Center of Percussion

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.

The net impulse of torques on the bat is equal to change of its angular momentum about C.

l’ Δ p + l Δ p’ = Δ L_CM = I_CM ω

where, is the angular velocity acquired by the bat during collision.

I_CM = Mk², where k is radius of gyration of the bat.

Now, the velocity of S with respect to ground is,



therefore, the velocity of S with respect to ground is,

v_S = v_S,C + v_C

or, v_S = v_C – ω l           (towards X-axis)

If the point S is rigidly pivoted, then v_S = 0, or v_C = ω l. Hence, we have

Δ p – Δ p’ = M ω l

And, l’ Δ p + l Δ p’ = Mk² ω

Eliminating ω, we find



That is, if we are holding (or pivoting) a bat at distance l above the center-of-mass C, then the bat produces an impulse of force Δ p’ on the pivot (towards X-axis), when the ball strikes at distance l’ below C. Obviously, if k² < ll’, then Δ p’ is – ve, or the impulse on the pivot is towards – ve X-axis.

That means, if we are rigidly holding the bat, our hand would get a jerk (impulse) towards right (or left) depending upon whether k² > ll’ (or k² < ll’).

Interestingly, if k² = ll’, then

Δ p’ = 0

That is, the point S remains automatically at rest; the bat transfers no momentum to the pivot point and we feel no jerk when the ball strikes at distance l’ below the CM, where



The point A at distance l’ = k²/l from C is called the center of percussion.

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