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

  • Thread starter Thread starter JazzCarrot
  • Start date Start date
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
Messages
5
Reaction score
0
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.
 
Physics news on Phys.org


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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top