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Chapter 11: Problem 52

A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 \mathrm{~m}\) that can rotate freely about its center like a merry-goround. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach- disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the kinetic energy?

Short Answer

Expert verified
(a) \(\omega_f \approx 0.347 \text{ rad/s}\); (b) \(\frac{K}{K_0} > 1\); (c) Internal work by cockroach.

Step by step solution

01

Calculate Initial Moment of Inertia

We start by calculating the initial moment of inertia of the system. The moment of inertia of the disk about its center is given by \(I_d = \frac{1}{2} M R^2\) where \(M = 4m\). Since the cockroach lies on the rim of the disk, its moment of inertia is \(I_c = m R^2\). Therefore, the total initial moment of inertia \(I_{i}\) is \[I_{i} = \frac{1}{2} (4m) R^2 + m R^2 = 2mR^2 + mR^2 = 3mR^2\].
02

Calculate Final Moment of Inertia

When the cockroach moves halfway to the center of the disk, its new distance from the center is \(\frac{R}{2}\). Thus, its new moment of inertia is \(I_c' = m (\frac{R}{2})^2 = \frac{1}{4} m R^2\). The moment of inertia of the disk remains as \(I_d = 2mR^2\). Therefore, the final total moment of inertia \(I_{f}\) is \[I_{f} = 2mR^2 + \frac{1}{4}mR^2 = 2.25mR^2\].
03

Use Conservation of Angular Momentum

Angular momentum \(L\) is conserved. Initially, \(L_i = I_i \omega_i = 3mR^2 \times 0.260 \). Let \(\omega_f\) be the final angular velocity. Using \(L_i = L_f\), we have \[3mR^2 \times 0.260 = 2.25mR^2 \times \omega_f\] which simplifies to \[\omega_f = \frac{3 \times 0.260}{2.25} = \frac{0.780}{2.25}\] Compute this to find \(\omega_f \approx 0.347 \text{ rad/s}\).
04

Calculate Initial and Final Kinetic Energies

The initial kinetic energy \(K_0\) is \[K_0 = \frac{1}{2} I_i \omega_i^2 = \frac{1}{2} \times 3mR^2 \times 0.260^2\] Compute this to find \(K_0\). The final kinetic energy \(K\) is \[K = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \times 2.25mR^2 \times 0.347^2\] Calculate this to find \(K\).
05

Calculate the Ratio of Kinetic Energies

Find the ratio \(\frac{K}{K_0}\): \[\frac{K}{K_0} = \frac{\frac{1}{2} \times 2.25mR^2 \times (0.347)^2}{\frac{1}{2} \times 3mR^2 \times (0.260)^2}\]Cancel terms and simplify to compute the exact ratio \(\frac{K}{K_0}\).
06

Explain the Change in Kinetic Energy

The change in kinetic energy is accounted for by the internal work done by the cockroach as it moves closer to the center. The reduction in the radius leads to a higher angular speed, resulting in a change in kinetic energy despite constant angular momentum. This transformation occurs due to the non-conservative force applied during the cockroach's movement.

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Key Concepts

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

Moment of Inertia
In rotational dynamics, the moment of inertia plays a similar role as mass in linear motion. It measures how difficult it is to change the rotational motion of an object. The distribution of mass from the axis of rotation greatly affects this quantity.
For a uniform disk, the moment of inertia is given by the formula: \[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass and \( R \) is the radius of the disk. When there is an additional mass, such as a cockroach on the rim, this mass adds its own contribution to the moment of inertia because it also resists changes in motion.
  • The cockroach's contribution when on the rim is \( m R^2 \).
  • As it moves inward, its distance to the center decreases, leading to a lower moment of inertia \( m (\frac{R}{2})^2 \).
Thus, the moment of inertia decreases as the cockroach moves inward, influencing the system's angular velocity.
Kinetic Energy
Kinetic energy in rotational motion depends not only on the speed of rotation but also on the moment of inertia. It can be calculated using the formula: \[ K = \frac{1}{2} I \omega^2 \] Here, \( I \) is the moment of inertia and \( \omega \) the angular velocity. When the cockroach starts at the rim and then moves inward, the system's kinetic energy changes.
  • Initially, with both on the rim, there is some kinetic energy \( K_0 \).
  • As the cockroach moves inward, the reduction in moment of inertia raises the angular velocity, potentially altering the kinetic energy.
It鈥檚 fascinating that even with conserved angular momentum, kinetic energy can change because the mass re-positioning works as internal work within the system.
Rotational Dynamics
In the context of rotational dynamics, an important principle is the conservation of angular momentum. When no external torques act on a system, its angular momentum remains constant. Angular momentum \( L \) is calculated as: \[ L = I \omega \]
  • Initially, the system has a certain angular momentum based on the initial moment of inertia and angular velocity.
  • As the cockroach moves, conservation of angular momentum ensures that any decrease in moment of inertia is compensated by an increase in angular velocity.
This internal change is key to understanding the rotation: the system increases in speed without any external inputs simply due to the cockroach moving closer to the center. The fascinating outcome of such dynamics is the change in kinetic energy even while the angular momentum remains unchanged.

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Most popular questions from this chapter

Force \(\vec{F}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a pebble with position vector \(\vec{r}=(0.50 \mathrm{~m}) \hat{\mathrm{j}}-(2.0 \mathrm{~m}) \hat{\mathrm{k}}\) relative to the origin. In unit-vector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m})\) ?

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In 1980 , over San Francisco Bay, a large yo-yo was released from a crane. The \(116 \mathrm{~kg}\) yo-yo consisted of two uniform disks of radius \(32 \mathrm{~cm}\) connected by an axle of radius \(3.2 \mathrm{~cm} .\) What was the magnitude of the acceleration of the yo-yo during (a) its fall and (b) its rise? (c) What was the tension in the cord on which it rolled? (d) Was that tension near the cord's limit of \(52 \mathrm{kN}\) ? Suppose you build a scaled-up version of the yo-yo (same shape and materials but larger). (e) Will the magnitude of your yo-yo's acceleration as it falls be greater than, less than, or the same as that of the San Francisco yo-yo? (f) How about the tension in the cord?

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

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