Question

Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 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 × 10−8 N m rad−1. The angular frequency of the oscillations in 𝑛 × 10−3 rad s−1. The value of 𝑛 is _____.

Answer 1

m1 = 30 gm m2 = 20 gm
Moment of inertia about the axis of rotation is
I = m1r12+ m2r22
Clearly r1 = 4 cm
And r2 = 6 cm
∴ I = (30 × 10–3 × 16 × 10–4) + (20 × 10–3 × 36 × 10–4)
⇒ I = 1200 × 10–7 kg m2
If the system is rotated by small angle ‘ θ , the restoring torque is τ (R) = –kθ

And $\frac{d^2θ}{dt^2}=\frac{-k}{l}θ = -w^2θ= \frac{-1.2\times10^-8}{1200 \times10^-7}θ$

$⇒ w^2= 10^{-4}\ rad\ s^{-1}$
$⇒ w= 10^{-2}\ rad\ s^{-1}$
$⇒ w=10 \times 10^{-3}\ rad\ s^{-1}$
Given that, the angular frequency of the oscillations in 𝑛 × 10−3 rad s−1
$⇒ n= 10$

So, the answer is $10$.