Inertia of a hollow cylinder and hollow sphere

  • Thread starter GayYoda
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Homework Statement


Consider a hollow cylinder of mass M with an outer radius R_out = 10 cm and an unknown inner radius R_in. If the hollow cylinder is to roll down an incline in the same time as a spherical shell of the same mass and the same outer radius, calculate R_in.

Homework Equations


I_cyl = MR^2/2 .
I_shell = 3/5(MR^2)

The Attempt at a Solution


I think the inertia of each are equal but i'm not sure
 

Answers and Replies

  • #2
PeroK
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You've given the moment of inertia of a solid cylinder. What you need is the inertia of a hollow cylinder of finite thickness.
 
  • #3
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I=M/2[(R_out)^2-(R_in)^2]?
 
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PeroK
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I=M/2[(R_out)^2-(R_in)^2]?
What happens in your formula if ##R_{in}## is close to ##R_{out}##?

To calculate this inertia is not that easy. I suggest you consider the density of the cylinder.
 
  • #5
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i got it now its I=M[(R_out)^2+(R_in)^2]/2 as the density becomes M/[pi*h*[(R_out)^2-(R_in)^2] and when you sub it back in to the intergral it becomes [(R_out)^4-(R_in)^4]/[(R_out)^2-(R_in)^2] = [(R_out)^2+(R_in)^2] because of difference of 2 squares
 

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