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A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(10.0 m\) that can rotate freely about its center like a merry-go-round. Initially the cockroach and disk rotate together with a rotational velocity of \(\omega_{1}\). Then the cockroach walks halfway to the center of the disk. (a) What is the change \(\Delta \omega\) in the rotational velocity of the cockroach-disk system? (b) What is the ratio \(K_{2} / K_{1}\) 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
The change in rotational velocity \(\Delta \omega\) is \(0.1429 \omega_1\). The ratio \(\frac{K_2}{K_1}\) is \1.143\). The kinetic energy changes due to the reduction in the moment of inertia.

Step by step solution

01

Identify system properties

The system includes a disk of mass 10m and a cockroach of mass m. The disk rotates about its center with an initial angular velocity \(\omega_1\).
02

Establish conservation of angular momentum

Since no external torques act on the system, the angular momentum is conserved. The initial angular momentum \(L_1\) is given by \({I_1} \omega_1\) where \(I_1\) is the initial moment of inertia.
03

Compute initial moment of inertia

The initial moment of inertia \(I_1\) is the sum of the moment of inertia of the disk and the cockroach. \(I_1 = I_{disk} + I_{cockroach} = \frac{1}{2}(10m)R^2 + mR^2 = 5mR^2 + mR^2 = 6mR^2\).
04

Compute final moment of inertia

When the cockroach walks halfway to the center, the final moment of inertia \(I_2\) is: \(I_2 = I_{disk} + I_{cockroach, halfway} = \frac{1}{2}(10m)R^2 + m(\frac{R}{2})^2 =5mR^2 + \frac{1}{4}mR^2 = 5.25mR^2\).
05

Apply conservation of angular momentum

Using conservation of angular momentum, \({I_1} \omega_1 = {I_2} \omega_2\), solve for the new angular velocity \(\omega_2\). \(\omega_2 = \omega_1 \frac{I_1}{I_2} = \omega_1 \frac{6mR^2}{5.25mR^2} = \omega_1 \frac{6}{5.25} = 1.1429 \omega_1\).
06

Determine the change in angular velocity \(\Delta \omega\)

The change in angular velocity is \(\Delta \omega = \omega_2 - \omega_1 = 1.1429 \omega_1 - \omega_1 = 0.1429 \omega_1\).
07

Compute initial kinetic energy

Kinetic energy is given by \(K = \frac{1}{2}I \omega^2\). The initial kinetic energy is \(K_1 = \frac{1}{2}I_1 \omega_1^2 = \frac{1}{2}(6mR^2) \omega_1^2 = 3mR^2 \omega_1^2\).
08

Compute new kinetic energy

The new kinetic energy is \(K_2 = \frac{1}{2}I_2 \omega_2^2 = \frac{1}{2}(5.25mR^2)(1.1429 \omega_1)^2 = \frac{1}{2}(5.25mR^2)1.306 \omega_1^2 = 3.43mR^2 \omega_1^2\).
09

Determine the ratio of kinetic energies \(\frac{K_2}{K_1}\)

The ratio of the new kinetic energy to the initial kinetic energy is \(\frac{K_2}{K_1} = \frac{3.43mR^2 \omega_1^2}{3mR^2 \omega_1^2} = 1.143\).
10

Explain the change in kinetic energy

The change in kinetic energy is due to the cockroach moving inward, which reduces the moment of inertia and increases the angular velocity, leading to an increase in kinetic energy.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a critical parameter in rotational dynamics. It measures how mass is distributed relative to the axis of rotation. For a rigid body, it is expressed as the sum of the products of the mass of each particle and the square of its distance from the rotation axis. Mathematically, it can be represented as: \[ I = \frac{1}{2} m R^2 \] for a disk rotating about its center and for point masses at a distance R, \[ I = m R^2 \]. In our exercise, initially, the disk and the cockroach together, have a combined moment of inertia of \[ I_1 = (5m + m) R^2 = 6m R^2 \]. When the cockroach moves halfway, new moment of inertia becomes \[ I_2 = 5m R^2 + \frac{1}{4} m R^2 = 5.25 m R^2 \]. This reduction in moment of inertia signifies that the mass is moving closer to the rotation axis, impacting the system's rotational characteristics.
Rotational Kinematics
Rotational kinematics describe the motion of rotating objects, including how their angular velocity changes over time. When you're dealing with rotating bodies, angular velocity (\( \omega \)) is a fundamental concept. It represents how fast an object rotates or spins around an axis. In this problem, the conservation of angular momentum is vital. Since no external torque acts on the cockroach-disk system, angular momentum remains constant: \[ I_1 \omega_1 = I_2 \omega_2 \]. Given the moments of inertia before and after the cockroach moves, we use them to find the final angular velocity: \[ \omega_2 = \omega_1 \frac{I_1}{I_2} = \omega_1 \frac{6}{5.25} = 1.1429 \omega_1 \]. This shows an increase in angular velocity, even though the cockroach moved toward the center, demonstrating how distributing mass affects rotational speed.
Kinetic Energy Changes
Kinetic energy in rotational motion is the energy due to the rotation of an object and is given by: \[ K = \frac{1}{2}I \omega^2 \]. Initially, the kinetic energy of the system is calculated using its initial moment of inertia and angular velocity: \[ K_1 = \frac{1}{2}I_1 \omega_1^2 = 3mR^2 \omega_1^2 \]. After the cockroach moves, the new kinetic energy is: \[ K_2 = \frac{1}{2}I_2 \omega_2^2 = 3.43mR^2 \omega_1^2 \]. When comparing the two kinetic energies, the ratio is: \[ \frac{K_2}{ K_1} = \frac{3.43mR^2 \omega_1^2}{3mR^2 \omega_1^2} = 1.143 \]. This increase in kinetic energy is due to the cockroach moving closer to the center, decreasing the moment of inertia and causing the system to rotate faster. Even in the absence of external torque, this change in mass distribution causes a higher rotational kinetic energy.

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Most popular questions from this chapter

A cockroach of mass \(m\) runs counterclockwise around the rim of a lazy Susan (a circular dish mounted on a vertical axle) of radius \(R\) and rotational inertia \(I\) and having frictionless bearings. The cockroach's speed (relative to the ground) is \(v\), whereas the lazy Susan turns clockwise with rotational speed \(\omega_{1}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the rotational speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved?

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its translational acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

What torque about the origin acts on a particle moving in the \(x y\) plane, clockwise about the origin, if the particle has the following magnitudes of rotational momentum about the origin: (a) \(4.0 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (b) \(\left(4.0 \frac{1}{s}\right) t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (c) \(\left(4.0 \frac{1}{s^{1}}\right) \sqrt{t} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (d) \(\left(4.0 \mathrm{~s}^{2}\right) / t^{2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) ?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? What are (c) the total kinetic energy of the sphere and (d)) the speed of its center of mass after it has moved \(1.0 \mathrm{~m}\) up along the incline from its initial position?

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

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