/*! 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 88 A hoop of radius \(R\) and mass ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 88

A hoop of radius \(R\) and mass \(m\) rotating with an angular velocity \(\omega_{0}\) is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity if the centre of the loop ceases to slip? (A) \(\frac{r \omega_{0}}{3}\) (B) \(\frac{r \omega_{0}}{2}\) (C) \(r \omega_{0}\) (D) \(\frac{r \omega_{0}}{4}\)

Short Answer

Expert verified
(A) \(\frac{R\omega_{0}}{3}\)

Step by step solution

01

Consider rolling without slipping condition

When the hoop ceases to slip, the rolling condition without slipping is valid. In that case, we can relate the linear velocity of the center of mass (v) and the angular velocity (蠅) of the hoop as follows: \[v = R蠅\]
02

Analyze conservation of angular momentum

When there is no slipping, the hoop's net torque about its center of mass will be zero. Hence, the angular momentum about the center of mass will be conserved throughout the motion. The initial angular momentum of the hoop about the center of mass can be obtained using the initial angular velocity: \[L_i = I_{cm}蠅_0\] Here, \(I_{cm}\) is the moment of inertia of the hoop about its center of mass, which can be calculated as: \[I_{cm} = mR^2\]
03

Calculate the final angular momentum

When the hoop stops slipping, the final angular momentum can be obtained using the final linear velocity (v) and angular velocity (蠅): \[L_f = I_{cm}蠅 + mvR\] As we have related the angular and linear velocities in Step 1, we can write the final angular momentum as: \[L_f = I_{cm}(v/R) + mvR\]
04

Apply conservation of angular momentum

As the angular momentum is conserved, we can equate the initial and final angular momenta. \[I_{cm}蠅_0 = I_{cm}(v/R) + mvR\]
05

Solve for the linear velocity (v)

Substituting the expression for the moment of inertia of the hoop, we get: \[mR^2蠅_0 = mR^2(v/R) + mvR\] Now, we solve for the linear velocity (v): \[v = \frac{R蠅_0}{3}\] Comparing the answer with the provided options, we see that the correct answer is: (A) \(\frac{R\omega_{0}}{3}\)

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 91影视!

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 force of \(-F \hat{k}\) acts on \(\mathrm{O}\), the origin of the coordinates system. The torque about the point \((1,-1)\) is [2006] (A) \(F(\hat{i}+\hat{j})\) (B) \(-F(\hat{i}-\hat{j})\) (C) \(F(\hat{i}-\hat{j})\) (D) \(-F(\hat{i}+\hat{j})\)

Three marbles roll down on three different smooth tracks of same vertical height. Track (i) is inclined at \(75^{\circ}\) to the ground, track (ii) is inclined at \(60^{\circ}\) to the ground and track (iii) is inclined at \(30^{\circ}\) to ground. These marbles reach the ground with respective velocities \(u_{1}, u_{2}\) and \(u_{3}\) (A) \(u_{1}\) is greatest of all (B) \(u_{3}\) is least of all (C) \(u_{1}=u_{2}\) (D) \(u_{2}=u_{3}\)

Consider a uniform square plate of side \(a\) and mass \(m\). The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is (A) \(\frac{5}{6} m a^{2}\) (B) \(\frac{1}{12} m a^{2}\) (C) \(\frac{7}{12} m a^{2}\) (D) \(\frac{2}{3} m a^{2}\)

The initial angular velocity of a circular disc of mass \(M\) is. Then two small spheres of mass \(m\) are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc? [2002] (A) \(\left(\frac{M+m}{M}\right) \omega_{1}\) (B) \(\left(\frac{M+m}{m}\right) \omega_{1}\) (C) \(\left(\frac{M}{M+4 m}\right) \omega_{1}\) (D) \(\left(\frac{M}{M+2 m}\right) \omega_{1}\)

While the disc skids, its translational and angular acceleration are related as (A) \(a=R \alpha\) (B) \(a=\frac{1}{4} R \alpha\) (C) \(a=\frac{1}{2} R \alpha\) (D) \(a=2 R \alpha\)
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Wave Optics

Read Explanation

Physics of Motion

Read Explanation

Particle Model of Matter

Read Explanation

Electricity and Magnetism

Read Explanation

Famous Physicists

Read Explanation
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.