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The total kinetic energy of a body of ma...

The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with a velocity of 2 m/s without slipping is 32.8 J. The radius of gyration of the body is

A

0.25 m

B

0.2 m

C

0.5 m

D

0.4m

Text Solution

Verified by Experts

The correct Answer is:
D
KE of rolling body = `1/2 mv^2 + 1/2 I omega^2 , E = 1/2 mv^2 + 1/2 mk^2(v/R)^2`
`E = 1/2 mv^2 (1 + (k^2)/(R^2)), (1 + (k^2)/(R^2)) = (2E)/(mv^2)`
`(k^2)/(R^2) = (2E)/(mv^2) - 1 , k^2 = R^2 ((2E)/(mv^2) - 1) = (0.5)^2[(2 xx 32.8)/(10 xx (2)^2) - 1]`
`= (0.5)^2 [(65.6 - 40)/(40)] = 0.5^2 xx (25.6)/(40) " " therefore k = 0.4 m`
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