Work Energy Power

JEE Main level, single correct, medium

168 of 183

Question 168 of 183

In case of vertical circular motion of a particle by a thread of length rr if the tension in the thread is zero at an angle 30∘30^{\circ} shown in figure, the velocity at the bottom point (A)(A) of the circular path is (g=g= gravitational acceleration)

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Answer: (A) 72gr\sqrt{\frac{7}{2} g r}

At the point where tension T=0T = 0 and angle from vertical is 60°60° (i.e., 30°30° from horizontal):
T+mgcos⁡60°=mV2ℓT + mg\cos 60° = \frac{mV^2}{\ell}
Since T=0T = 0:
V2=gℓ2V^2 = \frac{g\ell}{2}
Using mechanical energy conservation from bottom (A) to the point where T=0T = 0:
12mu2=mg(ℓ+ℓcos⁡60°)+12mV2\frac{1}{2}mu^2 = mg(\ell + \ell\cos 60°) + \frac{1}{2}mV^2
u2=2gℓ(1+12)+gℓ2=3gℓ+gℓ2=7gℓ2u^2 = 2g\ell\left(1 + \frac{1}{2}\right) + \frac{g\ell}{2} = 3g\ell + \frac{g\ell}{2} = \frac{7g\ell}{2}
u=7gℓ2u = \sqrt{\frac{7g\ell}{2}}

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