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A thin uniform circular ring is rolling ...

A thin uniform circular ring is rolling down an inclined plane of inclination `30^(@)` without slipping. Its linear acceleration along the inclined plane will be

A

g

B

`(g)/(2)`

C

`(g)/(3)`

D

`(g)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
In case of pure rolling,
Acceleration, `a=(g sin theta)/(1+(I)/(MR^(2))), "for ring" (I)/(MR^(2))=1`
`:. a =(g)/(4)" "[because theta=30]`.
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