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A mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased, then at which point the string will break :
A
A
B
B
C
C
D
D
Detailed Solution
At the lowest point B: $T_B - mg = \dfrac{mv_B^2}{r} \Rightarrow T_B = mg + \dfrac{mv_B^2}{r}$
At the highest point A: $T_A + mg = \dfrac{mv_A^2}{r} \Rightarrow T_A = \dfrac{mv_A^2}{r} - mg$
At B, the tension has to support the weight as well as provide the centripetal force, and the speed is also maximum there. So the tension in the string is maximum at the lowest position B.
So if the speed is increased, the string will break at B.
At the highest point A: $T_A + mg = \dfrac{mv_A^2}{r} \Rightarrow T_A = \dfrac{mv_A^2}{r} - mg$
At B, the tension has to support the weight as well as provide the centripetal force, and the speed is also maximum there. So the tension in the string is maximum at the lowest position B.
So if the speed is increased, the string will break at B.