Rolling Motion

  • Thread starter BlackMamba
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  • #1
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Well I'm back. LOL Yet another question regarding my homework problem.

Here's the problem: A ball of radius 0.370m rolls along a horizontal table top with a constant linear speed of 3.20m/s. The ball rolls off the edge and falls a vertical distance of 2.70m before hitting the floor. What is the angular displacement of the ball while it is in the air? (Assume there is no slipping of the surfaces in contact during the rolling motion.)

So I assumed it would be helpful to determine the time the ball is in the air which I did. Calculating t = 0.742s.

Since the linear speed is constant, both omega initial and omega final are equal to 3.20m/s. Am I correct in thinking this?

This is where I've reached a halt. I'm unsure of what to do next. Any help to point me in the right direction would be greatly appreciated.

Thanks in advance for any help provided. :smile:
 

Answers and Replies

  • #2
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Nobody can help me? :(
 
  • #3
Pyrrhus
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Well wouldn't the ball go in to projectile motion when it falls from the table? think about it.

Also

[tex] v = r \omega [/tex]
 
  • #4
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I forgot to metion that I already have the figure for v = rw. Giving me w = 8.65. Now would that number be rad/s?

So the w I just found: That becomes the new omega initial for the time in the air correct?

Wouldn't I also need to find alpha as well? If so would it be correct to use the equation alpha = a/r. a = -9.80m/s^2
 
Last edited:
  • #5
Doc Al
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BlackMamba said:
I forgot to metion that I already have the figure for v = rw. Giving me w = 8.65. Now would that number be rad/s?
Yes, omega has units of rad/s.

So the w I just found: That becomes the new omega initial for the time in the air correct?
Yes.

Wouldn't I also need to find alpha as well? If so would it be correct to use the equation alpha = a/r. a = -9.80m/s^2
No. What makes you think the rate of spinning changes when it's in the air? Ask yourself: What is required to create an angular acceleration?
 
  • #6
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Yeah I don't know why I asked that last question. I think I was just grasping at straws there. But I've reached a final answer of 3.21 rad. I would just submit it into webassign, but my submissions are limited at this point and I still have one more problem to go.

To get that answer of 3.21 rad, I used the equation:

Theta = 1/2(Omega initial + Omega final)t

Omega final being 0

the equation thus becomes 1/2(Wo)t

Did I do that correctly?
 
  • #7
Doc Al
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BlackMamba said:
To get that answer of 3.21 rad, I used the equation:

Theta = 1/2(Omega initial + Omega final)t

Omega final being 0
Why is omega final = 0?

the equation thus becomes 1/2(Wo)t

Did I do that correctly?
No. You are still thinking that the spinning rate is affected by the fall. Why would it be?
 
  • #8
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What else would omega final be? When it hits the floor doesn't it supposedly stop?

I am completely lost then where to go next with this problem.

I have t and Omega initial. Apparently now I have to find an Omega final....and I still don't have the slightest clue how to go about finding the angular displacement.
 
  • #9
Doc Al
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BlackMamba said:
What else would omega final be? When it hits the floor doesn't it supposedly stop?
I suppose it does stop after it hits the floor and stops rolling around. :smile: But that's no relevant. (By that logic, the final velocity of every falling body is zero!) But while it falls, does omega change?
I have t and Omega initial. Apparently now I have to find an Omega final....and I still don't have the slightest clue how to go about finding the angular displacement.
This problem is MUCH easier than you think. I'm still waiting for an answer to my question: Does the rotational speed of the ball change as it falls? (Not after it hits the ground!)
 
  • #10
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Well, I would think that that the rotational speed does not change as it falls. But at the rate I'm going my answer for that is also probably wrong.
 
  • #11
Doc Al
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constant omega as it falls

In order for the rotational speed to change, there would have to be a torque on the ball. Which there isn't. So omega stays constant as it falls. Now just calculate the angle the balls rotates as it falls.
 
  • #12
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Which is what i thought I did earlier with my equation:

Theta = 1/2 (Omega inital + Omega final) time

Should I not be using a rotational kinematic equation for this?

Or should I be doing something like:

theta = cos^-1 (Omega of the ball on the table/Omega of the ball in the air) ?
 
  • #13
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Ok I get why I shouldn't be using that kinematic equation. That would say that the rotational speed of the ball changes which we have clearly have discussed that it doesn't.
 
  • #14
Doc Al
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BlackMamba said:
Ok I get why I shouldn't be using that kinematic equation. That would say that the rotational speed of the ball changes which we have clearly have discussed that it doesn't.
Well, you could use that kinematic equation---just set omega final = omega initial. Or just use: [itex]\theta = \omega t[/itex] .
 
  • #15
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Are you kidding, that's it? God! LOL OK so I've got an answer of 6.42 rad. Does that sound about right?

Btw, thank you for all your help. It really is greatly appreciated. :smile:
 
  • #16
Doc Al
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BlackMamba said:
Are you kidding, that's it? God! LOL OK so I've got an answer of 6.42 rad. Does that sound about right?
Assuming your values for t and omega are correct, you got it. (It won't hurt you to check them over.)

Btw, thank you for all your help. It really is greatly appreciated. :smile:
My pleasure.
 
  • #17
187
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Your absolutely correct. I will do a double check.

Thanks again. :smile:
 

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