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Chapter 11: Problem 46

The rotational inertia of a collapsing spinning star drops to \(\frac{1}{3}\) its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?

Short Answer

Expert verified
The new kinetic energy is three times the initial energy.

Step by step solution

01

Understanding Rotational Inertia and Kinetic Energy

Rotational inertia (moment of inertia) decreases to \(\frac{1}{3}\) of its initial value when the star collapses. The rotational kinetic energy is given by \(K = \frac{1}{2} I \omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. Since the system is isolated, angular momentum \(L = I\omega\) remains constant.
02

Conserving Angular Momentum

Given the angular momentum \(L\) is constant, initially \(L_0 = I_0 \omega_0\) and after collapse \(L = I_f \omega_f\). So, \(I_0 \omega_0 = I_f \omega_f\), meaning \(\omega_f = \frac{I_0}{I_f} \omega_0\). Since \(I_f = \frac{1}{3} I_0\), \(\omega_f = 3 \omega_0\).
03

Finding the New Rotational Kinetic Energy

The initial rotational kinetic energy is \(K_i = \frac{1}{2} I_0 \omega_0^2\). The new kinetic energy is \(K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \left(\frac{1}{3} I_0\right) (3 \omega_0)^2\). Simplifying this gives \(K_f = \frac{1}{2} \times \frac{1}{3} I_0 \times 9 \omega_0^2 = \frac{3}{2} I_0 \omega_0^2\).
04

Calculating the Ratio of Kinetic Energies

The ratio of the new to the initial kinetic energy is \(\frac{K_f}{K_i} = \frac{\frac{3}{2} I_0 \omega_0^2}{\frac{1}{2} I_0 \omega_0^2} = 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Inertia
Rotational inertia, also known as the moment of inertia, measures how difficult it is to change the rotational motion of an object. Imagine trying to spin a heavy solid disc compared to a lightweight hoop. The solid disc resists changes to its rotation more because of its higher rotational inertia. In the context of a collapsing star, as the star shrinks, its mass distribution changes. The rotational inertia decreases, which means it's easier for the star to spin faster. In our example, the rotational inertia drops to \(\frac{1}{3}\) of its initial value. This significant change influences the star's angular velocity, highlighting the importance of rotational inertia in rotational dynamics.
Angular Momentum Conservation
Angular momentum is a key player in rotational dynamics. It combines rotational inertia and angular velocity, described by the formula \(L = I \omega\). One amazing property of angular momentum is its conservation in isolated systems. This means, unless acted on by an external force, the total angular momentum remains constant over time. For the collapsing star, its initial angular momentum \(I_0 \omega_0\) is the same as the final angular momentum \(I_f \omega_f\). When the star's rotational inertia decreases to \(\frac{1}{3} I_0\), its angular velocity must increase to conserve angular momentum. This results in \(\omega_f = 3 \omega_0\). Thus, a smaller inertial system spins much faster to balance out angular momentum.
Rotational Kinetic Energy
Rotational kinetic energy is the energy due to the rotation of an object. It is calculated using the equation \(K = \frac{1}{2} I \omega^2\). This formula shows that kinetic energy depends not only on how fast something spins (angle velocity) but also on how the mass is spread out from the axis of rotation (rotational inertia). For the star, we had to find the ratio of the new rotational kinetic energy to its initial energy. Initially, the energy \(K_i\) was \(\frac{1}{2} I_0 \omega_0^2\). After collapsing, the energy \(K_f\) becomes \(\frac{1}{2} \times \frac{1}{3} I_0 \times (3 \omega_0)^2 = \frac{3}{2} I_0 \omega_0^2\). The ratio \(\frac{K_f}{K_i} = 3\) indicates that the rotational kinetic energy has tripled. As the star spins faster, its energy of rotation greatly increases, demonstrating the effects of its enhanced angular velocity and reduced inertia.

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Most popular questions from this chapter

A uniform block of granite in the shape of a book has face dimensions of \(20 \mathrm{~cm}\) and \(15 \mathrm{~cm}\) and a thickness of \(1.2 \mathrm{~cm} .\) The density (mass per unit volume) of granite is \(2.64 \mathrm{~g} / \mathrm{cm}^{3} .\) The block rotates around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that axis is \(0.104 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its rotational kinetic energy about that axis?

Two \(2.00 \mathrm{~kg}\) balls are attached to the ends of a thin rod of length \(50.0 \mathrm{~cm}\) and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (Fig.\(11-57),\) a \(50.0 \mathrm{~g}\) wad of wet putty drops onto one of the balls, hitting it with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle will the system rotate before it momentarily stops?

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. \(11-48\) ). A toy train of mass \(m\) is placed on the track and, with the system initially at rest,the train's electrical power is turned on. The train reaches speed \(0.15 \mathrm{~m} / \mathrm{s}\) with respect to the track. What is the wheel's angular speed if its mass is \(1.1 \mathrm{~m}\) and its radius is \(0.43 \mathrm{~m} ?\) (Treat it as a hoop, and neglect the mass of the spokes and hub.)

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. \(11-62\) ). The radius of the axle is \(0.200 \mathrm{~m}\), and the rotational inertia of the wheel-axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m},\) what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

Force \(\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \mathrm{j}\) acts on a particle with position vector \(\vec{r}=(3.0 \mathrm{~m}) \mathrm{i}+(4.0 \mathrm{~m}) \hat{j}\) What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of \(\vec{r}\) and \(\vec{F} ?\)
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