1.

A thin uniform circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linearacceleration along the inclined plane is2(3) 93

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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