In the given figure F=10N , R=1 m , mass of the body is 2 kg and moment of inertia of the body about an axis passing through O and perpendicular to plane of body is `4 kg-m^(2)` .O is the centre of mass of the body.

If ground is sufficiently rough , what is kinetic energy of the body now in the given time interval ?

Options

A. `18.75J`
B. `25.67J`
C. `16.67J`
D. none of these

Answer

C

Explanation

When frictional force is absent, the rigid body does not translate `SigmavecF_("external")=0` therefore `veca_(CM)=0`
Taking torque about `CM`
`Fxx2R-FxxR=I_(CM)alpha`
`10xx1=4alpha`
`alpha=5/2rad//s^(2)`
`omega=alphat=5rad//s`
`KE_("total")=1/2I_(CM)omega^(2)=1/2xx4xx5xx5=50J`
Taking torque about `IC` (bottom most point)
`Fxx4R-Fxx3R=[I_(CM)+M(2R)^(2)]alpha`
`=10xx1=[4+2xx(2xx1)^(2)]alpha`
`alpha=10/2rad//s^(2)`
`omega=alphat=10/12xx2=5/2rad//s`
`KE_("total")=1/2[I_CM+M(2R)^(2)]omega^(2)`
`=1/2[4+2xx(2xx2)^(2)]xx(5/3)^(2)`
`=16.67J`