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A uniform ring of radius R is given a ba...

A uniform ring of radius `R` is given a back spin of angular velocity `V_(0)//2R` and thrown on a horizontal rough surface with velocity of centre to be `V_(0)`. The velocity of the centre of the ring when it starts pure rolling will be

A

`V_(0)/2`

B

`V_(0)/4`

C

`(3V_(0))/4`

D

0

Text Solution

Verified by Experts

The correct Answer is:
B
Here, `u =V_(0), omega_(0) = -V_(0)/(2R)` Let `F_(t)` be the friction forces acting in backward direction.
`a=-(F_(f))/m, alpha = tau/I = (F_(f)R)/(mR^(2))`
At pure rolling: `V=V_(0) -(F_(f))/mt` and `V/R = -V_(0)/(2R) + (F_(t))/(mR) t` (in pure rolling, `V= Romega`)
`rArr V_(0) -V = V + V_(0)/2 , 2V = V_(0)/2 rArr V = V_(0)/4`
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