A spherical object with moment of inertia 0.46mr 2 rolls without slipping down a

Question

A spherical object with moment of inertia 0.46mr2rolls without slipping down an incline. At the bottom of the incline, what fraction of its total kinetic energy is rotational kinetic energy? (Answer in percent)

If you could please show me the steps to this I would appreciate it! Thank you! A spherical object with moment of inertia 0.46mr2rolls without slipping down an incline. At the bottom of the incline, what fraction of its total kinetic energy is rotational kinetic energy? (Answer in percent)

If you could please show me the steps to this I would appreciate it! Thank you!

Explanation / Answer

rotational kinetic energy=0.5*I*w^2=0.5*0.46*mv^2

Translational kinetic energy=0.5*mv^2

Totalkinetic energy=0.5*1.46*mv^2

fraction of rotational kinetic energy=0.46/1.46*100=31.5 %

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