In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body.
If the ground is sufficiently rough to ensure rolling, what is the kinetic energy of the body now in the given time interval of `2s`?

Options

A. `18.75J`
B. `16.67J`
C. `5.55J`
D. cannot be found

Answer

B

Explanation

`tau=tau_A/I_A=(F4R-F3R)/(I_0+m(2R)^2)`

`tau=(FR)/(4+2(2x1)^2)=(10xx1)/12=5/6rad//s^2`
After `2s: omega=alphat=5/6xx2=5/3rad//s`
`KE=1/2mv^2+1/2I_0(omega^2)`
`=1/2xx2(omega2R)^2+1/2 4(omega)^2`
`=1/2xx2xx(5/3xx2xx2)^2+1/2xx4(5/3)^2`
`=150/9=16.67J`