Rolling without slipping


by ph123
Tags: rolling, slipping
ph123
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#1
May7-07, 08:08 PM
P: 41
A solid sphere rolls without slipping down a ramp that is at an angle of 32 above horizontal. The magnitude of the acceleration of the center of mass of the sphere as it rolls down the ramp is?


sum of torques = I(alpha)

rmgsin32 = (2/5)mr^2(a_tan/r)

the radii drop out as one would expect with no radius given. the masses also drop out since they weren't provided.

gsin32 = (2/5)a_tan

(9.8 m/s^2)sin32 = (2/5)a_tan

a = 12.98 m/s^2

This result clearly makes no sense because it is greater than the accelertion due to gravity. But that was the only approach I could think to use since I was only given the angle of the incline. Any ideas?
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Doc Al
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#2
May7-07, 08:12 PM
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Hints: Does gravity exert a torque on the sphere? What other force acts on the sphere that exerts the torque?
ph123
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#3
May7-07, 08:14 PM
P: 41
friction. but how can I calculate friction without mass or coefficient of friction?

Doc Al
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May7-07, 08:19 PM
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Rolling without slipping


Quote Quote by ph123 View Post
friction. but how can I calculate friction without mass or coefficient of friction?
Just try it--maybe you won't need that information.

Apply Newton's law for translation and rotation. And the condition for rolling without slipping.
ph123
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#5
May7-07, 08:43 PM
P: 41
omg duh. i forgot about the forces in the x-direction. i always do that. thanks.
Weimin
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#6
May7-07, 09:33 PM
P: 38
Quote Quote by ph123 View Post
sum of torques = I(alpha)

rmgsin32 = (2/5)mr^2(a_tan/r)
The second line is not correct. On the left hand side, you wrote the torque around the center which is the point where the sphere touches the slope, so you need to use inertial momentum I corresponding to the axis through above point.

On the right-hand side, it should be

I=2/5mr^2+mr^2=7/5 mr^2 (axis-parallel theorem, or Steiner's theorem)
Doc Al
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#7
May8-07, 08:10 AM
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That's perfectly OK as well: Since the sphere rolls without slipping, you can view it as being in pure instantaneous rotation about the point of contact. With this approach, you need to use torques and rotational inertia about the point of contact, not about the center. Note that gravity does exert a torque about the point of contact. (I find the two step approach--analyzing translation and rotatation separately--to be more instructive. But it's all good! )
Weimin
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#8
May8-07, 08:30 PM
P: 38
Yes Doc Al, that's what I meant.


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