A uniform circular disc placed on a rough horizontal surface has initially a velocity `v_(0)` and an angular velocity `omega_(0)` as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then `(v_(0))/(r omega_(0))` is.
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Options

A. `1//2`
B. 1
C. `3//2`
D. 2

Answer

A

Explanation

for translational motion `v = v_(0) + at`
`0 = v_(0) - (f)/(m) t rArr t = (mv_(0))/(f)`….(1)
for rotational motion
`omega = omega_(0) + alpha t`
`omega = omega_(0) + (tau)/(I) t`
`0 = omega_(0) - (Rf)/(I) t`
`t = (omega_(0) I)/(Rf)`
for disc `I = (1)/(2) MR^(2)`
`t = (1)/(2) (mR^(2). omega_(0))/(Rf) rArr t = (mR omega_(0))/(2 f)` ....(2)
from (1) & (2)
`(mv_(0))/(f) = (mR omega_(0))/(2 f) rArr (v_(0))/(R omega_(0)) = (1)/(2)`.
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