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Chapter 10: Problem 20

A man stands on a frictionless platform that is rotating with an angular speed of \(1.22 \mathrm{rev} / \mathrm{s} ;\) his arms are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man, the weights, and the platform is \(6.13 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the weights the man decreases the rotational inertia to \(1.97 \mathrm{~kg} \cdot \mathrm{m}^{2}\), what is the resulting angular speed of the platform?

Short Answer

Expert verified
After calculating the above expression, the resulting angular speed of the platform, when the man moves the weights and decreases the rotational inertia, is approximately 19.22 rad/s.

Step by step solution

01

Establish the angular momentum conservation equation

According to the conservation of angular momentum, the initial angular momentum of the system equals its final momentum, represented as \(L_{i}=L_{f}\). Angular momentum is calculated as the product of rotational inertia and angular speed, then this equation can be written as \[I_{i} \cdot \omega_{i} = I_{f} \cdot \omega_{f}\], whereby \(\omega_{i}\) represents the initial angular speed, \(I_{i}\) represents the initial rotational inertia, \(\omega_{f}\) represents the final angular speed and \(I_{f}\) represents the final rotational inertia.
02

Substitute the given values

Substitute the provided values: \(I_{i}=6.13~kg.m^2\), \(\omega_{i}=1.22~rev/s\), \(I_{f}=1.97~kg.m^2\). The rotational speed was given in revolutions per second, but this needs to be converted to radians per second because the standard units in physics for angular speed are radians per second. To do so, recall that \(1~rev = 2\pi~rad\). Thus, \(\omega_{i}=1.22~rev/s = 1.22\cdot2\pi~rad/s\). Substituting the given values into the angular momentum conservation equation gives \[6.13~kg.m^2\cdot1.22\cdot2\pi~rad/s = 1.97~kg.m^2\cdot\omega_{f}\].
03

Solve the equation for \(\omega_{f}\)

Solving for \(\omega_{f}\) gives \(\omega_{f} = \frac{6.13~kg.m^2\cdot1.22\cdot2\pi~rad/s}{1.97~kg.m^2}\) .

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Most popular questions from this chapter

A girl of mass \(50.6 \mathrm{~kg}\) stands on the edge of a frictionless merry- go-round of mass \(827 \mathrm{~kg}\) and radius \(3.72 \mathrm{~m}\) that is not moving. She throws a \(1.13-\mathrm{kg}\) rock in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(7.82 \mathrm{~m} / \mathrm{s}\). Calculate \((a)\) the angular speed of the merry-go-round and \((b)\) the linear speed of the girl after the rock is thrown. Assume that the merry-go-round is a uniform disk.

In a playground there is a small merry-go-round of radius \(1.22 \mathrm{~m}\) and mass \(176 \mathrm{~kg} .\) The radius of gyration (see Exercise \(9-20\) ) is \(91.6 \mathrm{~cm}\). A child of mass \(44.3 \mathrm{~kg}\) runs at a speed of \(2.92 \mathrm{~m} / \mathrm{s}\) tangent to the rim of the merry-go-round when it is at rest and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round and find the angular speed of the merry-go-round and child.

Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

The time integral of a torque is called the angular impulse. ( \(a\) ) Starting from \(\overrightarrow{\boldsymbol{\tau}}=d \overrightarrow{\mathbf{L}} / d t\), show that the resultant angular impulse equals the change in angular momentum. This is the rotational analogue of the linear impulse-momentum relation. (b) For rotation around a fixed axis, show that $$ \int \tau d t=F_{\mathrm{av}} r(\Delta t)=I\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right) $$ where \(r\) is the moment arm of the force, \(F_{\mathrm{av}}\) is the average value of the force during the time it acts on the object, and \(\omega_{\mathrm{i}}\) and \(\omega_{\mathrm{f}}\) are the angular velocities of the object just before and just after the force acts.

Astronomical observations show that from 1870 to 1900 the length of the day increased by about \(6.0 \times 10^{-3}\) s. (a) What corresponding fractional change in the Earth's angular velocity resulted? \((b)\) Suppose that the cause of this change was a shift of molten material in the Earth's core. What resulting fractional change in the Earth's rotational inertia could account for the answer to part \((a) ?\)
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