Question

Find the velocity of the centre of mass $C$ and the angular velocity of the system about the centre of mass after the collision.

• A. $\frac{2mv}{M+m},\frac{3Mv}{(M+2m)L}$
• B. $\frac{Mv}{M+m},\frac{6Mv}{(M+4m)L}$
• C. $\frac{2Mv}{M+m},\frac{3mv}{(M+4m)L}$
• D. $\frac{mv}{M+m},\frac{6Mv}{(M+4m)L}$

Answer 1

The correct option is D $\frac{mv}{M+m},\frac{6Mv}{(M+4m)L}$

Initially center of mass of the rod lies at O.

Let center of mass of the system after the point mass has stuck on the rod be Point C.

$(m+M)x=2m\left(0\right)+m\left(\frac{L}{2}\right)$

$⟹x=(\frac{m}{M+m})\frac{L}{2}$ and $BC=(\frac{M}{M+m})\frac{L}{2}$ ..............(1)

After the collision, the whole system moves with a velocity $v^{\prime }$ translationally and rotates about C at an angular velocity $w$.

For conservation of linear momentum, $P_i=P_f$

$mv+0=(M+m)\left(v^{\prime }\right)⟹v^{\prime }=\frac{mv}{M+m}$ ...................(2)

$I_{CM}=I_C^{\prime }$ (rod) + $I_C^{\prime \prime }$ (point mass)

$I_{CM}=[\frac{1}{12}\left(M\right)L^2+\left(M\right)x^2]+m(BC{)}^2$

$I_{CM}=(\frac{M+4m}{M+m})\frac{M{L}^2}{12}$ ................................(3)

Also, angular momentum about B is conserved.

$L_i=L_f⟹0=I_{CM}w-(M+m)\left(v^{\prime }\right)BC$

$0=(\frac{M+4m}{M+m})\frac{M{L}^2}{12}w-(M+m)\frac{mv}{M+m}\times (\frac{M}{M+m})\frac{L}{2}$

$⟹w=\frac{6mv}{L(M+4m)}$