Table Top Physics - Snooker/pool - left side

Ball A is sent toward ball B, initially A has clockwise spin and B has no spin, they hit head-on. Due to the rotation and friction, the impulse is at a different angle to the approach velocity and to surface normal. This impulse tends to reduce the angular speed of A and to cause B to rotate in the opposite direction. If we assume that the initial angular velocity is w, if we assume that the contact is perfect and all the angular velocity is transferred to ball B, and B is rotating in the opposite direction. So the angular momentum of A added to the angular momentum of B will change from +Iw to -Iw.


Since angular momentum of a closed system is conserved and we have a change of angular momentum of 2*I*w, where does this angular momentum go?

I think the answer is that we must measure the total angular velocity of the whole system, so let us take the CM of A at the point of collision to measure the total angular velocity. Since A is travelling directly to, and directly from, this point we only need to take into account its rotation. The velocity of Bs CM is offset, so we need to include this.


so initial angular velicity = final angular velocity

Iw = -Iw + m v offset

2 I w = m v offset

but for a sphere I = m * 2/5 r2

therefore for angular momentum to be conserved:

offset = 2*I*w / m*v

offset = 2*m * (2/5) r2 *w / m*v

offset = (4/5) r2 *w / v


So we can calulate the offset, and from that we can calculate the angle that ball B is deflected.

Can anyone check these calculations and finish it off?

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