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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 is :

A

g

B

`(g)/(2)`

C

`(g)/(3)`

D

`(g)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
`a=(g sin theta)/(1+(K^(2))/(r^(2)))=(gxx1//2)/(1+(r^(2))/(r^(2)))=(g)/(4)`
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