Question

A solid cylinder of mass $2$ Kg and radius $0.2$ m is rotating about its own axis without friction with angular velocity $3$ rad/s. A particle of mass $0.5$ Kg and moving with a velocity of $5$m/s strikes the cylinder and sticks to it as shown in. The velocity of the system after the particle sticks it will be

• A. $0.3$ radians/sec
• B. $5.3$ radians/sec
• C. $10.3$ radians/sec
• D. $8.3$ Radians/sec

Answer 1

Answer: (C)

$10.3$ radians/sec

Using the principle of conservation of angular momentum we have-
$\frac{1}{2}{m}_c{r}^2{\omega }_1+m_{p}vr=(\frac{1}{2}{m}_c{r}^2+m_p{r}^2){\omega }_2$
or
$\frac{1}{2}\left(2\right){0.2}^2\left(3\right)+0.5\left(5\right)\left(0.2\right)=\left(\frac{1}{2}\right(2){0.2}^2+0.5(0.2{)}^2){\omega }_2$
or
$0.12+0.5=(0.04+0.02){\omega }_2$
or
${\omega }_2=10.3rad/s$