A disc of radius 20, cm and mass half kg is r | vedclass

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Question

Consider the following figure.

From the above figure,

$Mg \sin 	heta- f = ma _{ cm }$

As we know that,

$	au_{ cm }= I _{ cm } \alpha$

Vedclass · Accepted Answer

$\frac{5}{3 \sqrt{2}} N$

Vedclass · Answer

Consider the following figure.

From the above figure,

$Mg \sin 	heta- f = ma _{ cm }$

As we know that,

$	au_{ cm }= I _{ cm } \alpha$

And,

$f =\frac{ MR ^{2}}{2}\left(\frac{ a _{ cm }}{ R ^{2}}ight)$

$=\frac{ Ma _{ cm }}{2}$

$a_{c m}=\frac{2 f_{R}}{M}$

Therefore,

$mg \sin 	heta- f = M \left(\frac{2 f }{ M }ight)$

$f =\frac{ mg \sin 	heta}{3}$

Substitute the values.

$f =\frac{1}{2}\left(\frac{10}{3}ight)\left(\frac{1}{\sqrt{2}}ight)$

$=\frac{5}{3 \sqrt{2}} N$

A disc of radius $20\, cm$ and mass half $kg$ is rolling on an inclined plane. Find out friction force so that disc performs pure rolling.

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