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

Question

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

Vikas TU · Answer

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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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

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