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A ball rolls without slipping. The radiu...

A ball rolls without slipping. The radius of gyration of the ball about about an axis passing through its center of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy be

A

`(K^(2)+R^(2))/(R^(2))`

B

`(K^(2))/(K^(2)+R^(2))`

C

`(K^(2))/(R^(2))`

D

`(R^(2))/(K^(2)+R^(2))`

Text Solution

Verified by Experts

The correct Answer is:
b
`E_("rot") = (1)/(2) I omega^(2) = (1)/(2)MK^(2) ((V)/(R))^(2)`
`E_("rot") = E_("tran") + E_("rot")`
`= (1)/(2) mv^(2) + (1)/(2) MK^(2) ((V)/(R))^(2) = (1)/(2) MV^(2) (1 + (K^(2))/(R^(2)))`
`(E_("rot"))/(E_("tot")) = ((1)/(2)mv^(2) ((K^(2))/(R^(2))))/((1)/(2)mv^(2) (1 + (K^(2))/(R^(2)))) = (K^(2))/(K^(2) + R^(2))`
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Knowledge Check

  • Moment of inertia of a cube about an axis passing through the center of mass and perpendicula to face is

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