Ring and solid sphere rolling down an incline - rotation problem

  • #1
Krushnaraj Pandya
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Homework Statement


A solid sphere, hollow sphere, disk and ring are released simultaneously from top of a incline. Friction is sufficient to prevent slipping of hollow sphere- what will reach the bottom first?

Homework Equations


a in pure rolling down an incline=gsinθ/(1 + I/mR^2)

The Attempt at a Solution


So first I determined which objects are pure-rolling and which are slipping. Lets say I/mR^2=n which is just the coefficient with the mr^2 term for convenience. Now from the above formula acceleration for a body in pure rolling is given by gsinθ/(1+n). Now, mgsinθ-frictionforce=mgsinθ/(1+n) therefore frictional force=nmgsinθ/(1+n). Using frictional force<=μN and N=mgcosθ we get μ>=ntanθ/1+n. So minimum coefficient of friction required for hollow sphere to roll is 2tanθ/5, plugging n for the other bodies we see that all the bodies except the ring will be pure rolling. Now when all 4 are pure rolling, solid sphere has highest value of acceleration so reaches first, but now ring is slipping so the competition is between the solid sphere and the ring.
Using s=(1/2)at^2 (s is same for both bodies) t for solid sphere is sqrt(14s/5gsinθ). I'm getting very confused finding the t for the ring though, any help would be appreciated-thank you
Edit: So, the acceleration for the ring is g(sinθ-μcosθ), plugging the minimum value of μ we found for hollow sphere, we get t=sqrt(10s/3gsinθ) which indicates the solid sphere will reach first. Is that correct?
 

Answers and Replies

  • #2
haruspex
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minimum coefficient of friction required for hollow sphere to roll is 2tanθ/5
Is that right? What value of n are you plugging in?
Friction is sufficient to prevent slipping of hollow sphere-
Should this say "just sufficient"?
 
  • #3
Krushnaraj Pandya
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Is that right? What value of n are you plugging in?
n is 2/3 for a hollow sphere. (2/3)/(1+(2/3)) gives 2/5 so 2/5tanθ.
Should this say "just sufficient"?
It doesn't say that. But knowing the way around this book, I can safely say that was most probably the meaning. Otherwise there'd be two cases- one where all are rolling and one where only the ring isn't; in any case the solid sphere seems to win both races
 
  • #6
haruspex
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So, the acceleration for the ring is g(sinθ-μcosθ), plugging the minimum value of μ we found for hollow sphere, we get t=sqrt(10s/3gsinθ) which indicates the solid sphere will reach first. Is that correct?
Yes, this all looks right, but I think you can make it a bit simpler by comparing accelerations instead of times: (5/7)g sin(θ) for the solid sphere, (3/5)g sin(θ) for the ring.
 
  • #7
Krushnaraj Pandya
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Yes, this all looks right, but I think you can make it a bit simpler by comparing accelerations instead of times: (5/7)g sin(θ) for the solid sphere, (3/5)g sin(θ) for the ring.
Alright thank you
 
  • #8
haruspex
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Should this say "just sufficient"?
On reflection, it does not need to. Increasing it will slow the ring further but not affect the others.
 
  • #9
Krushnaraj Pandya
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On reflection, it does not need to. Increasing it will slow the ring further but not affect the others.
ah! right.
 

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