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

Question

In case of vertical circular motion of a particle by a thread of length

r

if the tension in the thread is zero at an angle

30^{\circ}

shown in figure, the velocity at the bottom point

(A)

of the circular path is (

g=

gravitational acceleration)
In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an

Quantrex Academy · Accepted Answer

Correct answer: A. (7)/(2) g r

At the point where tension T = 0 and angle from vertical is 60° (i.e., 30° from horizontal): T + mg 60° = (mV^2)/( ) Since T = 0 : V^2 = (g )/(2) Using mechanical energy conservation from bottom (A) to the point where T = 0 : (1)/(2)mu^2 = mg( + 60°) + (1)/(2)mV^2 u^2 = 2g (1 + (1)/(2)) + (g )/(2) = 3g + (g )/(2) = (7g )/(2) u = (7g )/(2)

In case of vertical circular motion of a particle by a thread of...

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