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A uniform spherical shell rolls down a fixed inclined pla ne without sliping. Find the ratio of rotational kinetic energy to translational kinetic energy as is reaches lowest point of the incline. |
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Answer» Solution :As there is no SLIP `v=Romega` 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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