/*! 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 54 Shows an overhead view of a ring... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 by approximately 7.68 J.

Step by step solution

01

Define the moment of inertia

First, we need to find the moment of inertia for the ring. The ring has two parts - one from radius 0 to \(R_1\), which is empty, and another from \(R_1\) to \(R_2\), which carries the mass \(M\). The moment of inertia \(I\) for a ring rotating about its center is calculated as \(I = \frac{1}{2} M (R_1^2 + R_2^2)\). Since \(R_1 = \frac{R_2}{2}\), substitute this into our equation to get \(I = \frac{1}{2} M \left(\left(\frac{R_2}{2}\right)^2 + R_2^2\right)\).
02

Calculate the cat's moment of inertia initially

The cat initially is at radius \(R_2\). Its moment of inertia is given by \(I_{cat, initial} = m R_2^2\) where \(m = \frac{M}{4}\). Substitute \(M = 8 \text{ kg}\) into this formula, giving us \(I_{cat, initial} = \left(\frac{8}{4}\right) (0.8^2) = 1.28 \text{ kg} \cdot \text{m}^2\).
03

Apply conservation of angular momentum

Since there is no external torque, the angular momentum of the system is conserved. The initial angular momentum \(L_{initial} = (I + I_{cat, initial}) \omega\), and \(\omega = 8 \text{ rad/s}\). Compute the initial value of \(L_{initial}\).
04

Calculate the cat's moment of inertia when moved to \(R_1\)

Now, the cat moves to the inner radius \(R_1\). The new moment of inertia for the cat is \(I_{cat, final} = m R_1^2 = \frac{M}{4} \left(\frac{R_2}{2}\right)^2 = \frac{8}{4} \left(\frac{0.8}{2}\right)^2 = 0.32 \text{ kg} \cdot \text{m}^2\).
05

Calculate the final angular momentum

With the cat moved, the final angular momentum \(L_{final} = (I + I_{cat, final}) \omega_{final}\). Using conservation \(L_{initial} = L_{final}\), solve for the new angular speed \(\omega_{final}\).
06

Calculate initial and final kinetic energy

The initial kinetic energy is \(KE_{initial} = \frac{1}{2} (I + I_{cat, initial}) \omega^2\). The final kinetic energy is \(KE_{final} = \frac{1}{2} (I + I_{cat, final}) \omega_{final}^2\). Compute these values using the previously calculated \(\omega_{final}\).
07

Find the increase in kinetic energy

The increase in kinetic energy after the cat moves is \(\Delta KE = KE_{final} - KE_{initial}\). Substitute the calculated kinetic energies from Step 6 to find \(\Delta KE\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Moment of inertia
The moment of inertia is crucial in understanding rotational motion. Imagine it as the rotational counterpart of mass in linear motion. It tells us how much an object resists changes to its rotational state. For the exercise with the ring and the cat, the ring's moment of inertia accounts for its mass distribution around the center. Since the ring is not a solid disk but has an empty center from 0 to \(R_1\), the moment of inertia calculation slightly differs. The formula used is \(I = \frac{1}{2} M (R_1^2 + R_2^2)\). You substitute \(R_1 = \frac{R_2}{2}\) to fit this specific ring's situation.
Understanding this helps explain why the cat, while moving, changes the moment of inertia based on its position on the ring, from the outer to the inner edge. Key takeaways about the moment of inertia:
  • It's dependent on mass distribution.
  • Objects with mass farther from the axis have higher moments of inertia (harder to spin).
  • It plays a significant role in rotational motion calculations.
Rotational motion
Rotational motion deals with objects moving in a circular path around a central point. Similar to how linear motion describes straight-line movement, rotational motion specifically involves angles and spins. In the exercise involving the ring and the cat, the primary concept is the conservation of angular momentum. Angular momentum \(L\) is calculated as the product of the moment of inertia \(I\) and angular velocity \(\omega\), expressed as \(L = I \omega\). The principle of conservation of angular momentum states that if no external torques are acting on a system, the total angular momentum remains constant. As the cat moves from the outer radius \(R_2\) to the inner radius \(R_1\), no external force acts on the ring-cat system. Hence, the initial angular momentum \(L_{initial}\) equals the final angular momentum \(L_{final}\).Key concepts in rotational motion include:
  • Angular velocity, which measures how fast an object rotates.
  • The effect of changes in mass distribution on rotation speed due to angular momentum conservation.
  • The relationship between torque and changes in rotational speed, although not directly applied in this exercise.
Kinetic energy change
Kinetic energy is the energy an object possesses due to its movement. For rotational motion, kinetic energy \(KE\) depends on both the object's moment of inertia and its angular velocity. The formula used is \(KE = \frac{1}{2} I \omega^2\). In the context of the problem, initially, the system consisting of the ring and the cat spinning together has certain kinetic energy. When the cat moves to a different radius, the distribution of the mass changes the system's moment of inertia, affecting the overall kinetic energy.While the system's angular momentum stays the same due to conservation laws, the kinetic energy changes because the new angular velocity \(\omega_{final}\) differs slightly from the initial speed \(\omega\). By calculating both the initial and final kinetic energies, we find the increase in kinetic energy, which is quantified in the problem as \(\Delta KE = KE_{final} - KE_{initial}\).Concept highlights for kinetic energy in rotational systems:
  • It's influenced by changes in angular speed and mass distribution.
  • The total kinetic energy can change even though angular momentum is conserved.
  • Understanding how kinetic energy varies with rotational motion helps in evaluating the dynamics of systems like the cat-ring scenario.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A car travels at \(80 \mathrm{~km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Each tire has a diameter of \(66 \mathrm{~cm} .\) Relative to a woman riding in the car and in unit-vector notation, what are the velocity \(\vec{v}\) at the (a) center, (b) top, and (c) bottom of the tire and the magnitude \(a\) of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity \(\vec{v}\) at the (g) center, (h) top, and (i) bottom of the tire and the magnitude \(a\) of the acceleration at the (j) center, (k) top, and (I) bottom of each tire?

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.

An overhead view of a thin uniform rod of length \(0.800 \mathrm{~m}\) and mass \(M\) rotating horizontally at angular speed \(20.0 \mathrm{rad} / \mathrm{s}\) about an axis through its center. A particle of mass \(M / 3,00\) initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed \(v_{p}\) is \(6.00 \mathrm{~m} / \mathrm{s}\) greater than the speed of the rod end just after ejection, what is the value of \(v_{p} ?\)

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\) and a rotational inertia of \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.