Explain Center of Percussion

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Let us consider the following example:

A cricket bat of mass M is suspended, so that it can swing around a pivot at point S (near one end of the handle) in vertical plane. If a ball moving horizontally strikes the bat, at point A with an impulsive force F for time Δt, how much impulse is applied to the pivot by the bat during collision?

The ball, on collision with the bat, transfers a momentum Δ p to the bat given by,

Δ p = F Δ t = F Δ t i

where , Δ t is the time the ball remains in contact with the bat. The bat is consequently pushed towards X-axis. Being pivoted at S, suppose the bat applies an impulse Δ p’ on the pivot; pivot in turn applies an impulse (– Δ p’) on the bat. The net impulse on the bat gives a translation to the center-of-mass C of the bat:

Δ p – Δ p’ = Mvc

where v_c is the velocity acquired by CM of the bat immediately after collision (towards X-axis).

Also, the force F provides an anti-clockwise impulsive torque about CM of the bat given by

(r_A,C  × F) Δ t = l’ (F Δ t) k = l’ Δ p k

where l’ = |r_A,C| is the normal distance between points A and C.

Similarly, the impulse (– Δ p’) by pivot also provides the anti-clockwise torque.

r_S,C  × (– Δ p’) = l Δ p’ k

where l = | r_S,C | is the normal distance between points S and C.