Maximum Angle for Non-Null Initial Velocity of Billiard Ball Hit by Cue

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The discussion revolves around determining the maximum angle at which a billiard ball can be struck without resulting in null initial velocity, considering both translational and rotational motion. Participants express confusion over the implications of the impulse applied at an angle, the role of static friction, and the conditions under which the ball may not move. Key points include the need to distinguish between applied and effective impulses, as well as the significance of the coefficient of static friction in relation to the angle of impact. The conversation highlights the complexity of the problem, particularly in understanding how the angle affects both translational and rotational velocities post-impact. Ultimately, the participants seek clarity on the mathematical relationships governing these dynamics to solve the problem effectively.
  • #31
You have four conditions that must be satisfied if "null velocity" means no translational velocity and no rotational velocity:
(1) Net horizontal linear impulse = 0
(2) Net rotationall impulse = 0
(3) Net vertical impulse = 0
(4) ##f \leq \mu K##

The first two already restrict ##\theta## to one value.
Then, imposing the last two conditions restricts the allowed values of ##\mu##.

The fact that the problem statement makes no mention of a restriction on ##\mu## is just another oddity of the wording of the problem.
 
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  • #32
TSny said:
You have four conditions that must be satisfied if "null velocity" means no translational velocity and no rotational velocity:
(1) Net horizontal linear impulse = 0
(2) Net rotationall impulse = 0
(3) Net vertical impulse = 0
(4) ##f \leq \mu K##

The first two already restrict ##\theta## to one value.
Then, imposing the last two conditions restricts the allowed values of ##\mu##.

The fact that the problem statement makes no mention of a restriction on ##\mu## is just another oddity of the wording of the problem.

So, by your logic, I have μ \geq sqrt(2) /(2K)
But I know what (1) -> Jsinθ=K, hence μ \geq 1 /J (2)
All the values of μ satisfying (2) are good, and there's only one possible value for θ.
 
  • #33
anon90 said:
So, by your logic, I have μ \geq sqrt(2) /(2K)

How did you get this? Note that ##\mu## should be dimensionless.
 
  • #34
TSny said:
How did you get this? Note that ##\mu## should be dimensionless.
Right.
f≤μsK=μsJsinθ

Together with θ= 45° and Jcosθ−f=0, I get μ \geq 1
 
  • #35
OK, that's what I got, too.
 
  • #36
TSny said:
OK, that's what I got, too.
So, in the end, the two conditions should be
μ≥1
θ=45°
So the trick with this kind of problems is to figure out which forces can be though as impulsive in that laps of time, and then you just draw the fbd and solve it. Am I forgetting about anything else?
 
  • #37
I think that's essentially it.
 
  • #38
TSny said:
I think that's essentially it.
I see, thank you for your help and merry Christmass.
 
  • #39
Thanks. Merry Christmas to you, too.
 

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