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Rolling without slipping 
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#1
May707, 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? 


#2
May707, 08:12 PM

Mentor
P: 41,306

Hints: Does gravity exert a torque on the sphere? What other force acts on the sphere that exerts the torque?



#3
May707, 08:14 PM

P: 41

friction. but how can I calculate friction without mass or coefficient of friction?



#4
May707, 08:19 PM

Mentor
P: 41,306

Rolling without slipping
Apply Newton's law for translation and rotation. And the condition for rolling without slipping. 


#5
May707, 08:43 PM

P: 41

omg duh. i forgot about the forces in the xdirection. i always do that. thanks.



#6
May707, 09:33 PM

P: 38

On the righthand side, it should be I=2/5mr^2+mr^2=7/5 mr^2 (axisparallel theorem, or Steiner's theorem) 


#7
May807, 08:10 AM

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P: 41,306

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 approachanalyzing translation and rotatation separatelyto be more instructive. But it's all good! )



#8
May807, 08:30 PM

P: 38

Yes Doc Al, that's what I meant.



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