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PhyMedAdv 2023 • Paper 1

Angular Frequency of Torsional Pendulum with Two Point Masses

Two point-like objects of masses 20 gm20\text{ gm} and 30 gm30\text{ gm} are fixed at the two ends of a rigid massless rod of length 10 cm10\text{ 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−11.2 \times 10^{-8}\text{ N m rad}^{-1}. The angular frequency of the oscillations in n×10−3 rad s−1n \times 10^{-3}\text{ rad s}^{-1}. The value of nn is ________.

Question Diagram 1
Official Numerical Answer10

Topics & Concepts

OscillationsSHMCenter of MassMoment of Inertia

Step-by-Step Solution

To find the angular frequency of the torsional pendulum, we first determine the moment of inertia of the system about the vertical axis passing through its center of mass.

1. Moment of Inertia of the System: Let the two masses be: m1=30 gm=30×10−3 kgm_1 = 30\text{ gm} = 30 \times 10^{-3}\text{ kg} m2=20 gm=20×10−3 kgm_2 = 20\text{ gm} = 20 \times 10^{-3}\text{ kg} The length of the rod is: L=10 cm=0.1 mL = 10\text{ cm} = 0.1\text{ m}

The reduced mass μ\mu of the two-body system is given by: μ=m1m2m1+m2=30×2030+20 gm=12 gm=1.2×10−2 kg\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{30 \times 20}{30 + 20}\text{ gm} = 12\text{ gm} = 1.2 \times 10^{-2}\text{ kg}

The moment of inertia II of the system about an axis perpendicular to the rod and passing through its center of mass is: I=μL2=(1.2×10−2 kg)×(0.1 m)2=1.2×10−4 kg m2I = \mu L^2 = (1.2 \times 10^{-2}\text{ kg}) \times (0.1\text{ m})^2 = 1.2 \times 10^{-4}\text{ kg m}^2

2. Angular Frequency of Oscillation: The torsional constant of the wire is: k=1.2×10−8 N m rad−1k = 1.2 \times 10^{-8}\text{ N m rad}^{-1}

The angular frequency ω\omega of the torsional oscillations is given by: ω=kI\omega = \sqrt{\frac{k}{I}}

Substituting the values: ω=1.2×10−8 N m rad−11.2×10−4 kg m2=10−4 rad s−1=10−2 rad s−1\omega = \sqrt{\frac{1.2 \times 10^{-8}\text{ N m rad}^{-1}}{1.2 \times 10^{-4}\text{ kg m}^2}} = \sqrt{10^{-4}}\text{ rad s}^{-1} = 10^{-2}\text{ rad s}^{-1}

Expressing ω\omega in the form n×10−3 rad s−1n \times 10^{-3}\text{ rad s}^{-1}: ω=10×10−3 rad s−1\omega = 10 \times 10^{-3}\text{ rad s}^{-1}

Therefore, the value of nn is 10.

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