Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 10, Problem 40P

In Figure P10.40, the hanging object has a mass of m1 = 0.420 kg; the sliding block has a mass of m2 = 0.850 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is μk = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table. (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. (b) Find the angular speed of the pulley at the same moment.

Figure P10.40

Chapter 10, Problem 40P, In Figure P10.40, the hanging object has a mass of m1 = 0.420 kg; the sliding block has a mass of m2

(a)

Expert Solution
Check Mark
To determine

The speed of the lock after moved to a second point.

Answer to Problem 40P

The speed of the lock after moved to a second point is 1.59m/s_.

Explanation of Solution

According to law of conservation of energy the total energy of the system is remains constant.

Write the expression for the conservation of energy of the system.

  ΔK+ΔU+ΔEint=0(KfKi)+(UfUi)+ΔEint=0        (I)

Here, Kf is the final kinetic energy, Ki is the final kinetic energy, Uf is the final potential energy, Ui is the initial potential energy, ΔEint is the energy lost due to friction.

The initial kinetic energy involves the kinetic energy of the hanging block, the sliding block, rotational kinetic energy, and the final kinetic energy involves the final kinetic energy of hanging block, the sliding block, rotational kinetic energy.

Write the expression for the initial kinetic energy.

  Ki=12m1υi2+12m2υi2        (II)

Here, m1 is the mass of the hanging block, m2 is the mass of the sliding block, υi is the initial velocity.

Write the expression for the final kinetic energy.

  Kf=12m1υ2+12m2υ2        (III)

Here, υ is the final velocity.

Write the expression for the change in rotational kinetic energy.

  ΔKrot=12Iω212Iωi2        (IV)

Here, I is the moment of inertia, ωi is the initial angular velocity, ω is the final angular velocity.

Write the expression energy lost due to friction.

  ΔEint=fkd        (V)

Here, fk is the force due to kinetic friction, d is the distance moved by the mass m2.

Substitute, μm2g for fk in equation (V).

  ΔEint=μm2gd        (VI)

Here, μ is the coefficient of kinetic friction, g is the acceleration due to gravity.

Write expression for change in potential energy.

  UfUi=m1gym1gyi        (VII)

Substitute, equation (VII), (VI), (IV), (III), (II) in equation (I).

  ((12m1υ2+12m2υ2)(12m1υi2+12m2υi2))+(12Iω212Iωi2)+(m1gym1gyi)+μm2gd=012(m1+m2)(υ2υi2)+12I(ω2ωi2)+m1g(yy1)+μm2gd=0        (VIII)

Substitute, 12M(R12+R12) for I, υR for ω, υiR for ωi in equation (VIII).

12(m1+m2)(υ2υi2)+12[12M(R12+R22)]((υR)2(υiR)2)+m1g(d)+μm2gd=012(m1+m2)(υ2υi2)+12[12M(R12R22+1)](υ2υi2)=gd(m1μm2)12[(m1+m2)+12M(R12R22+1)](υ2υi2)=gd(m1μm2)υ={υi2+4gd(m1μm2)2(m1+m2)+12M(R12R22+1)}1/2        (IX)

Here, M is the mass of the hollow cylinder, R1 is the inner radius of the cylinder, R2 is the outer radius of the cylinder.

Conclusion:

Substitute, 0.820m/s for υi, 9.80m/s2 for g, 0.420kg for m1, 0.250 for μ, 0.850kg for m2, 0.350kg, 0.020m for R1, and 0.030m for R2, 0.700m for d in equation (IX).

υ={(0.820m/s)2+4(9.80m/s2)(0.700m)(0.420kg(0.250)(0.850kg))2(0.420kg+0.850kg)+12(0.350kg)((0.020m)2(0.030m)+1)}1/2=1.59m/s

Therefore, speed of the lock after moved to a second point is 1.59m/s_.

(b)

Expert Solution
Check Mark
To determine

The angular speed of the pulley.

Answer to Problem 40P

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

Explanation of Solution

Write the expression for angular speed of the pulley.

  ω=υr        (X)

Conclusion:

Substitute, 1.59m/s for υ, 0.030m for r in equation (X) to find ω.

  ω=1.59m/s0.030m=53.1rad/s

Therefore, the angular speed of the pulley is 53.1rad/s_.

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Chapter 10 Solutions

Principles of Physics: A Calculus-Based Text

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