Question

Suppose the particle of the previous problem has a mass $m$ and a speed $v$ 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.

Accepted Answer

(a) Let the velocity of the centre of mass $C$ of the system constituting "the rod plus the particle" be $v'$ .

By the conservation of momentum about the centre of mass,

$V_{com} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} \Rightarrow v' = \frac{M \times 0 + m v}{M + m} = \frac{m v}{M + m}$

(b) Let the velocity of the particle with respect to $C$ before the collision be $v_{pc}$ ,

$v_{pc} = v_{p} - v_{c} = v - v' \Rightarrow v_{pc} = v - \frac{m v}{M + m} \Rightarrow v_{pc} = \frac{M v}{M + m}$

(c) Let the velocity of the rod with respect to $m$ before the collision be $v$ ,

$v_{rc} = v_{r} - v_{c} = 0 - v' \Rightarrow v_{rc} = - \frac{m v}{M + m}$

(d) The position of the centre of mass, if the centre of the rod is at the origin just before the collision,

$r = \frac{M \times 0 + m \times L / 2}{M + m} = \frac{m L}{2 (M + m)}$

So, the distance of the particle from the centre of mass,

$r_{p} = \frac{L}{2} - \frac{m L}{2 (M + m)} = \frac{M L}{2 (M + m)}$

So, the angular momentum of the particle about the centre of mass before the collision will be $= m v_{pc} r_{p}$

$= m \frac{M v}{M + m} \times \frac{M L}{2 (M + m)} = \frac{m M^{2} v L}{2 {(M + m)}^{2}}$

The position of the centre of mass, if the centre of the rod is at the origin just before the collision,

$r = \frac{M \times 0 + m \times L / 2}{M + m} = \frac{m L}{2 (M + m)}$

So, the angular momentum of the rod about the centre of mass before the collision will be $= M v_{rc} r$

$= M \frac{m v}{M + m} \times \frac{m L}{2 (M + m)} = \frac{m^{2} M v L}{2 {(M + m)}^{2}}$

(e) The moment of inertia of the rod about the centre of mass,

$I_{r} = \frac{M L^{2}}{12} + M r^{2} I_{r} = \frac{M L^{2}}{12} + M {(\frac{m L}{2 (M + m)})}^{2}$

The moment of inertia of the particle about the centre of mass,

$I_{p} = m {r_{p}}^{2} = m {(\frac{M L}{2 (M + m)})}^{2}$

So, the total moment of inertia,

$I = I_{r} + I_{p} I = \frac{M L^{2}}{12} + M {(\frac{m L}{2 (M + m)})}^{2} + m {(\frac{M L}{2 (M + m)})}^{2}$

or, $I = \frac{M (M + 4 m) L^{2}}{12 (M + m)}$

(f) Since, there is no external force acting on the system, the velocity of the centre will be the same after the collision.

$v_{cm} = \frac{m v}{M + m}$

Also, there is no external torque on the system, hence, the angular momentum of the system about the centre of mass before collision is equal to the angular momentum of the system about the centre of mass before collision.

Angular momentum of the rod about the centre of mass before collision $+$ angular momentum of the particle about the centre of mass before collision $= I ω$

$\frac{m^{2} M v L}{2 {(M + m)}^{2}} + \frac{m M^{2} v L}{2 {(M + m)}^{2}} = \frac{M (M + 4 m) L^{2}}{12 (M + m)} ω$

$\Rightarrow ω = \frac{6 m v}{(M + 4 m) L}$

Important Questions on Rotational Mechanics

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