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

Options

A. `0.25m`
B. `0.2m`
C. `0.5m`
D. `0.4m`

Answer

D

Explanation

Total energy of rolling body
`E=(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))" "["where "V_(cm)=v]`
`therefore 32.8xx2=10xx(2)^(2) [1+(k^(2))/((0.5)^(2))]`
`therefore (65.6)/(10xx4)=1+(k^(2))/(0.25^(2))`
`therefore 1.64=1+(k^(2))/(0.25)`
`therefore 0.64=(k^(2))/(0.25)`
`therefore k^(2)=0.64xx0.25`
`therefore k=0.8xx0.5`
`therefore k=0.4m`