Two masses m and frac{m}{2} are connected at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. Locate the center of mass $(CM)$: Let the distance of mass $m$ from the suspension point be $x_1$ and mass $m/2$ be $x_2$. Given $x_1 + x_2 =

Vedclass · Accepted Answer

$\frac{k	heta_0^2}{l}$

Vedclass · Answer

(C) $1$. Locate the center of mass $(CM)$: Let the distance of mass $m$ from the suspension point be $x_1$ and mass $m/2$ be $x_2$. Given $x_1 + x_2 = l$. For $CM$,$m x_1 = (m/2) x_2 \implies x_2 = 2x_1$. Thus,$3x_1 = l \implies x_1 = l/3$ and $x_2 = 2l/3$.$2$. Moment of inertia $I$ about the suspension point: $I = m(l/3)^2 + (m/2)(2l/3)^2 = m(l^2/9) + (m/2)(4l^2/9) = ml^2/9 + 2ml^2/9 = ml^2/3$.$3$. Angular frequency $\omega$: $\omega = \sqrt{\frac{k}{I}} = \sqrt{\frac{k}{ml^2/3}} = \sqrt{\frac{3k}{ml^2}}$.$4$. Maximum angular velocity $\omega_{max}$: Using energy conservation,$\frac{1}{2}I\omega_{max}^2 = \frac{1}{2}k	heta_0^2 \implies \omega_{max} = 	heta_0 \sqrt{\frac{k}{I}} = 	heta_0 \omega$.$5$. Tension $T$ at mean position: The tension is provided by the centripetal force on either mass. For mass $m$ at distance $l/3$,$T = m \omega_{max}^2 (l/3) = m (	heta_0^2 \frac{k}{I}) (l/3) = m 	heta_0^2 \frac{k}{ml^2/3} \frac{l}{3} = \frac{k	heta_0^2}{l}$.

Similar Questions

Two oscillating systems,a simple pendulum and a vertical spring-mass system,have the same time period of motion on the surface of the Earth. If both are taken to the moon,then-

$A$ block is placed on a horizontal plank. The plank is performing $SHM$ along a vertical line with an amplitude of $40 \, cm$. The block just loses contact with the plank when the plank is momentarily at rest. Then:

The amplitude of a particle executing $SHM$ about $O$ is $10 \, cm.$ Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module