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 above question , distance travelled by centre of mass of the wheel before it stops is `-`

Options

A. `(v_(0)^(2))/(2mug)(1+(r^(2))/(k^(2)))`
B. `(v_(0)^(2))/(2mug)`
C. `(v_(0)^(2))/(2mug)(1+(k^(2))/(r^(2)))`
D. None of these

Answer

B

Explanation

Torque of friction about `A` is zero.
`a_(cm)=- mu g`
`0^(2)=v_(0)^(2)-2 mugsrArr S=(v_(0)^(2))/(2mug)`