Question

A uniform thin rod of mass, $M=1\text{kg}$ and length, $L=2\text{m}$ is hinged by a frictionless pivot at its one end $\text{O}$ as shown in figure. A bullet of mass, $m=50\text{g}$ moving horizontally with a velocity $v=1500{\text{ms}}^{-1}$ strikes the free end of the rod and gets embedded in it the angular velocity of the system about $\text{O}$ just after the collision is-

• A. $98{\text{rad s}}^{-1}$
• B. $95{\text{rad s}}^{-1}$
• C. $100{\text{rad s}}^{-1}$
• D. $76{\text{rad s}}^{-1}$

Answer 1

The correct option is A $98{\text{rad s}}^{-1}$

As no external force involved here, hence linear and angular momentum of the system are conserved.

Using, conservation of angular momentum about the hinge point $\text{O}$,

$L_i=L_f$

$\Rightarrow L_{bullet}+L_{rod}=L_{bullet+rod}$

$\Rightarrow mvL+0=I_{\text{(hinge point)}}\omega .....\left(1\right)$

Now, moment of inertia of system about hinge point $\text{O}$ is,

$I_{\text{(hinge point)}}=\frac{1}{3}M{L}^2+m{L}^2$

$=\frac{1}{3}\times 1\times 4+0.05\times 4=1.53{\text{kgm}}^2$

Using equation (1), we get

$0.05\times 1500\times 2=1.53\omega$

$\therefore \omega =\frac{150}{1.53}=98.03\approx 98{\text{rad s}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.