Question

In a horizontal smooth plane. If a particle sticks after collision to the rod, find the final angular velocity .

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is D

None of these

Lets say when the particle sticks to the rod, the system rotates with final angular velocity $\omega$

Change in linear momentum of rod

=$(m_2{V}_{CM}+m_1\omega \ell )-m_{1}u$

= $(m_2\omega +m_1\omega \ell )-m_{1}u$ [as the velocity of CM will be $\omega$ times its distance from hinge]

Also change in linear momentum of rod =

Impulse provided by the particle = - Impulse provided by the hinge

$\Rightarrow$ Impulse due to hinge =$-[m_2\omega +m_1\omega \ell -m_{1}u]$ - - - - - - (1)

Here also we do not know '$\omega$'

To find $\omega$:

If I take the whole rod and particle as a system then net T is zero and angular momentum about the hinge must be conserved.

$L_{abouthinge}initially=m_{1}u\ell$ [only of the particle]

$L_{abouthinge}final=I_{new}\times \omega$

$=(\frac{m_2{l}^2}{3}+m_1{l}^2)\omega$

$L_{initial}=L_{final}$

$m_{1}ul=(\frac{m_2{l}^2}{3}+m_1{l}^2)\omega$

$\Rightarrow \omega =\frac{3m_{1}u}{(m_2+3m_1)l}$

Substitute in equation 1 to get the exact answer