equation to find the linear speed.

  v=rω                                                                                                                 (III)

Here, ω is the angular speed of the disk.

Substitute equation (III) in (II).

  a=(rω)2r=ω2r                                                                                                        (IV)

Rewrite equation (IV) to get ω.

  ω=ar                                                                                                                (V)

Write the equation to find the angular momentum of the disk.

  L=Iω                                                                                                               (VI)

Here, L is the angular momentum.

Substitute (V) to (VI) to get L.

  L=Iar                                                                                                           (VII)

Conclusion:

Substitute 10.0kg for M and 34cm for r in equation (I) to get I.

  I=12(10.0kg)(34cm(1m102cm))2=0.578kgm2

Substitute 4.00m/s2 for a , 0.578kgm2 for I and 34cm for r in equation (VII) to get L.

  L=0.578kgm2(4.00m/s2)34cm(1m102cm)=1.98kgm2/s

Therefore, angular momentum of the disk is 1.98kgm2/s_.

(b)

Expert Solution
Check Mark
To determine

The time duration for which the force have to be applied so that the disk achieves the centripetal acceleration.

Answer to Problem 73PQ

The forces have to act for a time duration of 1.46s_.

Explanation of Solution

Torque is the rate of change of angular momentum.

Write the equation to find the torque acting on the disk.

  τ=ΔLΔt                                                                                                               (VIII)

Here, τ is the torque acting on the disk, ΔL is the change in angular momentum and Δt is the time duration.

Rewrite equation (VIII) to get Δt.

  Δt=ΔLτ                                                                                                                (IX)

Write the equation to find τ in both the sides of the disk.

  τ=2Fr

Here, F is the force acting on the disk.

Substitute above equation in (IX).

  Δt=ΔL2Fr                                                                                                             (X)

Change in angular momentum is the difference between the final angular momentum and initial angular momentum of the disk.

Rewrite equation (IX).

  Δt=LfLi2Fr                                                                                                       (XI)

Here, Lf is the final angular momentum and Li is the initial angular momentum.

Conclusion:

The disk was initially at rest. Therefore Li is zero.

Substitute 1.98kgm2/s for Lf, 0kgm2/s for Li, 2.00N for F , and 34cm for r in equation (XI) to get Δt.

  Δt=1.98kgm2/s2(2.00N)(34cm(1m102cm))=1.46s

Therefore, the forces have to act for a time duration of 1.46s_.

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Chapter 13 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

Ch. 13 - Rotational Inertia Problems 5 and 6 are paired. 5....Ch. 13 - A 12.0-kg solid sphere of radius 1.50 m is being...Ch. 13 - A figure skater clasps her hands above her head as...Ch. 13 - A solid sphere of mass M and radius Ris rotating...Ch. 13 - Suppose a disk having massMtot and radius R is...Ch. 13 - Problems 11 and 12 are paired. A thin disk of...Ch. 13 - Given the disk and density in Problem 11, derive...Ch. 13 - A large stone disk is viewed from above and is...Ch. 13 - Prob. 14PQCh. 13 - A uniform disk of mass M = 3.00 kg and radius r =...Ch. 13 - Prob. 16PQCh. 13 - Prob. 17PQCh. 13 - The system shown in Figure P13.18 consisting of...Ch. 13 - A 10.0-kg disk of radius 2.0 m rotates from rest...Ch. 13 - Prob. 20PQCh. 13 - Prob. 21PQCh. 13 - In Problem 21, what fraction of the kinetic energy...Ch. 13 - Prob. 23PQCh. 13 - Prob. 24PQCh. 13 - Prob. 25PQCh. 13 - A student amuses herself byspinning her pen around...Ch. 13 - The motion of spinning a hula hoop around one's...Ch. 13 - Prob. 28PQCh. 13 - Prob. 29PQCh. 13 - Prob. 30PQCh. 13 - Sophia is playing with a set of wooden toys,...Ch. 13 - Prob. 32PQCh. 13 - A spring with spring constant 25 N/m is compressed...Ch. 13 - Prob. 34PQCh. 13 - Prob. 35PQCh. 13 - Prob. 36PQCh. 13 - Prob. 37PQCh. 13 - Prob. 38PQCh. 13 - A parent exerts a torque on a merry-go-round at a...Ch. 13 - Prob. 40PQCh. 13 - Today, waterwheels are not often used to grind...Ch. 13 - Prob. 42PQCh. 13 - A buzzard (m = 9.29 kg) is flying in circular...Ch. 13 - An object of mass M isthrown with a velocity v0 at...Ch. 13 - A thin rod of length 2.65 m and mass 13.7 kg is...Ch. 13 - A thin rod of length 2.65 m and mass 13.7 kg is...Ch. 13 - Prob. 47PQCh. 13 - Two particles of mass m1 = 2.00 kgand m2 = 5.00 kg...Ch. 13 - A turntable (disk) of radius r = 26.0 cm and...Ch. 13 - CHECK and THINK Our results give us a way to think...Ch. 13 - Prob. 51PQCh. 13 - Prob. 52PQCh. 13 - Two children (m = 30.0 kg each) stand opposite...Ch. 13 - A disk of mass m1 is rotating freely with constant...Ch. 13 - Prob. 55PQCh. 13 - Prob. 56PQCh. 13 - The angular momentum of a sphere is given by...Ch. 13 - Prob. 58PQCh. 13 - Prob. 59PQCh. 13 - Prob. 60PQCh. 13 - Prob. 61PQCh. 13 - Prob. 62PQCh. 13 - A uniform cylinder of radius r = 10.0 cm and mass...Ch. 13 - Prob. 64PQCh. 13 - A thin, spherical shell of mass m and radius R...Ch. 13 - To give a pet hamster exercise, some people put...Ch. 13 - Prob. 67PQCh. 13 - Prob. 68PQCh. 13 - The velocity of a particle of mass m = 2.00 kg is...Ch. 13 - A ball of mass M = 5.00 kg and radius r = 5.00 cm...Ch. 13 - A long, thin rod of mass m = 5.00 kg and length =...Ch. 13 - A solid sphere and a hollow cylinder of the same...Ch. 13 - A uniform disk of mass m = 10.0 kg and radius r =...Ch. 13 - When a person jumps off a diving platform, she...Ch. 13 - One end of a massless rigid rod of length is...Ch. 13 - A uniform solid sphere of mass m and radius r is...Ch. 13 - Prob. 77PQCh. 13 - A cam of mass M is in the shape of a circular disk...Ch. 13 - Prob. 79PQCh. 13 - Consider the downhill race in Example 13.9 (page...Ch. 13 - Prob. 81PQ
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