/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A turntable of radius \(25 \math... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 42

A turntable of radius \(25 \mathrm{cm}\) and rotational inertia \(0.0154 \mathrm{kg} \cdot \mathrm{m}^{2}\) is spinning freely at 22.0 rpm about its central axis, with a \(19.5-\mathrm{g}\) mouse on its outer edge. The mouse walks from the edge to the center. Find (a) the new rotation speed and (b) the work done by the mouse.

Short Answer

Expert verified
The new rotation speed is \(2.754\) rad/s and the work done by the mouse is \(0.007619\) J.

Step by step solution

01

Convert the given units

The first step is to get all the units in standard SI form, to eliminate any unit mismatch. Convert the mass of the mouse to \(kg\) by dividing by \(1000\) to obtain \(0.0195\) \(kg\). Convert the radius of the turntable to \(m\) by dividing by \(100\), giving \(0.25\) \(m\). Convert the turntable's angular speed to \(rad/s\) by multiplying \(22\) rpm by \(\frac{2\pi}{60}\) to obtain \(2.309\) \(rad/s\).
02

Calculate initial moment of inertia and angular momentum

Next, use the formula for moment of inertia \(I\) of a system with the mouse at the edge: \(I=I_{turntable}+m_{mouse}r^2\)= \(0.0154\) \(kg\cdot m^2\) +\(0.0195\) \(kg\) X\((0.25) m)^2 = 0.01835 kg\cdot m^2\). Now calculate initial angular momentum using the formula \(L_{initial}= I_{initial}\times\omega_{initial}\)= \(0.01835 kg\cdot m^2\) X \(2.309 rad/s\) = \(0.042390 kg\cdot m^2/s\).
03

Calculate final moment of inertia and angular velocity

When the mouse reaches the center, the new moment of inertia \(I_{final}\) is just the rotational inertia of the turntable, \(0.0154 kg \cdot m^2\). Conservation of angular momentum implies that \(L_{initial}= L_{final}\) or \(I_{initial}\omega_{initial} = I_{final}\omega_{final}\), where \(\omega_{final}\) is the final angular velocity. Solving for \(\omega_{final}\) results in \(\omega_{final}= \frac{L_{initial}}{I_{final}}= \frac{0.042390 kg\cdot m^2/s}{0.0154 kg\cdot m^2}= 2.754 rad/s\).
04

Calculate work done by mouse

The work done by the mouse is given by the difference between the initial and final kinetic energy of the system. The kinetic energy of a rotating object is given by \(0.5\cdot I \cdot \omega^2\). Thus, the work done \(W\)= \(0.5\cdot I_{final}\cdot \omega_{final}^2 - 0.5\cdot I_{initial}\cdot\omega_{initial}^2\)= \(0.5\cdot 0.0154 kg\cdot m^2\cdot (2.754 rad/s)^2 - 0.5\cdot 0.01835 kg\cdot m^2\cdot (2.309 rad/s)^2 = 0.007619 J\).

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Most popular questions from this chapter

Jumbo is back! Jumbo is the 4.8 -Mg elephant from Example 9.4 This time he's standing at the outer edge of a 15 -Mg turntable of radius \(8.5 \mathrm{m},\) rotating with angular velocity \(0.15 \mathrm{s}^{-1}\) on frictionless bearings. Jumbo then walks to the center of the turntable. Treating Jumbo as a point mass and the turntable as a solid disk, find (a) the angular velocity of the turntable once Jumbo reaches the center and (b) the work Jumbo does in walking to the center.

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