Rotational MotionMedium
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
$$mg\frac{l}{2} = \frac{ml^{2}}{3}\alpha$$
$$\begin{array}{r} \alpha = \frac{3\text{ }g}{2l}\#(1) \end{array}$$
$${mg - T = {ma}_{c} }{T = mg - {ma}_{c}}$$
$$\begin{matrix} & \ = mg - m\left( \frac{l}{2}\alpha \right) \\ & \ = mg - m\left( \frac{l}{2} \cdot \frac{3\text{ }g}{2l} \right) \end{matrix}$$
$$T = \frac{mg}{4}$$
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