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 smooth, what is the total kinetic energy of the body after `2s`?A. `25J`B. `16.67J`C. `50J`D. `37.5J`

In the given figure `F=10N, R=1m`, mass of the body is `2kg` and momen

1.	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 smooth, what is the total kinetic energy of the body after `2s`?A. `25J`B. `16.67J`C. `50J`D. `37.5J`
Answer» Correct Answer - C 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`

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