A circular disc of radius R is removed from a | vedclass

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Question

Let the mass per unit area 

be $\sigma $.

Then the mass of the 

complete disc 

$ = \sigma \left[ {\pi {{\left( {2R} \r

Vedclass · Accepted Answer

$\;\frac{1}{3}$

Vedclass · Answer

Let the mass per unit area 

be $\sigma $.

Then the mass of the 

complete disc 

$ = \sigma \left[ {\pi {{\left( {2R} ight)}^2}} ight] = 4\pi \sigma {R^2}$

The mass of the removed disc $ = \sigma \left( {\pi {R^2}} ight) - \pi \sigma {R^2}$

Let us consider the above situation to be a complete disc of radius $2R$ on which a disc radius $R$ of negative mass is superimposed. Let $O$ be the origin. Then the above figure can be redrawn keeping in mind the concept of center of mass as :

$\begin{array}{l}
{x_{c.m}} = \frac{{\left( {4\pi \sigma {R^2}} ight) 	imes 0 + \left( { - \pi \sigma {R^2}} ight)R}}{{4\pi \sigma {R^2} - \pi \sigma {R^2}}}\
	herefore \,{x_{c.m}} = \frac{{ - \pi \sigma {R^2} 	imes R}}{{3\pi \sigma {R^2}}}\,\,\,\,	herefore {x_{c.m}} =  - \frac{R}{3} \Rightarrow \alpha  = \frac{1}{3}
\end{array}$

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\frac{\alpha}{R}$ form the centre of the bigger disc. The value of a is $\alpha $ is

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