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A uniform spherical sheel rolls down a fixed inclined plane without sliping. Find the ratio of rotational kinetic energy to translational kinetic energy as it reaches lowest point of the incline. |
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Answer» Solution :As there is no slip, `v = R omega` is valid and rotational kinetic energy `K_(R )=(1)/(2)I_(cm)omega^(2)=(1)/(2)mk^(2)omega^(2)` and translational kinetic energy `K_(T)=(1)/(2)MV^(2)=(1)/(2)MR^(2)omega^(2)` therefore `(K_(R ))/(K_(T))=(K^(2))/(R^(2))=(2)/(3)` |
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