A solid cylinder of mass M and radius R rolls | vedclass

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Question

Acceleration of a rolling body down an inclined plane is

$a=\frac{g \sin 	heta}{1+\frac{k^{2}}{R^{2}}}$

For solid cylinder, $\mathrm{k}^{2}=\fr

Vedclass · Accepted Answer

$\frac{2}{3}\,g\, \sin \, 	heta$

Vedclass · Answer

Acceleration of a rolling body down an inclined plane is

$a=\frac{g \sin 	heta}{1+\frac{k^{2}}{R^{2}}}$

For solid cylinder, $\mathrm{k}^{2}=\frac{\mathrm{R}^{2}}{2}$

$a=\frac{g \sin 	heta}{1+\frac{1}{2}}=\frac{2}{3} g \sin 	heta$

A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle $\theta $ with the horizontal. then its acceleration is

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