Question

A rod of mass $20$ kg & length $10$ m is hinged at A & hanging vertically. A bullet of mass $5$ kg moving with velocity $10$ m/s sticks to one end of rod. Find angular velocity of rod just after the collision particle sticks to it.

• A. $\frac{7}{3}$ rad/sec
• B. $\frac{3}{7}$ rad/sec
• C. $\frac{3}{10}$ rad/sec
• D. $3$ rad/sec

Answer 1

The correct option is B $\frac{3}{7}$ rad/sec

Given speed of the $u_B=10m/s$,mass of the bullet $m_B$=5 kg,mass of rod$M$= 20 kg and length of rod $l=10m$

Angular momentum of the system before collision $L_1=\frac{1}{3}M{L}^2\left(v\right)+m_B{u}_{B}l$

as speed of the rod is 0 before collision, thus the above equation reduces to

$L_1=500$

Let $\omega$ be the angular velocity of the rod after collision thus the angular momentum of the system will be

$L_2=\left(\frac{1}{3}M{L}_2+m_B{L}^2\right)\omega$

$=\left(\frac{1}{3}\right(20){10}^2+5\times {10}^2)\omega$

$L_2=\left(\frac{3500}{3}\right)\omega$

Now as per law of conservation of momentum

$L_1=L_2$

$500=\left(\frac{3500}{3}\right)\omega$

$\omega =\frac{3}{7}rad/s$