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A solid cylinder of mass 5.0 kg and radius 0.12 m rolls without slipping on a level floor, with its…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solid cylinder of mass $5.0\ \text{kg}$ and radius $0.12\ \text{m}$ rolls without slipping on a level floor, with its centre moving at $3.0\ \text{m/s}$ . Find its angular velocity, its kinetic energy of moving along and of turning, and its total kinetic energy.

Need a reminder? The key ideaShow

Rolling without slipping: $v = \omega R$ . Kinetic energy: $\tfrac12 Mv^2\left(1 + \dfrac{I}{MR^2}\right)$ . Down a slope: $a = \dfrac{g\sin\theta}{1 + I/MR^2}$ and $v = \sqrt{\dfrac{2gh}{1 + I/MR^2}}$ ; static friction $f = \dfrac{I/MR^2}{1 + I/MR^2}\,mg\sin\theta$ does no work.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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