Rotational Kinematics- sphere

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  • #1
mattmannmf
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A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 4.2 m down a q = 26° incline. The sphere has a mass M = 5.1 kg and a radius R = 0.28 m.

a) Of the total kinetic energy of the sphere, what fraction is translational?


b) Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

I figured out that KEtran is 65.8 j. I figured out that the phere reaches the bottom of the ramp at a speed of 5.08. the magnitude of friction forceon the sphere is 6.25 N. I just don't know what to do for a or b...any help?
 

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  • #2
Doc Al
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I just don't know what to do for a or b...any help?
If there's no friction, what fraction of the total KE is translational?
 
  • #3
mattmannmf
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all?
 
  • #4
Doc Al
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all?
Right. Without the torque due to friction, it will slide down the incline without rolling.
 
  • #5
mattmannmf
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ok so that would be for b. I am still kind of stuck on a.
 
  • #6
mattmannmf
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im confused on how to find the total kinetic energy
 
  • #7
Doc Al
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im confused on how to find the total kinetic energy
Assuming that you posted the problem completely, for a you don't need to find the actual total energy, just the fraction that's translational. (Don't plug in any specific numbers.)

Express the translational and rotational KE and see how they relate for a sphere rolling without slipping. Hint: How does ω relate to translational speed?
 

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