Question

Suppose the particle of the previous problem has a mass m and a speed \[\nu\] before the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the centre of mass C of the system constituting "the rod plus the particle". (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle and of the rod about the centre of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision. (f) Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.

Answer 1

(a) It is given that no external torque and force is applied on the system.

Applying the law of conservation of momentum, we get

\[m\nu = \left( M + m \right)  \nu'\]

\[\Rightarrow \nu' = \frac{m\nu}{M + m}\]

(b) Velocity of the particle w.r.t. centre of mass (COM) C before the collision = \[v_c  = v - v'\]

\[\Rightarrow  v_c  = v - \frac{mv}{M + m} = \frac{Mv}{M + v}\]

(c) Velocity of the particle w.r.t. COM C before collision \[=  - \frac{M\nu}{M + m}\]

(d) Distance of the COM from the particle,

\[x_{cm}  = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\]

\[ \Rightarrow r = \frac{M \times \frac{L}{2} + m \times 0}{M + m}\]

\[ \Rightarrow r = \frac{ML}{2\left( M + m \right)}\]

∴ Angular momentum of body about COM

\[= mvr\]

\[ = m \times \frac{Mv}{\left( M + m \right)} \times \frac{ML}{2\left( M + m \right)}\]

\[ = \frac{M^2 mvL}{2 \left( M + m \right)^2}\]

∴ Angular momentum of rod about COM

\[= M \times \left( \frac{mv}{\left( M + m \right)} \right) \times \frac{1}{2}\frac{mL}{\left( M + m \right)}\]

\[ =   \frac{M m^2 vL}{2  \left( M + m \right)^2}\]

(e) Moment of inertia about COM = I

\[=  I_1  +  I_2\]

\[I = m \left[ \frac{ML}{2\left( M + M \right)} \right]^2  + \frac{M L^2}{12} + M \left[ \frac{ML}{2\left( m + M \right)} \right]^2   \]

\[   = \frac{m M^2 L^2}{4 \left( m + M \right)^2} + \frac{M L^2}{12} + \frac{M m^2 L^2}{4 \left( M + m \right)^2}\]

\[   = \frac{3m M^2 L^2 + M \left( m + M \right)^2 L^2 + 3M m^2 L^2}{12 \left( m + M \right)^2}\]

\[   = \frac{M\left( M + 4m \right) L^2}{12\left( M + m \right)}\]

(f) About COM,

\[V_{cm}  = \frac{m\nu}{M + m}\]

\[ \therefore   I\omega = mvr = mv \times \frac{ML}{2  \left( M + m \right)}\]

\[ \Rightarrow   \omega = \frac{m\nu ML}{2  \left( M + m \right)} \times \frac{12  \left( M + m \right)}{M  \left( M + 4m \right) L^2}\]

\[= \frac{6m\nu}{\left( M + 4m \right)L}\]