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Question

A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is


A
25
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B
27
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C
35
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D
37
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Solution

The correct option is B $$\cfrac{2}{7}$$

$${KE}_{Total}$$ = $$\cfrac{1}{2} IW^{2}+\cfrac{1}{2} MV^{2}$$
$$=\cfrac{1}{2}\times\cfrac{2}{5}MR^{2}  W^{2}+\cfrac{1}{2}MV^{2}$$
$$=\cfrac{1}{5}MR^{2}W^{2}+\cfrac{1}{2}MV^{2}$$    =$$\cfrac{1}{5}MV^{2}+\cfrac{1}{2}MV^{2}$$
$$=\cfrac{7}{10}MV^{2}$$

$${ KE }_{ rot }=\cfrac { 1 }{ 2 } IW^{ 2 }=\cfrac { 1 }{ 5 } M{ V }^{ 2 }$$

$$\cfrac { {KE}_{rot} } { {KE}_{Total} } = \cfrac { \cfrac { 1 }{ 5 } MV^{ 2 } }{ \cfrac { 7 }{ 10 } MV^{ 2 } } $$
$$=\cfrac{2}{7}$$.

Physics

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