COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 111QAP
To determine

The angular velocity of the skater when he pulls his parallel to his trunk.

Expert Solution & Answer
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Answer to Problem 111QAP

The angular velocity of the skater is, 642.786rpm.

Explanation of Solution

Given information:

The mass distribution of a human body can be considered as, 80% for the trunk and legs 13%, for the arms and hands, and 7% for the head.

The spinning skater with hands outstretched can be modeled as a vertical cylinder with two uniform rods attached to it.

The mass of the skater is, 62kg and he is, 1.8m tall.

His trunk can be modeled as a cylinder with 35cm diameter.

The length of both arm and a hand is, 65cm.

The spinning speed of the skater with his arms outstretched is, 70rpm.

Formula used:

  I=m.r2

  I1ω1=I2ω2

Calculation:

Let us first consider the hand outstretched position.

  COLLEGE PHYSICS, Chapter 8, Problem 111QAP , additional homework tip  1

The moment of inertia of the arms can be calculated as follows.

The mass of an arm and a hand is,

  62kg×13100×12

  4.03kg

Since the arms are modeled as uniform rods, let us assume that they have a uniform linear density λ.

  λ=4.030.65

  λ=6.2

Let us consider the moment of inertia of small particle of the rod, which is x length away from its center of rotation.

Using the given formula, its moment of inertia can be written as,

  I=x2dm

  I=x2d(λx)

  I=λx2dx

Applying the limits,

  I=λ0.351x2dx

  I=6.23(130.353)

  I=1.978

The moment of inertia for the both arms is, 3.956kg.m2.

Let us now consider the head and the trunk.

The mass of the trunk and head is,

  62kg×87100

  53.94kg

Assuming the cylinder to be uniform, let us denote its density by ρ.

  ρ=53.94π×0.352×1.8

  ρ=77.8668

It can be modeled as a solid cylinder.

  COLLEGE PHYSICS, Chapter 8, Problem 111QAP , additional homework tip  2

Using the given formula, its moment of inertia can be written as,

  I=r2dm

  I=r2d(ρV)

  I=ρr2dV

But, dV=2πrLdr.

  I=2πρLr3dr

Applying the limits;

  I=2πρL00.175r3dr

  I=(2π×77.8668×1.8)00.175r3dr

  I=0.2065kg.m2

Considering the whole system, the moment of inertia when the hands are outstretched is,

  I1=3.956kg.m2+0.2065kg.m2

  I1=4.1625kg.m2

Let us now consider the case when skater is pulling his arms towards him.

The moment of inertia of the head and the trunk remains the same.

But moment of inertia of the hands and arms change.

Since there is no mass distribution perpendicular to the axis of rotation, the moment of inertia can be directly calculated.

  I=mr2+mr2I=2mr2I=2×4.03×0.1752I=0.2468kg.m2

Therefore, the moment of inertia for the whole system will be,

  I2=0.2468kg.m2+0.2065kg.m2

  I2=0.4533kg.m2

Now let us consider the conservation of angular momentum;

  I1ω1=I2ω2

  4.1625×70=0.4533×ω2ω2=642.786rpm

Conclusion:

The angular speed of the skater when he pulls his arms towards him is, 642.786rpm.

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Chapter 8 Solutions

COLLEGE PHYSICS

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