Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 11, Problem 61CP

(a)

To determine

The angular speed of the disk once pure rolling takes place.

(a)

Expert Solution
Check Mark

Answer to Problem 61CP

The angular speed of the disk once pure rolling takes place is 13ωi_.

Explanation of Solution

For the particle under the net force model the net force will be the rate of change of linear momentum of the body.

Write the expression for net force on the disc .

  fk=m(vfvi)Δt

Here, m is mass of the disc, vf is final linear velocity of the disc, Δt is time of contact, vi is initial linear velocity of the disc and fk is for net force on the disc.

The initial velocity of the disc is zero.

Substitute 0 for vi in above expression as.

  fk=mvfΔt                                                                                                       (I)

For the particle under the net torque model the net torque will be the rate of change of angular momentum of the body.

Write the expression for net torque on the disc as.

  fkR=I(ωfωi)Δt

Simplify the above expression for fk as.

  fk=I(ωfωi)RΔt                                                                                         (II)

Here, ωi is the initial angular speed of the disc, ωf is the final angular speed of the disc, I is a moment of inertia of the disc about a rotational axis and R is the radius of the disc.

Substitute mvfΔt for fk in equation (II).

  mvfΔt=I(ωfωi)RΔt

Simplify the above-obtained expression for R,

  R=I(ωiωf)mvf                                                                                           (III)

Write the expression for final linear velocity of the disc.

  vf=Rωf                                                                                                    (IV)

Substitute Rωf for vf in equation (IV).

  R=I(ωiωf)mRωf                                                                                           (V)

Simplify equation (V) for ωf .

  R=I(ωiωf)mRωf

Re-arrange the terms.

  mR2ωf=IωiIωfωf(I+mR2)=Iωi

Simplify the above expression for ωf .

  ωf=IωiI+mR2                                                                                            (VI)

Write the expression for moment of inertia of the disc as.

  I=12mR2                                                                                                 (VII)

Conclusion:

Substitute 12mR2 for I in equation (VII).

  ωf=(12mR2)ωi(12mR2)+mR2

Simplify the above expression for ωf it gives.

  ωf=12mR2ωi32mR2=13ωi

Thus, the angular speed of the disk once pure rolling takes place is 13ωi_.

(b)

To determine

The fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs.

(b)

Expert Solution
Check Mark

Answer to Problem 61CP

The fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs 23_.

Explanation of Solution

Write the expression for the rotational kinetic energy of the disc as.

  (KE)R=12Iω2                                                                                        (VIII)

Here, (KE)R is the rotational kinetic energy of the disc and ω is the corresponding angular speed of the disc.

Write the expression for kinetic energy due to the linear velocity of the disc as

  EL=12mvCM2                                                                                              (IX)

Here, EL is kinetic energy due to the linear velocity of the disc and vCM linear velocity of the center of mass of the disc.

The change in kinetic energy is the difference between initial and final kinetic energy.

Initially, the disc is only rotating so initial kinetic energy is rotational kinetic energy. In the final condition, the disc has linear and angular velocities so the kinetic energy becomes the sum of linear and rotational kinetic energy.

Write the expression for the change in kinetic energy of the disc as.

  ΔE=12Iωf2+12mvCM212Iωi2                                                                  (X)

Here, ΔE is the change in kinetic energy.

Divide equation (X) by equation (IX) it gives.

  ΔEEL=12Iωf2+12mvCM212Iωi212Iωi2                                                              (XI)

Conclusion:

Substitute ωi3 for ωf, 12mR2 for I and Rωi3 for vCM in equation (XI)

  ΔEEL=12(12mR2)(ωi3)2+12m(Rωi3)212(12mR2)ωi212(12mR2)ωi2=(12mR2ωi2)[(12)(19)+(19)(12)]12(12mR2)ωi2=(13)(12)

Simplify the above expression for ΔEEL it gives.

  ΔEEL=23

Thus, the fractional change in kinetic energy from the moment the disk is set down until pure rolling occurs 23_.

(c)

To determine

The time interval after setting the disk down before pure rolling motion begins.

(c)

Expert Solution
Check Mark

Answer to Problem 61CP

The time interval after setting the disk down before pure rolling motion begins is Rωfμg_.

Explanation of Solution

Write the expression for time interval after setting the disk down before pure rolling motion begins as.

  Δt=Δpf                                                                                                   (XII)

Here, Δp change in momentum, f is friction force between disc and plane surface.

Initially, the linear velocity of the disc is zero therefore the initial linear momentum of the disc becomes zero. As the initial momentum is zero, the change in linear momentum equal to the final linear momentum of the disc.

Write the expression for change in linear momentum as.

  Δp=mvf                                                                                                  (XIII)

Write the expression for friction force between disc and plane surface as.

  f=μmg                                                                                                  (XIV)

Here, μ is coefficient of friction between disc and plane surface and g gravitational acceleration.

Substitute mvf for Δp, μmg for f and in equation (XII).

  Δt=mvfμmg                                                                                                (XV)

Conclusion:

Substitute Rωf for vf in equation (XV).

  Δt=m(Rωf)μmg

Simplify the above obtained expression for Δt as.

  Δt=Rωfμg

Thus, the time interval after setting the disk down before pure rolling motion begins is Rωfμg_.

(d)

To determine

The distance travel by the disc before pure rolling begins.

(d)

Expert Solution
Check Mark

Answer to Problem 61CP

The distance travel by the disc before pure rolling begins is R2ωi218μg_.

Explanation of Solution

Write the expression for average velocity of the disc as.

  vavg=vivf2

Initial linear velocity of the disc is zero.

Substitute 0 for vi in above expression as.

  vavg=vf2

Substitute Rωf for vf in above expression.

  vavg=Rωf2                                                                                               (XVI)

Here, vavg average speed of the disc.

Write the expression for the disc under constant acceleration as.

  Δx=vavgΔt                                                                                              (XVII)

Here, Δx is distance travelled by the disc before pure rolling begins.

Conclusion:

Substitute Rωf2 for vavg and Rωfμg for Δt in equation (XVII).

  Δx=Rωf2(Rωfμg)

Substitute ωi3 for ωf in above expression as.

  Δx=R(ωi3)2(R(ωi3)μg)=Rωi6(Rωi3μg)

Simplify the above expression for Δx as.

  Δx=R2ωi218μg

Thus, the distance travel by the disc before pure rolling begins is R2ωi218μg_.

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Chapter 11 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

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