Sorry here is another question that I am not sure of(adsbygoogle = window.adsbygoogle || []).push({});

1. A cylindrical hoop rests on a rough uniform incline. It is released and rolls

without slipping through a vertical distance h0. It then continues up a perfectly

smooth incline. What height does it reach?

Here is my solution

Because of the conservasion of energy

mgh0=1/2mv^2+1/2IOmega^2

since Omega=v/r and the moment of inertia of a cylindrical hoop is I=mr^2

therefore

mgh0=1/2mv^2+1/2mr^2*(v/r)^2

=mv^2

therefore final velocity is v=Square root(gh0)

when goin up the perfect smooth surface it looses the ability to roll, so the only energy is the translaional energy

so let the maximum height it reached be h

mgh=1/2mv^2=1/2mgh0

so h=1/20

Is this right? Thanks

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# Homework Help: Another Solution check Please. (cylinder rolling down a slope)

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