In the case of vertical circular motion of a | vedclass

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(B) Let the particle be at a point $P$ where the thread makes an angle of $30^{\circ}$ with the horizontal. The angle with the vertical is $	heta = 9

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$\sqrt{\frac{7}{2}gr}$

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(B) Let the particle be at a point $P$ where the thread makes an angle of $30^{\circ}$ with the horizontal. The angle with the vertical is $	heta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.At this point, the forces acting along the radial direction are the tension $T$ and the component of gravity $mg \cos 60^{\circ}$.The equation of motion is $T + mg \cos 60^{\circ} = \frac{mv^2}{r}$.Given that the tension $T = 0$ at this point, we have $mg \cos 60^{\circ} = \frac{mv^2}{r}$.Since $\cos 60^{\circ} = \frac{1}{2}$, we get $mg(\frac{1}{2}) = \frac{mv^2}{r}$, which simplifies to $v^2 = \frac{gr}{2}$.Now, apply the law of conservation of mechanical energy between the bottom point $A$ and point $P$.The height of point $P$ above point $A$ is $h = r + r \sin 30^{\circ} = r + r(\frac{1}{2}) = \frac{3r}{2}$.Let $u$ be the velocity at point $A$. Then, $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh$.Substituting the values: $\frac{1}{2}mu^2 = \frac{1}{2}m(\frac{gr}{2}) + mg(\frac{3r}{2})$.$u^2 = \frac{gr}{2} + 3gr = \frac{7gr}{2}$.Therefore, the velocity at the bottom point is $u = \sqrt{\frac{7}{2}gr}$.

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

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