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A spherical shell and thin cylindrical shell with the same mass and radius roll at the...

A spherical shell and thin cylindrical shell with the same mass and radius roll at the same velocity side by side on a level surface without slipping. If the two objects subsequently approach and roll up an inclined plane without slipping, how much higher will the cylindrical shell go before they both stop?

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Answer #1

By conservation of energy

At the bottom of the ramp, both objects will have translational and rotational KE

At the height, they will both have PE

Ke rotational = .5Iw2

I for the cylindrical shell = MR2 and w = v/r, so...

KE rotational = .5(mr2)v2/r2 which simplifies to .5mv2

Thus for the cylindrical shell by conservation of energy...

mgh = .5mv2 + .5mv2 (mass cancels)

gh = v2

For the spherical shell, I = (2/3)(mr2), so

KE = .5(2/3)(mr2)v2/r2) = (1/3)(mv2)

So for the conservation of energy here, we have...

mgh = (1/3)(mv2) + .5mv2 (mass cancels)

gh' = 5/6(v2)

Set the ratio...

gh/gh' = v2/(5/6)v2

h/h' = 6/5

Thus the cylinderical shell will go 6/5 ths higher (that is 1.2 times higher)

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