Two point-like objects of masses 20 text{ g} | vedclass

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(C) The angular frequency $\omega$ of a torsional pendulum is given by $\omega = \sqrt{\frac{C}{I}}$,where $C$ is the torsional constant and $I$ is th

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) The angular frequency $\omega$ of a torsional pendulum is given by $\omega = \sqrt{\frac{C}{I}}$,where $C$ is the torsional constant and $I$ is the moment of inertia about the axis of rotation.First,we find the center of mass $(CM)$ of the system. Let the $30 	ext{ g}$ mass be at $x = 0$ and the $20 	ext{ g}$ mass be at $x = 10 	ext{ cm}$. The position of the $CM$ is $x_{cm} = \frac{(30 	imes 0) + (20 	imes 10)}{30 + 20} = \frac{200}{50} = 4 	ext{ cm}$ from the $30 	ext{ g}$ mass.Thus,the distances of the $30 	ext{ g}$ and $20 	ext{ g}$ masses from the axis of rotation $(CM)$ are $r_1 = 4 	ext{ cm} = 0.04 	ext{ m}$ and $r_2 = 10 - 4 = 6 	ext{ cm} = 0.06 	ext{ m}$,respectively.The moment of inertia $I$ is $I = m_1 r_1^2 + m_2 r_2^2 = (30 	imes 10^{-3} 	ext{ kg}) 	imes (0.04 	ext{ m})^2 + (20 	imes 10^{-3} 	ext{ kg}) 	imes (0.06 	ext{ m})^2$.$I = 0.03 	imes 0.0016 + 0.02 	imes 0.0036 = 0.000048 + 0.000072 = 0.00012 	ext{ kg m}^2 = 1.2 	imes 10^{-4} 	ext{ kg m}^2$.Given $C = 1.2 	imes 10^{-8} 	ext{ N m rad}^{-1}$,the angular frequency is $\omega = \sqrt{\frac{1.2 	imes 10^{-8}}{1.2 	imes 10^{-4}}} = \sqrt{10^{-4}} = 10^{-2} 	ext{ rad s}^{-1}$.Comparing this with $n 	imes 10^{-3} 	ext{ rad s}^{-1}$,we get $n 	imes 10^{-3} = 10 	imes 10^{-3}$,so $n = 10$.

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