Rotational Momentum

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What about the following situation:
2 spheres with the same mass and radius are released on an incline, one of the spheres is hallow and the other one is solid. Which sphere would reach first to the bottom?
Which sphere would travel larger distance (after the incline) before it stops?
 

Doc Al

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What about the following situation:
2 spheres with the same mass and radius are released on an incline, one of the spheres is hallow and the other one is solid. Which sphere would reach first to the bottom?
The two spheres have different rotational inertias, of course. You can see which will roll down the incline faster by extending the reasoning I gave in post #12.
Which sphere would travel larger distance (after the incline) before it stops?
What's stopping them?
 
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For each case, the total kinetic energy of each ball, assuming equal mass and uniform density, and assuming rolling without sliding, is

E = (1/2)mv2 + (2/5)mv2 = (7/10)mv2 = mgh

The first term on the left is due to translation of the center of mass, and the second to the rotational kinetic energy. here I have used wR = v, where w is the angular velocity, moment of inertia I = (2/5)mR2, and R is the raduis.

In the above equation, m cancels out, (as does R), so the velocity of each ball is

v=sqrt(10 gh/7)

This is independed of ball radius and mass. Note that if g has units meters per sec^2 and h has units meters, then v has units meters per sec, so both balla have the same velocity, depending only on g and h.
 
thank you bob,
same velocity, same energy
this makes sense to me, such as it ought too.
 

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