A uniform rod of mass m and length l is 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) [figure]
- mg/3
- mg/4
- mg
- mg/2
Answer
(2) mg/4
Solution
Just after string B is cut, the rod begins to rotate about end A (its angular velocity is still zero).
Torque about A: mg l2 = IA = ml23
= 3g2l
Acceleration of the centre of mass: acm = l2 = 3g4 (downward)
For the translation of the rod: mg - T = macm
T = mg - 3mg4 = mg4
Answer: (2)