Inclined Plane Physics: Cylinder vs. Superconductor

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cragar
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Lets say I have an inclined plane and I roll a solid cylinder down it and slide a brick down it.
And let's say I slide the brick down on oil,Or better yet I slide a superconductor down and have It float above bar magnets. The superconductor moving down the plane will always beat the cylinder. We let them go from rest. Is this because some of the Gravitational potential energy goes into rotating the cylinder and not just sliding it down the ramp.
 
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cragar said:
Is this because some of the Gravitational potential energy goes into rotating the cylinder and not just sliding it down the ramp.
Exactly. Whenever something rolls down an incline, some of the potential energy must be used for rotational kinetic energy instead of translational.
 
ok, thanks for your response. And it is easier to rotate something when the mass is distributed near the center of mass.
 
Right. You won't be able to roll anything without some friction. And if there's friction, something sliding will be slowed as well.
 
If rotational kinetic energy is the lesson to be communicated, I think a good example is the fact that a solid cylinder will reach the bottom of the incline before a hollow cylinder that has the same mass. The hollow cylinder has more of its mass distributed a greater distance from the center, therefore a greater moment of inertia, therefore greater final rotational KE, therefore less final translational KE.
 
sophiecentaur said:
I could be difficult and say that you may not be making a 'fair' comparison here. Under conditions of no friction, the cylinder wouldn't actually start rolling, would it? :biggrin:
Would the cylinder just slide then.
As long as the friction was low enough but not zero the sliding object would win.
Or maybe we should say that, there would be a point where if the friction was low enough the sliding object would beat the rolling object.
 
With no friction there would be no difference. With high friction , only the cylinder would get there. There would, presumably be a value of coefficient for which they will reach the bottom at the same time.
 
ok that makes sense. thanks for your answer.