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

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SUMMARY

The forum discussion centers on determining the maximum angle at which a billiard ball, struck by a cue, maintains a non-null initial velocity. Key equations discussed include the impulse-momentum relationship (J = Δp) and the static friction condition (|F| ≥ μN). Participants express confusion regarding the implications of the angle θ on both translational and rotational velocities of the ball, particularly in relation to the coefficient of static friction (μ) and the moment of inertia (I). The consensus is that the problem's wording may lead to misunderstandings about the conditions for achieving zero translational velocity.

PREREQUISITES
  • Understanding of impulse-momentum theory (J = Δp)
  • Knowledge of static friction and its coefficient (μ)
  • Familiarity with rotational dynamics and moment of inertia (I)
  • Basic principles of angular momentum and torque
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  • Study the relationship between impulse and angular momentum in rotational dynamics.
  • Learn about the conditions for static versus kinetic friction in motion analysis.
  • Explore the implications of angle θ on translational and rotational motion in physics.
  • Investigate torque equations and their applications in determining motion outcomes.
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  • #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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