Question 223: 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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