Question

A uniform rod having mass $m_{1}$ and length $L$ lies on a smooth horizontal surface. A particle of mass $m_{2}$ moving with speed $u$ on the horizontal surface strikes the free rod perpendicularly at an end and it sticks to the rod.

$(a)$ Calculate the velocity of the centre of mass $C$ of the system constituting the rod plus the particle.

$(b)$ Calculate the velocity of the particle with respect to $C$ before the collision.

$(c)$ Calculate the velocity of the rod with respect to $C$ before the collision.

$(d)$ Calculate the angular momentum of the particle and of the rod about the centre of mass $C$ before the collision.

$(e)$ Calculate the moment of inertia of the rod plus particle about the vertical axis through the centre of mass $C$ after the collision.

$(f)$ Calculate the velocity of the centre of mass $C$ and the angular velocity of the system about the centre of mass after the collision.

Accepted Answer

$(a)$ The velocity of the centre of mass $C$

$V_{cm} = \frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1} + m_{2}} = \frac{m_{1} \times 0 + m_{2} u}{m_{1} + m_{2}} = \frac{m_{2} u}{m_{1} + m_{2}}$ .

$(b)$ The velocity of the particle with respect to $C$ before the collision

$V_{2 C} = u_{2} - V_{cm} = u - \frac{m_{2} u}{m_{1} + m_{2}} = \frac{m_{1} u}{m_{1} + m_{2}}$ .

$(c)$ The velocity of the rod with respect to $C$ before the collision

$V_{1 C} = u_{1} - V_{cm} = 0 - \frac{m_{2} u}{m_{1} + m_{2}} = \frac{- m_{2} u}{m_{1} + m_{2}}$ .

$(d)$ To calculate the angular momentum of the particle about the com $C$ .

First, calculate the perpendicular distance $r$ of the particle from com $C$

here, $r = B C = (\frac{L}{2}) - O C = (\frac{L}{2}) - \frac{m_{2} (\frac{L}{2})}{(m_{1} + m_{2})} = \frac{m_{1} L}{2 (m_{1} + m_{2})}$ .

Now required angular momentum,

$L_{2 C} = m_{2} V_{2 C} r = (m_{2}) (\frac{m_{1} u}{m_{1} + m_{2}}) (\frac{m_{1} L}{2 (m_{1} + m_{2})}) = \frac{{m_{1}}^{2} m_{2} u L}{2 {(m_{1} + m_{2})}^{2}}$ .

To calculate the angular momentum of the rod about the com $C$ .

First, calculate the perpendicular distance $r$ of centre of a rod from com $C$

here, $r = O C = \frac{m_{2} (\frac{L}{2})}{(m_{1} + m_{2})} = \frac{m_{2} L}{2 (m_{1} + m_{2})}$ .

Now required angular momentum,

$L_{1 c} = m_{1} V_{1 C} r = (m_{1}) (\frac{m_{2} u}{m_{1} + m_{2}}) (\frac{m_{2} L}{2 (m_{1} + m_{2})}) = \frac{{m_{2}}^{2} m_{1} u L}{2 {(m_{1} + m_{2})}^{2}}$ .

$(e)$ Using parallel axis theorem, the moment of inertia of the rod about com $C$

$I_{1 C} = \frac{m_{1} L^{2}}{12} + m_{1} \times O C^{2} = \frac{m_{1} L^{2}}{12} + m_{1} {(\frac{m_{2} L}{2 (m_{1} + m_{2})})}^{2} \dots (i)$

Similarly, the moment of inertia of the particle about com $C$

$I_{2 C} = m_{2} \times B C^{2} = m_{2} {(\frac{m_{1} L}{2 (m_{1} + m_{2})})}^{2} \dots (ii)$

Now the moment of inertia of the particle plus rod about com $C$

$I_{C} = I_{1 C} + I_{2 C} = \frac{m_{1} L^{2}}{12} + m_{1} {(\frac{m_{2} L}{2 (m_{1} + m_{2})})}^{2} + m_{2} {(\frac{m_{1} L}{2 (m_{1} + m_{2})})}^{2}$

$\Rightarrow I = \frac{m_{1} L^{2}}{12} + \frac{m_{1} m_{2} L^{2}}{4 (m_{1} + m_{2})} = \frac{m_{1} L^{2} (m_{1} + m_{2}) + 3 m_{1} m_{2} L^{2}}{12 (m_{1} + m_{2})} = \frac{m_{1} (m_{1} + 4 m_{2}) L^{2}}{12 (m_{1} + m_{2})}$ .

$(f)$ As there is no external force, the velocity of com $V_{cm}$ will not change after the collision

$\Rightarrow V_{cm} = \frac{m_{2} u}{m_{1} + m_{2}}$ .

To calculate angular velocity about $C$

$ω_{C} = \frac{L_{C}}{I_{C}} = \frac{L_{1 C} + L_{2 C}}{I_{C}} = [\frac{{m_{2}}^{2} m_{1} u L}{2 {(m_{1} + m_{2})}^{2}} + \frac{{m_{1}}^{2} m_{2} u L}{2 {(m_{1} + m_{2})}^{2}}] \times \frac{12 (m_{1} + m_{2})}{m_{1} (m_{1} + 4 m_{2}) L^{2}} = \frac{6 m_{2} u}{(m_{1} + 4 m_{2}) L}$ .

Important Questions on Centre of Mass, Momentum and Collisions

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