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Figure 11-51 shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?

Short Answer

Expert verified
The cat increases the kinetic energy of the system by 38.91 J.

Step by step solution

01

Define Given Values

Let's start by organizing the given information. The outer radius is \( R_2 = 0.800 \, \text{m} \), inner radius is \( R_1 = \frac{R_2}{2.00} = 0.400 \, \text{m} \), mass of the ring \( M = 8.00 \, \text{kg} \), initial angular speed \( \omega_0 = 8.00 \, \text{rad/s} \), and mass of the cat \( m = \frac{M}{4.00} = 2.00 \, \text{kg} \).
02

Calculate Moment of Inertia

The moment of inertia of a ring about its center is given by \( I = \frac{1}{2} M (R_1^2 + R_2^2) \). Substituting the given values, we find \( I = \frac{1}{2} \times 8.00 \times ((0.400)^2 + (0.800)^2) \equiv \frac{1}{2} \times 8.00 \times (0.16 + 0.64) = 3.20 \, \text{kg} \, \text{m}^2 \).
03

Calculate Initial Kinetic Energy

The initial kinetic energy \( K_i \) of the system is given by \( K_i = \frac{1}{2} I \omega_0^2 + \frac{1}{2} m v_0^2 \), where \( v_0 = \omega_0 R_2 \). Then, \( v_0 = 8.00 \times 0.800 = 6.40 \, \text{m/s} \). Therefore, \( K_i = \frac{1}{2} \times 3.20 \times (8.00)^2 + \frac{1}{2} \times 2.00 \times (6.40)^2 = 102.4 + 40.96 = 143.36 \, \text{J} \).
04

Calculate Final Angular Speed

Using conservation of angular momentum, the initial and final angular momenta must be equal. Thus, \( I_\text{totalini} \omega_0 = I_\text{totalfin} \omega_f \). Here, \( I_\text{totalini} = I + m R_2^2 \) and \( I_\text{totalfin} = I + m R_1^2 \). Thus, \( (3.20 + 2.00 \times 0.800^2) \times 8.00 = (3.20 + 2.00 \times 0.400^2) \times \omega_f \), leading to \( 4.48 \times 8.00 = 3.52 \times \omega_f \). Therefore, \( \omega_f = \frac{35.84}{3.52} = 10.18 \, \text{rad/s} \).
05

Calculate Final Kinetic Energy

The final kinetic energy \( K_f \) is given by \( K_f = \frac{1}{2} I_\text{totalfin} \omega_f^2 = \frac{1}{2} (3.52) (10.18)^2 \), which results in \( K_f = 182.27 \, \text{J} \).
06

Calculate Kinetic Energy Increase

The increase in kinetic energy is the difference between the final and initial kinetic energies: \( \Delta K = K_f - K_i = 182.27 - 143.36 = 38.91 \, \text{J} \).

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Key Concepts

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

Angular Momentum Conservation
Angular momentum is an extremely important quantity in rotational dynamics. It is analogous to linear momentum but applies to rotational motion. In a closed system, angular momentum is conserved, meaning there is no net external torque acting on the system.

In our problem, the system consists of a rotating ring and a cat moving within the ring. The initial angular momentum of this closed system depends on the total initial moment of inertia and the initial angular velocity. As the cat moves radially inward or outward, the system readjusts by changing the angular velocity to keep the angular momentum constant.
- **Initial State**: Initial angular momentum is calculated using the sum of the inertias and the initial angular speed.- **Later State**: As the cat moves closer to the center, the new angular speed is computed such that the product of the new inertia and speed matches the initial momentum value.

This principle allows us to find the final angular speed after the cat has moved, by equating initial and final angular momentum: \[ I_{\text{initial}} \omega_{0} = I_{\text{final}} \omega_{f} \]
Moment of Inertia
The moment of inertia is a quantity expressing an object's tendency to resist angular acceleration. It is akin to mass in linear dynamics; however, unlike mass, it depends on how mass is distributed relative to the axis of rotation.

For a ring, the moment of inertia about its center is calculated using both its inner and outer radii. In our scenario:\[ I = \frac{1}{2} M (R_1^2 + R_2^2) \]This accounts for the geometry of the ring.

As the cat moves inward, its contribution to the moment of inertia changes. The new effective moment of inertia is then calculated by considering the cat's mass moving to the inner radius. Understanding how this value changes helps in determining how other properties of the system, like angular speed, must adjust to maintain angular momentum conservation.
Kinetic Energy
Kinetic energy in rotational dynamics is the energy an object possesses due to its rotation. It is calculated using the moment of inertia and angular velocity in this relation:
\[ K = \frac{1}{2} I \omega^2 \]

In the initial state, both the ring and the cat contribute to the kinetic energy. The initial kinetic energy includes both the rotational energy of the ring and the translational energy of the cat moving on the outside of the ring. As the cat moves inside, kinetic energy is recalculated using the new angular speed found from angular momentum conservation.
- **Initial Energy**: Computed using initial inertia and angular speed.- **Final Energy**: Final kinetic energy considers the new inertia and increased angular speed.

The change in kinetic energy illustrates how the system's total energy morphs as the internal mass distribution changes without external influences.
Angular Speed
Angular speed describes how quickly an object rotates around an axis and is often denoted by \( \omega \). It is a central aspect of rotational motion, much like speed is for linear motion.

Initially, the system rotates with a given angular speed. As the cat moves radially inward, the system spontaneously accelerates to preserve the angular momentum—a direct result of the conservation law.
- **Initial Speed**: Given as the starting point for the system.- **Final Speed**: Calculated after using angular momentum conservation, showcasing an increase due to a decreased moment of inertia.

This scenario reveals a fundamental principle: for systems with constant total angular momentum, reductions in the moment of inertia must result in increased angular speeds, maintaining balance. This principle is applicable in various physical contexts from mechanical systems to natural phenomena.

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Most popular questions from this chapter

A \(140 \mathrm{~kg}\) hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of \(0.150 \mathrm{~m} / \mathrm{s}\). How much work must be done on the hoop to stop it?

A particle is to move in an \(x y\) plane, clockwise around the origin as seen from the positive side of the \(z\) axis. In unit-vector notation, what torque acts on the particle if the magnitude of its angular momentum about the origin is (a) \(4.0 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), (b) \(4.0 t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), (c) \(4.0 \sqrt{t} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), and (d) \(4.0 / t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) ?

A man stands on a platform that is rotating (without friction) with an angular speed of \(1.2 \mathrm{rev} / \mathrm{s} ;\) his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the bricks the man decreases the rotational inertia of the system to \(2.0\) \(\mathrm{kg} \cdot \mathrm{m}^{2}\), what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?

The rotor of an electric motor has rotational inertia \(I_{m}=\) \(2.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia \(I_{p}=12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about this axis. Calculate the number of revolutions of the rotor required to turn the probe through \(30^{\circ}\) about its central axis.

A particle is acted on by two torques about the origin: \(\vec{\tau}_{1}\) has a magnitude of \(2.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the positive direction of the \(x\) axis, and \(\vec{\tau}_{2}\) has a magnitude of \(4.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the negative direction of the \(y\) axis. In unit-vector notation, find \(d \vec{\ell} / d t\), where \(\vec{\ell}\) is the angular momentum of the particle about the origin.

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