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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/2 (b) g/3 (c) g/4 (d) 2g/3

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/2 (b) g/3
(c) g/4 (d) 2g/3

Grade:12

1 Answers

Vikas TU
14149 Points
2 years ago
Acceleration on an inclined plane 
a= g sinθ/ (1 + I/MR²)
for circular rings: I= MR²
so by putting the value in above equation we get
a= g sinθ/ 2
a = g sin30/2
a= g/4

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