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 centre of mass, as shown in the figure. The resu

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 centre of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 \times 10^{- 8} {N m rad}^{- 1} .$ The angular frequency of the oscillations is $n \times 10^{- 3} {rad s}^{- 1} .$ The value of $n$ is ______. [2023]

Smart Achievers · Accepted Answer

(10)

$Time period of torsional pendulum$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{I}{C}}$

$∴$ $ω = \sqrt{\frac{C}{I}}$

Where $I =$ moment of inertia

$I = (30) {(4)}^{2} + (20) {(6)}^{2} = 1200 {g cm}^{2}$

or, $I = 1.2 \times 10^{- 4} {kg m}^{2}$ $= \sqrt{\frac{C}{I}} ω = \sqrt{\frac{1.2 \times 10^{- 8}}{1.2 \times 10^{- 4}}}$

$[∵ Torsional constant of the wire C = 1.2 \times 10^{- 8} {N m rad}^{- 1} given]$

$∴$ $ω = \sqrt{10^{- 4}} = 10^{- 2}$

$∴$ $n \times 10^{- 3} = 10^{- 2}$

$∴$ $n = 10$

Solution

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Solution

Ans. (10) Time period of torsional pendulum T=2πω=2πIC ∴ ω=CI Where I= moment of inertia I=(30)(4)2+(20)(6)2=1200 g cm2 or, I=1.2×10-4 kg m2=CIω=1.2×10-81.2×10-4 [∵ Torsional constant of the wire C=1.2×10-8 N m rad-1 given] ∴ ω=10-4=10-2 ∴ n×10-3=10-2 ∴ n=10

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