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

The correct option is A g/4 For ring, #160; #160; I= MR^2 Acceleration of a body (pure) rolling on the inclined plane, #160; #160; a= ...

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Standard XII

Physics

Race of Rollers

A thin unifor...

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

g / 2

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g / 3

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g / 4

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2 g / 3

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Solution

The correct option is A $g / 4$
For ring, $I = M R^{2}$
Acceleration of a body (pure) rolling on the inclined plane, $a = \frac{g s i n θ}{1 + \frac{I}{M R^{2}}}$
$a = \frac{g s i n 30}{1 + \frac{M R^{2}}{M R^{2}}}$
$⟹ a = \frac{g}{4}$

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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

The correct option is A g/4For ring, I=MR2Acceleration of a body (pure) rolling on the inclined plane, a=gsinθ1+IMR2 a=gsin301+MR2MR2⟹a=g4

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

The correct option is A $g / 4$
For ring, $I = M R^{2}$
Acceleration of a body (pure) rolling on the inclined plane, $a = \frac{g s i n θ}{1 + \frac{I}{M R^{2}}}$
$a = \frac{g s i n 30}{1 + \frac{M R^{2}}{M R^{2}}}$
$⟹ a = \frac{g}{4}$