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 \(10\).

Answer 2

Given:
- Mass of the first object (\(m_1\)) = 20 g = 0.02 kg,
- Mass of the second object (\(m_2\)) = 30 g = 0.03 kg,
- Distance of the first object from the axis of rotation (\(r_1\)) = 4 cm = 0.04 m,
- Distance of the second object from the axis of rotation (\(r_2\)) = 6 cm = 0.06 m,
- Torsional constant (\(k\)) = \(1.2 \times 10^{-8}\) Nm rad\(^{-1}\).

The moment of inertia about the axis of rotation is given by:
\[ I = m_1r_1^2 + m_2r_2^2 \]
\[ = (0.02 \times (0.04)^2) + (0.03 \times (0.06)^2) \]
\[ = (0.02 \times 0.0016) + (0.03 \times 0.0036) \]
\[ = 0.00032 + 0.000108 \]
\[ = 0.000428 \text{ kg m}^2 \]

The restoring torque (\(\tau_R\)) when the system is rotated by a small angle \(\theta\) is given by:
\[ \tau_R = -k\theta \]

Using Newton's second law for rotational motion:
\[ \frac{d^2\theta}{dt^2} = \frac{-k}{I}\theta \]

Substituting the values, we get:
\[ \frac{d^2\theta}{dt^2} = \frac{-1.2 \times 10^{-8}}{0.000428}\theta \]
\[ \Rightarrow \omega^2 = 10^{-4} \text{ rad/s}^2 \]
\[ \Rightarrow \omega = 10^{-2} \text{ rad/s} \]
\[ \Rightarrow \omega = 10 \times 10^{-3} \text{ rad/s} \]

Since the angular frequency is given in the form \(n \times 10^{-3}\) rad/s, we have \(n = 10\).

Therefore, the value of \(n\) is \(\boxed{10}\).