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Byju's Answer

Standard XII

Physics

Pure Rolling

A ring of rad...

Question

A ring of radius R is rolling purely on the outer surface of a fixed pipe of radius 4R., At some instant,when ring has no angular acceleration the centre of the ring has a speed v. Then, the acceleration of the point on the ring which is in contact with the surface of the pipe is.

\frac{4 v^{2}}{5 R}

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\frac{3 v^{2}}{5 R}

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\frac{v^{2}}{5 R}

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Zero

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Solution

The correct option is A $\frac{4 v^{2}}{5 R}$
Acceleration of centre of ring $_{c} = \frac{v^{2}}{5 R} (-^j)$ ,Acceleration of point of contact w.r.t. centre of ring $_{c} = \frac{v^{2}}{R}^j$

$|_{c} | = |_{b c} +_{c} | = \frac{V^{2}}{R} - \frac{V^{2}}{5 R} = \frac{4}{5} \frac{V^{2}}{R}$

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Similar questions

A ring of radius R is rolling purely on the outer surface of a fixed pipe of radius 4R. At some instant, the centre of the ring has a constant speed = v as shown in fig.

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A ring of radius R is rolling purely on the outer surface of a fixed pipe of radius 4R., At some instant,when ring has no angular acceleration the centre of the ring has a speed v. Then, the acceleration of the point on the ring which is in contact with the surface of the pipe is.

The correct option is A 4v25RAcceleration of centre of ring →ac=v25R(−^j),Acceleration of point of contact w.r.t. centre of ring →ac=v2R^j|→ac|=|→abc+→ac|=V2R−V25R=45V2R

The correct option is A $\frac{4 v^{2}}{5 R}$
Acceleration of centre of ring $_{c} = \frac{v^{2}}{5 R} (-^j)$ ,Acceleration of point of contact w.r.t. centre of ring $_{c} = \frac{v^{2}}{R}^j$

$|_{c} | = |_{b c} +_{c} | = \frac{V^{2}}{R} - \frac{V^{2}}{5 R} = \frac{4}{5} \frac{V^{2}}{R}$