Question 226: Centre of Mass and Rotational Motion

Question

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 center of mass of the new disc is $\alpha R$ form the center of the bigger disc. The value of $\alpha $ is

Accepted 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}$
So mass of the remaining disc = $4\pi \sigma {R^2}$ - $\pi \sigma {R^2}$ = $3\pi \sigma {R^2}$
Let center of mass of $3\pi \sigma {R^2}$ mass is at x distance from origin O.
$	herefore$ ${{3\pi {R^2}\sigma .x + \pi {R^2}\sigma .R} \over {4\pi {R^2}\sigma }} = 0$
As center of mass of full disc is at Origin.
$	herefore$ $x =  - {R \over 3}$
According to the question, $x$ = $\alpha R$
$	herefore$ $\alpha  =  - {1 \over 3}$
$ \Rightarrow $ $\left| \alpha  ight| = {1 \over 3}$

Answer

$1/4$

Answer

$1/3$

Answer

$1/2$

Answer

$1/6$

Centre of Mass and Rotational Motion

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 center of mass of the new disc is \alpha R form the center of the bigger disc. The value of \alpha is

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