A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension in the string is $T$. If speed becomes $2\omega$ while keeping the same radius, the tension in the string becomes;

A $\frac{T}{4}$

B $\sqrt{2}T$

C $T$

D $4T$

The answer is D.

Solution

Centripetal force $F_c = mr\omega^2 = T$ Now $T' = mr(2\omega)^2 = 4mr\omega^2$ $T' = 4T$

Concept Tension dependence on angular speed in circular motion

72 questions from this chapter have appeared across 27 exam years.