Question:
a thin uniform,circular ring is rolling down an inclind plane of inclination 30*without slipping. what its acceleration along the inclined planewill be
Answer:
Acceleration on an inclined plane
a= g sinθ/ (1 + I/MR²)
for circular rings moment of inertia : I= MR²
so by putting the value in above equation we get
a= g sinθ/ 2
a = g sin30/2
a= g/4 = 2.45 m/sec2
(g = 9.8 and sin30 =1/2 )
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