A wheel is released on a rough horizontal floor after imparting it an initial horizontal velocity `v_(0)` and angular velocity` omega_(0)` as shown in the figure below. Point `O` is the centre of mass of the wheel and point `P` is its instantaneous point of contact with the groundl. The radius of wheel is `r` and its radius of gyration about `O` is `k`. Coefficient of friction between wheel and ground is `mu`. A is a fixed point on the ground.

If the wheel comes to permanent rest after sometimes, then `:`

Options

A. `v_(0)=omega_(0)r`
B. `v_(0)=(omega_(0)k^(2))/(r)`
C. `v_(0)=(omega_(0)r^(2))/(R)`
D. `V_(0)=omega_(0)(r+(k^(2))/(r))`

Answer

B

Explanation

Torque of friction about `A` is zero.
Anglar momentum conservation about point `A`.
`L_(i n)=mv_(0)r-mk^(2)omega`
` L_(f i n)=0`
`L_(f i n)=L_(i n )`
`rArr v_(0)=omegak^(2)//r`