Question

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

• A. 0.3 rad/s
• B. 5.3 rad/s
• C. 10.3 rad/s
• D. 89.3 rad/s

Answer 1

Answer: (C)

10.3 rad/s

Answer 2

Initial angular momentum of bullet + initial angular momentum of cylinder

= Final angular momentum of (bullet + cylinder) system

⇒ $mvr + I_{1}\omega = (I_{1} + I_{2})\omega'$

⇒ $mvr + I_{1}\omega$ = $\left( \frac{1}{2}Mr^{2} + mr^{2} \right)\omega'$

⇒ $0.5 \times 5 \times 0.2 + 0.12 =$ $\left( \frac{1}{2}2(0.2)^{2} + (0.5)(0.2)^{2} \right)\omega'$

∴ $\omega' = 10.3$ rad/sec.