Rotational MotionHard
Question
A uniform rod of mass m and length $l$ suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is
$\_\_\_\_$ . (g acceleration due to gravity)
Options
A.$mg/2$
B.$mg/d$
C.$mg/3$
D.mg
Solution
$$mg\frac{l}{2} = \frac{ml^{2}}{3}\alpha$$
$$\begin{array}{r} \alpha = \frac{3\text{ }g}{2l}\#(1) \end{array}$$
$${mg - T = {ma}_{c} }{T = mg - {ma}_{c}}$$
$$\begin{matrix} & \ = mg - m\left( \frac{l}{2}\alpha \right) \\ & \ = mg - m\left( \frac{l}{2} \cdot \frac{3\text{ }g}{2l} \right) \end{matrix}$$
$$T = \frac{mg}{4}$$
Create a free account to view solution
View Solution FreeMore Rotational Motion Questions
If the earth is a point mass of 6 × 1024 kg revolving around the sun at a distance of 1.5 × 108 km and in time...The moment of inertia of a square lamina about the perpendicular axis through its centre of mass is 20 kg-m2. Then, its ...The wheels of moving vehicles are made hollow in the middle and thick at the periphery, because...Radius of gyration of a body depends on : -...Two rotating bodies have same angular momentum but their moments of inertia are I1 and I2 respectively (I1 > I2). Whi...