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A circular disc rolls down on an inclined plane without slipping. What fraction of its total energy to translational? |
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Answer» `2/3` Rotational kinetic energy, `K_(R)=1/2Iomega^(2)` Total kinetic energy, K`=K_(T)+K_(R)` `=1/2Mv^(2)+1/2Iomega^(2)` `=1/2Mv^(2)+1/2(1/2MR^(2))(V^(2))/(R^(2))` (`becausev=Romega)` `=1/2Mv^(2)[1+1/2]` (`becauseI_("circular disc")=1/2MR^(2))` `THEREFORE(K_(T))/K=(1/2Mv^(2))/(1/2Mv^(2)[1+1/2])=1/([1+1/2])=2/3` |
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