A thin uniform circular ring is rolling down an inclined plane of induction 30° without slipping, its linear acceleration along the inclined plane will be

Question

A thin uniform circular ring is rolling down an inclined plane of induction 30° without slipping, its linear acceleration along the inclined plane will be (A)  (2/3)g (B)  (g/3) (C)  (g/4) (D)  (g/2)

Tardigrade Admin · Accepted Answer

a = (g sin θ/1 + (k2)R2) = (g sin 30°/1 + 1) = (g/2) × (1/2) = (g/4)

Q. A thin uniform circular ring is rolling down an inclined plane of induction 30 $^{\circ}$ without slipping, its linear acceleration along the inclined plane will be

$\frac{2}{3} g$

$\frac{g}{3}$

$\frac{g}{4}$

$\frac{g}{2}$

$a = \frac{g s in θ}{1 + \frac{k ^{2}}{R ^{2}}} = \frac{g s in 3 0 ^{\circ}}{1 + 1} = \frac{g}{2} \times \frac{1}{2} = \frac{g}{4}$

Q. A thin uniform circular ring is rolling down an inclined plane of induction 30∘ without slipping, its linear acceleration along the inclined plane will be

32​g

3g​

4g​

2g​