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Rotational And Translational Motion Of A Bowling Ball

  • Thread starter mrshappy0
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  • #1
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Homework Statement


A spherical bowling ball with mass m = 3.4 kg and radius R = 0.115 m is thrown down the lane with an initial speed of v = 8.5 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.27. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.3)How long does it take the bowling ball to begin rolling without slipping?



Homework Equations


Torque = I (angular acceleration)
f=ma

I also drew the body diagram. I believe it would be the same when it is sliding and when it is not sliding.


The Attempt at a Solution


I found the angular and linear acceleration when it is sliding but I am not sure how to figure out how long it slides for. I am thinking this has little to do with rotational dynamics and more to do with translational motion.
 

Answers and Replies

  • #2
Doc Al
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I found the angular and linear acceleration when it is sliding but I am not sure how to figure out how long it slides for. I am thinking this has little to do with rotational dynamics and more to do with translational motion.
You'll need both rotational and translational dynamics here. Hint: What's the condition for rolling without slipping?
 
  • #3
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I am not sure would this condition would be.
 
Last edited:
  • #4
Doc Al
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I am not sure would this condition would be.
You need to figure that out--it's the key to this problem. What must be the relationship between the translational and rotational speeds for there to be rolling without slipping?

(I suspect that your textbook discusses this condition.)
 
  • #5
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I am thinking this relationship would be velocity = r (w)
 
  • #6
Doc Al
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I am thinking this relationship would be velocity = r (w)
Exactly!

Now write expressions for the translational speed as a function of time and the rotational speed as a function of time. Then you can solve for when they meet that condition.
 
  • #7
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Okay, so I used (wf-wi)/t=ang.acc. and (vf-vi)/t=acc. I then solved for wf and vf and plugged them into v=r(w). I solved for t and realized there is no relation between wi and vi because that is when the ball is sliding.
 
  • #8
Doc Al
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Okay, so I used (wf-wi)/t=ang.acc. and (vf-vi)/t=acc. I then solved for wf and vf and plugged them into v=r(w).
Good.
I solved for t and realized there is no relation between wi and vi because that is when the ball is sliding.
The time t will be the time when rolling without slipping begins, which is what you need to find.
 
  • #9
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So the only way to solve this requires assuming that the initial rotational velocity is zero?
 
  • #10
Doc Al
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So the only way to solve this requires assuming that the initial rotational velocity is zero?
Yes, good point. Usually that's explicitly stated.
 
  • #11
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Hmm, I have vi/(-a+(ang.a)r)= t and it is still wrong.
 
  • #12
Doc Al
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Hmm, I have vi/(-a+(ang.a)r)= t and it is still wrong.
Express your answer in terms of the given quantities: m, μ, R, and vi.

What did you get for the accelerations (in terms of those quantities)?

Show how you solved for the time.
 
  • #13
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Okay, to shortcut. The only thing that was wrong was the negative sign in the above expression. so vf=vi-at. I don't know why it would be the other way around. Oh, is this because the translational acceleration vector points in the opposite direction?
 
  • #14
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Okay, last bothersome question. would the magnitude of the final velocity just be be vi-at?
 
  • #15
Doc Al
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Okay, to shortcut. The only thing that was wrong was the negative sign in the above expression. so vf=vi-at. I don't know why it would be the other way around. Oh, is this because the translational acceleration vector points in the opposite direction?
Right. The translational speed is decreasing while the rotational speed increases.
 
  • #16
Doc Al
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Okay, last bothersome question. would the magnitude of the final velocity just be be vi-at?
Yep.
 

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