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 Options: A . &nbspg2 B . &nbspg3 C . &nbspg4 D . &nbspg5

Answer: Option C : C a = g sin θ 1 + k 2 R 2 = g sin 30 ∘ 1 + 1 = g 4 [ A s k 2 R 2 = 1 a n d θ = 30 ∘ ]

A thin uniform circular ring is rolling down an inclined plane of inclination 30

Question

A thin uniform circular ring is rolling down an inclined plane of inclination

30^{\circ}

without slipping. Its linear acceleration along the inclined plane will be

Options:

A . g2

B . g3

C . g4

D . g5

Answer: Option C
:
C

a = \frac{g sin θ}{1 + \frac{k^{2}}{R^{2}}} = \frac{g sin 30^{\circ}}{1 + 1} = \frac{g}{4}

[A s \frac{k^{2}}{R^{2}} = 1 a n d θ = 30^{\circ}]

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