Question

A solid cylinder of mass $M=1kg$ and radius $R=0.5m$ is pivoted at its centre and has three particles of mass $m=0.1kg$ mounted at its perimeter in the vertical plane as shown in the figure.The system is initially at rest.Find the angular speed of the cylinder , when it has swung through ${90}_0$ in anticlockwise direction.[Take $g=10m/s^2$]

Answer 1

Potential energy of the system initially $P_i=+mgR+0-mgR=0$
Kinetic energy of the system initially $K.E_i=0$
Potential energy of system finally $P_f=0-mgR-0=-mgR$
Kinetic energy of system finally $K.E_f=\frac{1}{2}I{w}^2$
where $I=\frac{1}{2}M{R}^2+3\left(m{R}^2\right)$
Using conservation of total energy :
$K.E_f+P_f=K.E_i+P_i$
Or $\frac{1}{2}(\frac{M{R}^2}{2}+3m{R}^2)w^2-mgR=0$
Or $(\frac{M}{2}+3m)R{w}^2=2mg$
Or $(\frac{1}{2}+3\times 0.1)\left(0.5\right)w^2=2\times 0.1\times 10$
$⟹w=\sqrt{5}=2.24rad/s$