View Full Version : Why does a backspin in billiards stop the white ball?
4th May 2008, 08:14 PM
I'm pondering on an activity for my newsletter this week, while also thinking of going for a beer later and playing a game or two of pool. So...it struck me to combine the two and think of a science activity which involves pool balls!
Anyway, I started to think of explaining in basic terms how a newton's cradle works, but then remembered I'd already done something similar a while ago. This led to me thinking of demonstrating how backspin stops the cue ball as it connects with its target...except I'm now not sure of why this actually happens.
Any physics-minded souls have a clue they can lend me?
4th May 2008, 08:44 PM
Putting spin on a cue ball relies on the fact that the ball skims across the surface at high speed, but grips the surface at low speed. When putting backspin on a cue ball it must be hit relatively hard. It retains its backspin until it makes contact with the object ball, at which time most of its forward momentum is transferred to the object ball. Left without momentum it grips the surface, and the angular momentum of the backspin is transferred to linear momentum in the direction it came from. The amount of spin, and the strength with which the cue ball are struck control how much reverse speed it will have after hitting the object ball.
Since the collision is not perfectly elastic, without backspin the cue ball would continue with a small forward momentum, so applying only a small backspin at relatively low speed will stop the cue ball, whereas applying a lot of backspin at high speed will cause it to reverse its path and go back the way it came.
You can also add side spin or top spin, or combinations of side spin and top/back spin to produce different effects, including making the cue ball swerve a long way from its initial path without hitting any other balls. This is achieved by hitting down onto the ball with side spin, so that it grips the surface and the side spin alters its path as it is moving.
4th May 2008, 08:52 PM
Friction between the cue ball and felt. If the cue ball rotation is forward when it strikes the object ball head on it will tend to follow along. If it is still rotating backward when at the collision point it will pull back toward the shooter. On a frictionless surface, or if the cue ball has no rotation when it strikes the object ball, it will stop at the collision point.
Ignoring losses due to the elasticity of the balls, all momentum is transferred from cue to object with a direct collision, but very little of the rotation is transferred since since there is little friction between the balls.
4th May 2008, 09:02 PM
I believe a cue ball that's struck without spin can also be made to stop on contact with the object ball, because of the combination of a nearly elastic collision and friction between the table and ball (and a small amount of energy lost to heat, I suppose).
ETA: what BobK said.
4th May 2008, 09:31 PM
I believe a cue ball that's struck without spin can also be made to stop on contact with the object ball, because of the combination of a nearly elastic collision and friction between the table and ball (and a small amount of energy lost to heat, I suppose).Only if you hit it perfectly dead centre, perfectly straight, at relatively low speed.
4th May 2008, 10:04 PM
Slight derail, but I remember many moons ago when I played snooker (not well at all). One science geek friend in our group said "I like snooker because of the physics involved", another, very good player responded "I like snooker because I can beat the crap out of you".
4th May 2008, 10:17 PM
Wicked. You guys rock. Thanks for that - I realised it must be something fairly simple in terms of momentum and friction, but I just needed the confirmation of it. And cousin, you did a great job of explaining it. I just might pinch a line or two to use. ;)
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