Question 608: Centre of Mass and Rotational Motion

Question

Three point masses each of mass ' $m$ ' are kept at the corners of an equilateral triangle of side. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to $\left(\cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$

Accepted Answer

The situation can be shown below,
where, $O$ is the centroid of equivalent triangle.
The length $A O$ can be found by
$$\begin{aligned}
A O & =\sqrt{O D^2+A D^2} \
& =\sqrt{\left(\frac{1}{3} B Dight)^2+A D^2}=\sqrt{\frac{1}{9} 	imes\left(\frac{\sqrt{3}}{2} Light)^2+\frac{L^2}{4}} \
& =\sqrt{\frac{1}{12} L^2+\frac{L^2}{4}}=\frac{L}{\sqrt{3}}
\end{aligned}$$
$$\begin{aligned}
&	ext { The moment of inertia about } O 	ext { is given by }\
&\begin{aligned}
I & =M R^2 \
& =m 	imes\left(\frac{L}{\sqrt{3}}ight)^2=\frac{m L^2}{3}\quad	ext{.... (i)}
\end{aligned}
\end{aligned}$$
According to law of conservation of angular momentum,
$$\begin{aligned}
I \omega & =	ext { constant } \
I 	imes \frac{2 \pi}{T} & =	ext { constant } \
\Rightarrow \quad T & \propto I
\end{aligned}$$
From Eq. (i), we get
$$T \propto L^2$$

Answer

$\sqrt{L}$

Answer

$L^{3 / 2}$

Answer

$L$

Answer

$L^{-2}$

Centre of Mass and Rotational Motion

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