(C) Since the force applied to pull the string is radial,the torque acting on the object about the center of the circle is zero. Therefore,the angular

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$\left(\frac{r_1}{r_2}\right)^2$

(C) Since the force applied to pull the string is radial,the torque acting on the object about the center of the circle is zero. Therefore,the angular momentum of the object is conserved.$L = m v_1 r_1 = m v_2 r_2$$\therefore v_2 = v_1 \left(\frac{r_1}{r_2}\right)$Now,the ratio of the new kinetic energy $(K_2)$ to the original kinetic energy $(K_1)$ is:$\frac{K_2}{K_1} = \frac{\frac{1}{2} m v_2^2}{\frac{1}{2} m v_1^2} = \left(\frac{v_2}{v_1}\right)^2$Substituting the value of $v_2/v_1$ from the angular momentum equation:$\frac{K_2}{K_1} = \left(\frac{r_1}{r_2}\right)^2$

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

Medium

$A$ small object is attached to a light string which passes through a hollow tube. The tube is held by one hand and the string by the other. The object is set into rotation in a circle of radius $r_1$. The string is then pulled down,shortening the radius of the circle to $r_2$. The ratio of the new kinetic energy to the original kinetic energy is

A
$\frac{r_1}{r_2}$
B
$1$
C
$\left(\frac{r_1}{r_2}\right)^2$
D
$\left(\frac{r_2}{r_1}\right)^2$

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System of Particles and Rotational Motion

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Conservation of angular momentum (combined translation and rotational motion)

$A$ circular platform is mounted on a frictionless vertical axle. Its radius $R = 2\, m$ and its moment of inertia about the axle is $200\, kg\, m^2$. It is initially at rest. $A$ $50\, kg$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1\, m/s$ relative to the ground. Time taken by the man to complete one revolution is

MediumAIPMT 2012

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$A$ train is moving slowly at $2 \, m/s$ next to a railway platform. $A$ man,$1.5 \, m$ tall,alights from the train such that his feet are fixed on the ground. Taking him to be a rigid body,the instantaneous angular velocity (in $rad/s$) is

KVPY 2019

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$A$ rotating table completes one revolution in $10 \ s$. Its moment of inertia is $100 \ kg \cdot m^2$. $A$ man of mass $50 \ kg$ stands at the center of the rotating table. If the man moves $2 \ m$ away from the center,what will be the angular velocity of the table in $rad/s$?

Medium

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$A$ thin rod of mass $0.9\, kg$ and length $1\, m$ is suspended,at rest,from one end so that it can freely oscillate in the vertical plane. $A$ particle of mass $0.1\, kg$ moving in a straight line with velocity $80\, m/s$ hits the rod at its bottom-most point and sticks to it. The angular speed (in $rad/s$) of the rod immediately after the collision will be

DifficultJEE MAIN 2020

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$A$ 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:

MediumAIPMT 2009

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$A$ train is moving slowly at $2 \, m/s$ next to a railway platform. $A$ man,$1.5 \, m$ tall,alights from the train such that his feet are fixed on the ground. Taking him to be a rigid body,the instantaneous angular velocity (in $rad/s$) is

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