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. 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  __________.

Correct Answer: w = 10 × 10^-3 rad/s (corresponding to n = 10)

Step-by-Step Explanation:

1. Understanding the Given Data:

From the problem text and the attached diagram, we have:

• Mass of the first point-like object, _m1 = 30 gm = 30 × 10^-3 kg = 0.03 kg
• Mass of the second point-like object, _m2 = 20 gm = 20 × 10^-3 kg = 0.02 kg
• Length of the rigid rod, _L = 10 cm = 0.1 m
• Torsional constant of the suspending wire, _C = 1.2 × 10^-8 N m rad^-1

2. Finding the Position of the Center of Mass:

Let the wire be attached at the center of mass (CM) of the rod. Let _r1 be the distance of mass _m1 (30 gm) from the CM, and _r2 be the distance of mass _m2 (20 gm) from the CM.

Using the definition of the center of mass:

$m_1{r}_1=m_2{r}_2$

Since _{r1 + r2 = L}, we get:

$r_1=\frac{m_2}{m_1+m_2}L=\frac{20}{30+20}\times 10\text{cm}=4\text{cm}=0.04\text{m}$

$r_2=\frac{m_1}{m_1+m_2}L=\frac{30}{30+20}\times 10\text{cm}=6\text{cm}=0.06\text{m}$

3. Calculating the Moment of Inertia (I) about the Axis of Rotation:

The moment of inertia of the two point masses about the vertical wire passing through their center of mass is given by:

$I=m_1{r}_1^2+m_2{r}_2^2=\mu L^2$

where _μ is the reduced mass of the system:

$\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{30\times 20}{30+20}\text{gm}=12\text{gm}=12\times {10}^{-3}\text{kg}$

Now, calculate _I in SI units:

$I=(12\times {10}^{-3}\text{kg})\times (0.1\text{m}{)}^2=12\times {10}^{-3}\times {10}^{-2}\text{kg m}{2}^{}=1.2\times {10}^{-4}\text{kg m}{2}^{}$

4. Calculating Angular Frequency (ω):

The angular frequency of a torsional pendulum is given by the formula:

$\omega =\sqrt{\frac{C}{I}}$

Substituting the values of _C and _I:

$\omega =\sqrt{\frac{1.2\times {10}^{-8}}{1.2\times {10}^{-4}}}=\sqrt{{10}^{-4}}={10}^{-2}\text{rad s}{-1}^{}$

Expressing this in the form _{n × 10^-3 rad s^-1}:

$\omega =10\times {10}^{-3}\text{rad s}{-1}^{}$

Thus, the angular frequency is w = 10 × 10^-3 rad/s, and the value of _n is 10.