/*! 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 57 A uniform disk of mass \(10 m\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform disk of mass \(10 m\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

Short Answer

Expert verified
The final angular velocity is approximately 16.85 rad/s and the kinetic energy ratio is about 0.843.

Step by step solution

01

Calculate Initial Moment of Inertia

The moment of inertia of the system initially consists of the moment of inertia of the large disk and the small disk. The moment of inertia of a disk is given by \( I = \frac{1}{2} M R^2 \) where \( M \) is the mass and \( R \) is the radius. For the larger disk: \[ I_{\text{large}} = \frac{1}{2} \times 10m \times (3r)^2 = \frac{1}{2} \times 10m \times 9r^2 = 45mr^2 \]For the smaller disk: \[ I_{\text{small}} = \frac{1}{2} \times m \times r^2 = \frac{1}{2} m r^2 \]Initial total moment of inertia:\[ I_{\text{initial}} = 45mr^2 + \frac{1}{2}mr^2 = 45.5mr^2 \]
02

Calculate Final Moment of Inertia

When the smaller disk slides to the edge of the larger disk, the final moment of inertia changes.The small disk is now at a radius of \(3r\), so:\[ I_{\text{small,new}} = m \times (3r)^2 = 9mr^2 \]The moment of inertia of the large disk remains:\[ I_{\text{large}} = 45mr^2 \]Total final moment of inertia:\[ I_{\text{final}} = 45mr^2 + 9mr^2 = 54mr^2 \]
03

Apply Conservation of Angular Momentum

Angular momentum \( L \) is conserved, so\[ L_{\text{initial}} = L_{\text{final}} \]Since \( L = I\omega \), we can write\[ I_{\text{initial}} \cdot \omega_{\text{initial}} = I_{\text{final}} \cdot \omega_{\text{final}} \]\[ 45.5mr^2 \cdot 20 = 54mr^2 \cdot \omega_{\text{final}} \]Solve for \( \omega_{\text{final}} \):\[ \omega_{\text{final}} = \frac{45.5 \times 20}{54} = \frac{910}{54} \]\[ \omega_{\text{final}} \approx 16.85 \text{ rad/s} \]
04

Calculate Initial and Final Kinetic Energies

The initial kinetic energy is given by:\[ K_{0} = \frac{1}{2} I_{\text{initial}} \omega_{\text{initial}}^2 \]\[ K_{0} = \frac{1}{2} \times 45.5mr^2 \times (20)^2 \]\[ K_{0} = \frac{1}{2} \times 45.5 \times 400 \times mr^2 \]\[ K_{0} = 9100mr^2 \]The final kinetic energy is:\[ K = \frac{1}{2} I_{\text{final}} \omega_{\text{final}}^2 \]\[ K = \frac{1}{2} \times 54mr^2 \times (16.85)^2 \]\[ K \approx \frac{1}{2} \times 54 \times 284.1225 \times mr^2 \]\[ K \approx 7671.3mr^2 \]
05

Calculate Ratio of Kinetic Energies

The ratio \( K/K_{0} \) is given by:\[ \frac{K}{K_{0}} = \frac{7671.3mr^2}{9100mr^2} = \frac{7671.3}{9100} \]\[ \frac{K}{K_{0}} \approx 0.843 \]

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Key Concepts

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

Understanding Moment of Inertia
Moment of inertia is a crucial concept in rotational dynamics. It measures an object's resistance to changes in its rotational motion. Think of it as the rotational equivalent of mass in linear motion. The higher the moment of inertia, the harder it is to change the object's rotational speed.
For a disk, the moment of inertia is calculated using the formula: \[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass and \( R \) is the radius of the disk. This formula helps quantify how the mass is spread in relation to the axis of rotation—when mass is further from the axis, the moment of inertia increases.
In our exercise, we calculated the initial and final moments of inertia of two disks—one larger and one smaller. Initially, the smaller disk is centered and contributes less to the total moment. When it moves to the edge, it drastically increases the system's total moment of inertia. This is because the radius at which the mass acts is squared, making distance a significant factor. Increasing the radius at which the small mass acts causes a larger change in the overall system's inertia.
Kinetic Energy in Rotational Motion
Kinetic energy in rotational motion is akin to that in linear motion but involves rotational variables. It is the energy that an object possesses due to its spin by virtue of its rotational velocity. The formula to calculate rotational kinetic energy is:\[ K = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
In this exercise, we start with the initial kinetic energy of the system when both disks are rotating together at a given speed. We then calculate what happens to this energy when the smaller disk moves outward. We find that the system's final kinetic energy decreases compared to the initial kinetic energy.
This happens due to conservation of angular momentum. As the smaller disk moves outward, it causes the system to rotate slower. Although the moment of inertia increases, making it harder to spin, the decrease in angular velocity leads to a net reduction in the kinetic energy of the system.
Exploring Rotational Dynamics
Rotational dynamics explores the principles governing the rotation of bodies and is governed primarily by Newton’s second law for rotation. This involves understanding how torque, angular momentum, and angular velocity interplay.
One central principle is the conservation of angular momentum, which states that if no external torque acts on a system, the total angular momentum remains constant. This idea is crucial in our exercise. The initial angular momentum of both disks together is preserved when the smaller disk moves outward.
This conservation law enables us to connect the initial and final states of the system despite the redistribution of mass. It allows us to solve for the system's final angular velocity using:\[ L_{\text{initial}} = L_{\text{final}} \]\[ I_{\text{initial}} \cdot \omega_{\text{initial}} = I_{\text{final}} \cdot \omega_{\text{final}} \]where \( L \) denotes angular momentum. By maintaining this balance, we predict the behavior of rotating bodies, explaining why changing the position of masses in a system affects rotational speed.

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Most popular questions from this chapter

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