Solving Rolling & Slipping Homework Problem

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SUMMARY

The discussion centers on a physics homework problem involving a 0.18-kg billiard ball with a radius of 2.8 cm, initially moving at 3.9 m/s after being struck by a cue stick. The coefficient of kinetic friction between the ball and the table is 0.64. Key equations include translational kinetic energy, rotational kinetic energy, and the relationship between linear and angular motion. The participant ultimately resolves the problem by recognizing that the velocity at the point of contact becomes zero when the ball transitions from sliding to rolling without slipping.

PREREQUISITES
  • Understanding of translational and rotational kinetic energy
  • Familiarity with Newton's laws of motion
  • Knowledge of frictional forces and coefficients
  • Basic concepts of angular motion and acceleration
NEXT STEPS
  • Study the relationship between linear velocity and angular velocity in rolling motion
  • Learn about the conservation of energy in systems involving friction
  • Explore the equations of motion for rotating bodies
  • Investigate the effects of different coefficients of friction on rolling behavior
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to understand practical applications of rotational dynamics and friction in real-world scenarios.

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Homework Statement


"A 0.18-kg billiard ball whose radius is 2.8 cm is given a sharp blow by a cue stick. The applied force is horizontal and the line of action of the force passes through the center of the ball. The speed of the ball just after the blow is 3.9 m/s and the coefficient of kinetic friction between the ball and the billiard table is 0.64."
How long does the ball slide before it begins to roll without slipping?
How far does it slide?
What is its speed once it begins rolling without slipping?

Homework Equations


translational kinetic=(1/2)mv^2 Rotational kinetic = (1/2)Iω^2
(when rotating without slipping) v=rω a=rα
Force of friction = force normal * coefficient of friction
Iα=r x F
L= r x p = rmvsin θ (i'm not sure if that is useful yet)

The Attempt at a Solution


I know that it's losing kinetic energy to friction
Force of friction*distance to point where it begins rotating
so I know that after that amount of energy loss... v=rω a=rα will hold true
but I don't know what to do with this...

The net force should just be the force of friction right? so a= Ff/m
so vf=vi+at

I'm confused as to what to do with any of this... it seems like I'm missing something...

I have (1/2)mVi^2=Ff*Δx+.7mVf^2 (added rotational and kinetic assuming that it is rotating)
and that's about all I can do I rearranged somethings and got about .7 seconds for the time. . . but I'm pretty sure that is wrong...

direction please?
 
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Hope you don't mind me bumping it (I'll compensate by adding some more ideas I have...I'm still working on it and failing haha)
I was thinking about the point of contact...when it begins to roll without slippage, the velocity at that point is 0 right? but with slipping...the sum of the vectors is NOT 0...it's v translation+r(omega) right? could I just use the angular acceleration I have to fine when r(omega) = v?
 
I solved my own problem =DDDD sorry guys...I'm stupid... maybe next time when it's really hard you guys will be interested enough to help? haha
 

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