Question 281: Centre of Mass and Rotational Motion

Question

The disc of mass M with uniform surface mass density $$\sigma$$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $${x \over 3}{a \over \pi },{x \over 3}{a \over \pi }$$ where x is _____________. (Round off to the Nearest Integer).[a is an area as shown in the figure]

Accepted Answer

$$dm = \sigma {1 \over 2}R 	imes Rd	heta $$
$$dm = {{\sigma {R^2}d	heta } \over 2}$$
$${x_{cm}} = {{\int {x\,dm} } \over {\int {dm} }} = {{\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d	heta \left( {{{2R} \over 3}\cos 	heta } ight)} } \over {\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d	heta } }}$$
$$ = {{2R} \over 3}{{\int\limits_0^{\pi /2} {\cos 	heta d	heta } } \over {\int\limits_0^{\pi /2} {d	heta } }}$$
$$ = {{2R} \over 3}\left( {{2 \over \pi }} ight)$$
$$ = {{4R} \over {3\pi }}$$
So the value of x = 4

Centre of Mass and Rotational Motion

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