Question

A mass $m$ is attached to a pulley through a cord as shown in the figure. The pulley is a solid disk with radius $R$. The cord does not slip on the disk. The mass is released from rest at a height $h$ from the ground and at the instant the mass reaches the ground, the disk is rotating with angular velocity $\omega$. The mass of the disk is $M=xm\left(\begin{array}{c}\frac{2gh}{R^2{\omega }^2}-1\end{array}\right)$. Find $x$.

Answer 1

Let the tension be T

now $mg-T=ma$..(1)

$TR=\frac{M{R}^2\alpha }{2}$....(2)

$a=\alpha R$...(3)

from above equations we get $a=\frac{mg}{\frac{M}{2}+m}$

now final velocity $V^2=2ah=\frac{2mgh}{\frac{M}{2}+m}$

Also $V^2={\omega }^2{R}^2$

so $M=2m(\frac{2gh}{R^2{\omega }^2}-1)$