Question 743: Centre of Mass and Rotational Motion

Question

Two point-like objects of masses $20 ~\mathrm{gm}$ and $30 ~\mathrm{gm}$ are fixed at the two ends of a rigid massless rod of length $10 \mathrm{~cm}$. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 	imes 10^{-8} \mathrm{~N} \mathrm{~m}  ~\mathrm{rad}^{-1}$. The angular frequency of the oscillations in $n 	imes 10^{-3} ~\mathrm{rad} ~\mathrm{s}^{-1}$. The value of $n$ is _________ .

Accepted Answer

$\begin{aligned} & m_1=30 \mathrm{gm} \\ & m_2=20 \mathrm{gm}\end{aligned}$
Moment of inertia about the axis of rotation is
$$
I=m_1 r_1^2+m_2 r_2^2
$$
Clearly $r_1=4 \mathrm{~cm}$
And $r_2=6 \mathrm{~cm}$
$$
\begin{aligned}
& 	herefore I=\left(30 	imes 10^{-3} 	imes 16 	imes 10^{-4}ight)+\left(20 	imes 10^{-3} 	imes 36 	imes 10^{-4}ight) \\
& \Rightarrow I=1200 	imes 10^{-7} \mathrm{~kg} \mathrm{~m}^2
\end{aligned}
$$
If the system is rotated by small angle ' $	heta$ ', the restoring torque is $	au_{(R)}=-k 	heta$
And $\frac{d^2 	heta}{d t^2}=\frac{-k}{l} \cdot 	heta=-\omega^2 	heta=\frac{-1.2 	imes 10^{-8}}{1200 	imes 10^{-7}} \cdot 	heta$
$$
	herefore \omega^2=10^{-4}
$$
$\begin{aligned} & 	ext { So, } \omega=\frac{1}{100} \mathrm{rad} / \mathrm{s} \\ & \Rightarrow \omega=10 	imes 10^{-3} \mathrm{rad} / \mathrm{s}\end{aligned}$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type