Finding the Ratio of Initial Speed and Angular Speed for a Struck Snooker Ball

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SUMMARY

The discussion focuses on calculating the ratio of initial linear speed (V0) to angular speed (w0) for a struck snooker ball. The problem states that the cue tip travels horizontally at a distance h above the table, leading to confusion regarding the relationship between h and the ball's radius (r). Participants clarify that the cue's motion should be interpreted as occurring in a vertical plane through the ball's center, implying no side spin is applied. This understanding allows for a clearer approach to solving the problem.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the mechanics of motion, particularly in sports contexts like snooker.

JazzCarrot
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So, we're playing snooker...

Homework Statement



A (uniform) snooker ball of radius r, at rest on a table, is struck by a cue at a point a distance h above the table. Assume the cue tip is traveling horizontally, in a plane through the centre of the ball. As a result, the ball begins to move with an initial linear speed V0 and angular speed w0.

Consider the cue as acting with a large force F for a short time. Ignoring the effects of friction between the ball and the table for this time, find an expression for the ratio V0/w0 in terms of h.

Homework Equations



Once I've sussed the question, I can identify which.

The Attempt at a Solution



Well, I'm confused due to the sentence; "Assume the cue tip is traveling horizontally, in a plane through the centre of the ball". This to me means, that h = r? Therefore, if friction is to be ignored this time, the ball will not (initially) rotate?

If someone could clear this up, I'm sure I can crack on with the next few parts of the question.
 
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JazzCarrot said:
Well, I'm confused due to the sentence; "Assume the cue tip is traveling horizontally, in a plane through the centre of the ball". This to me means, that h = r? Therefore, if friction is to be ignored this time, the ball will not (initially) rotate?
I suspect that it's just a sloppily worded problem, and that the 'in a plane through the center of the ball' should be ignored. (Otherwise your interpretation is correct, but then the problem makes little sense.)

(This isn't from some textbook, I hope. If it is, give a reference.)
 


I would assume it means in a vertical plane through the centre of the ball. In other words, you are not putting any side spin on it.
 


Stonebridge said:
I would assume it means in a vertical plane through the centre of the ball. In other words, you are not putting any side spin on it.
Excellent. That's it.
 


I suspect that it's just a sloppily worded problem

Probably, my Mechanics Lecturer isn't the greatest at writing them.

I would assume it means in a vertical plane through the centre of the ball. In other words, you are not putting any side spin on it.

Awesome, I'll try it like that. Thanks!
 

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