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

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AI Thread Summary
The discussion centers on a physics problem involving a snooker ball struck by a cue, where the goal is to find the ratio of initial linear speed (V0) to angular speed (w0) in terms of height (h). Participants express confusion over the phrasing of the problem, particularly the statement about the cue tip traveling horizontally through the ball's center, leading to interpretations that h may equal the radius (r). Clarifications suggest that the problem likely assumes no side spin is applied, indicating a vertical plane of impact. The consensus is that the problem's wording may be unclear, but the key point is to proceed with the assumption of no initial rotation. This clarity allows for further progress in solving the problem.
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!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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