Rolling Motion

  • Thread starter chrisfnet
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  • #1
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Homework Statement



http://img688.imageshack.us/img688/5277/loopb.jpg [Broken]

A cylindrical shell is released from rest and rolls down an incline without slipping. Find speed of shell at the top of the loop.

h(incline) = 2.0m
h(loop) = 1.3m

The Attempt at a Solution



vf = sqrt(g * yi)
vf = sqrt(9.8m/s^2 * 2.0m)
vf = 4.42719m/s = vi(loop)

KEi = PEf + KEf
1/2mvi^2 = mgy + 1/2mvf^2 + 1/2mr^2(v^2/r^2)
1/2vi^2 = gy + 1/2vf^2 + 1/2vf^2
1/2(4.42719m/s)^2 = -9.8m/s^2(1.3m) + 1/2vf^2 + 1/2vf^2
vf = 1.71464m/s


Would this be correct?
 
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Answers and Replies

  • #2
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No. First, it would be [tex]v_{f}=\sqrt{2gh}[/tex] but you can't use that equation anyway because in addition to translational kinetic energy, there is rotational kinetic energy.

Start off by looking up or deriving the moment of inertia of the cylinder for the rotation. Use your conservation of energy equation with change in potential energy on one side and change in rotational and translational kinetic energy on the other. Because the cylinder is rolling without slipping, you can find the angular velocity in terms of the linear velocity. Now you can solve for linear velocity.
 
  • #3
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Actually, sqrt(2gh) would be right if it were a block going down the incline. However, it's not. Since it's a cylindrical shell, the equation sqrt(gh) is correct.

I'd use:

mgh = 1/2mvf^2 + 1/2Iwf^2

Which simplifies to vf = sqrt(gh) after substitution.

This would give me the velocity at the bottom of the incline.
 

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