(C) In the absence of any external torque,the angular momentum of the system remains conserved.$L_{initial} = L_{final}$$I \omega = I' \omega'$Initial

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) In the absence of any external torque,the angular momentum of the system remains conserved.$L_{initial} = L_{final}$$I \omega = I' \omega'$Initially,the moment of inertia of the ring about its axis is $I = MR^2$.When two objects of mass $m$ are placed at opposite ends of a diameter,they are at a distance $R$ from the axis of rotation. The new moment of inertia becomes $I' = MR^2 + mR^2 + mR^2 = (M + 2m)R^2$.Using the conservation of angular momentum:$MR^2 \omega = (M + 2m)R^2 \omega'$$\omega' = \frac{MR^2 \omega}{(M + 2m)R^2}$$\omega' = \frac{\omega M}{M + 2m}$

Class 11 Physics System of Particles and Rotational Motion Conservation of angular momentum (combined translation and rotational motion)

Difficult

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are gently placed at opposite ends of a diameter of the ring. The new angular velocity of the ring will be

A
$\frac{\omega M}{M + m}$
B
$\frac{\omega (M - 2m)}{M + 2m}$
C
$\frac{\omega M}{M + 2m}$
D
$\frac{\omega (M + 2m)}{M}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

System of Particles and Rotational Motion

Subtopic

Conservation of angular momentum (combined translation and rotational motion)

$A$ thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity:

MediumJEE MAIN 2021

View Solution

$A$ ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the ring. Because of this,the kinetic energy loss will be

DifficultJEE MAIN 2013

View Solution

$A$ ring of mass $m$ and radius $r$ is placed on a rough surface with an initial angular velocity $\omega_0$. The initial velocity of the center of the ring is zero. What will be the velocity of its center when the ring stops slipping?

Medium

View Solution

$A$ mass is whirled in a circular path with constant angular velocity and its linear velocity is $v$. If the string is now halved keeping the angular momentum same,the linear velocity is

MediumMHT CET 2019

View Solution

When a body is spinning on its axis in the absence of any external torque,choose the wrong statement.

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are gently placed at opposite ends of a diameter of the ring. The new angular velocity of the ring will be

Similar Questions

$A$ thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity:

$A$ ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the ring. Because of this,the kinetic energy loss will be

$A$ ring of mass $m$ and radius $r$ is placed on a rough surface with an initial angular velocity $\omega_0$. The initial velocity of the center of the ring is zero. What will be the velocity of its center when the ring stops slipping?

$A$ mass is whirled in a circular path with constant angular velocity and its linear velocity is $v$. If the string is now halved keeping the angular momentum same,the linear velocity is

When a body is spinning on its axis in the absence of any external torque,choose the wrong statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module