A uniform circular disc of radius `r` is placed on a rough horizontal surface and given a linear velocity `v_(0)` and angular velocity `omega_(0)` as shown. The disc comes to rest after moving some distance to the right. It follows that

Options

A. `3v_(0)=2omega_(0)r`
B. `2v_(0)=omega_(0)r`
C. `v_(0)=omega_(0)r`
D. `2v_(0)=3omega_(0)r`

Answer

B

Explanation

Since the disc comes rest, it stops rotation and translating simultaneously `v= 0` and `omega=0`. That means the agular momentum about the instantaneous point to contact just after the time of stopping is zero. We know that the agular momentum of the disc about `P` remains constant because frictional force `F, N` and `mg` pass through point `P` and thus produce no torque about this point
`L_("initial")=L_("final")impliesmvr-l_(0)omega_(0)=0`
`implies mvr=1/2mr^(2)omega_(0)implies2v_(0)=omega_(0)r`