Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 10, Problem 53P

(a)

To determine

The speed of the block.

(a)

Expert Solution
Check Mark

Answer to Problem 53P

The speed of the block is 1.59 m/s.

Explanation of Solution

The change in kinetic energy of the system is equal to the net work done on the system. Only friction on the block of mass m2 and gravitation force on block of mass m1 have non-zero work. Speed of a pulley of a hollow cylinder shape has a moment of inertia.

Write the expression for the conservation of energy.

  ΔK+ΔU+ΔEint=0                                                                                      (I)

Here, ΔK is the change in kinetic energy, ΔU is the change in potential energy and ΔEint is the internal kinetic energy.

Change in kinetic energy is sum of rotational kinetic energy and translational kinetic energy.

Write the expression for change in kinetic energy.

    ΔK=ΔΚtrans+ΔΚrotational

Here, ΔKtrans is the change in translational kinetic energy and ΔKrotational is the rotational kinetic energy.

Write the expression for change in translational kinetic energy.

    ΔKtrans=KfKi

Here, Kf is the final kinetic energy and Ki is the initial kinetic energy.

Write the expression for the change in potential energy.

    ΔU=UfUi

Here, Uf is the final potential energy and Ui is the initial potential energy.

Substitute (KfKi)+ΔKrotational for ΔK and UfUi for  ΔU in equation (I).

    (KfKi)+ΔKrotational+(UfUi)+ΔEint=0                                             (II)

Write the expression for final kinetic energy.

    Kf=12m2(vf2vi2)

Here, m2 is the mass of second block and vf is the final speed and vi is initial speed for second block.

Write the expression for initial kinetic energy.

    Ki=12m1(vf2vi2)

Here, m1 is the mass of first block.

Write the expression for change in rotational kinetic energy.

    ΔKrotational=12I(ωf2ωi2)                                                                          (III)

Here, ωf is final angular speed, I is the inertia of the pulley and ωi is initial angular speed.

Write the expression for inertia of pulley.

    I=12M(R12+R22)

Here, M is the mass of pulley, R1 is the inner radius of pulley and R2 is the outer radius of pulley.

Substitute 12M(R12+R22) for I in equation (III).

  ΔKrotational=12(12M(R12+R22))(ωf2ωi2)

Write the expression for change in potential energy.

    ΔU=m1g(yfyi)

Here, g is acceleration under gravity, yf is final height and yi is initial height.

Write the expression for energy of the second block.

    ΔEint=fkd

Here, fk is frictional force and d is the distance covered by the block.

Substitute 12m1(vf2vi2) for Kf, 12m1(vf2vi2) for Ki, 12(12M(R12+R22))(ωf2ωi2) for ΔKrotational, m1g(yfyi) for UfUi and fkd for ΔEint in equation (II).

    [12m2(vf2vi2)+12m1(vf2vi2)+12(12M(R12+R22))(ωf2ωi2)+m1g(yfyi)+fkd]=0                         (IV)

Write the expression for frictionof second block.

    fk=μkm2g

Here, μk is coefficient of friction.

Angular speed of the pulley is related to the speed of the objects.

Write the expression for the final angular speed.

  ωf=vfR2

Write the expression for the initial angular speed.

  ωi=viR2

Substitute μm2g for fk, vfR2 for ωf, viR2 and d for (yfyi) in equation (IV).

    [12m2(vf2vi2)+12m1(vf2vi2)+12(12M(R12+R22)((vfR2)2(viR2)2))+m1g(d)+μm2gkd]=0

Rearrange the above expression for vf.

    vf=vi2+4gd(m1μkm2)2(m1+m2)+M(1+R12R22)                                                        (V)

Conclusion:

Substitute 0.820 m/s for vi, 9.80 m/s2 for g, 0.700 m for d, 0.420 kg for m1, 0.250  for μk, 0.850 kg for m2, 0.350 kg for M, 0.0200 m for R1 and 0.0300 m for R2 in equation (III).

    vf=(0.820 m/s)2+4(9.80 m/s2)(0.700 m)(0.420 kg0.250 (0.850 kg))2(0.420 kg+0.850 kg)+0.350 kg(1+(0.0200 m)2(0.0300 m)2)=0.6724 (m/s)2+5.6938 kg(m/s)23.0455 kg=0.6724 (m/s)2+1.8695 (m/s)2 =1.59 m/s

Thus, the speed of the block is 1.59 m/s.

(b)

To determine

The angular speed of the pulley.

(b)

Expert Solution
Check Mark

Answer to Problem 53P

The angular speed of the pulley is 53.1 rad/s.

Explanation of Solution

Write the expression for the angular speed of the pulley.

    ω=vfR2                                                                                                        (VI)

Conclusion:

Substitute 1.59 m/s for v and 0.0300 m for R2 in equation (VI).

    ω=1.59 m/s0.0300 m=53.1 rad/s

Thus, the angular speed of the pulley is 53.1 rad/s.

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Chapter 10 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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