A heavy particle is attached to one end of a string 1m long whose other end is fixed at O. It is projected from its lowest position horizontally with a velocity V:

Moderate

A

If V² > 5 g the particle will describe complete circular motion in the vertical plane
B

If V² = 3.5 g the tension in the string will become zero after the string has turned through 120^o
C

If V² = 2 g, the tension in the string becomes zero the velocity of the particle also becomes zero
D

If V² = g the velocity of the particle becomes zero after the string turns through 60°

Solution

$\begin{matrix} - mgl (1 - \cos θ) = \frac{1}{2} {mv}^{2} - \frac{1}{2} m u^{2} . . . . . . . (i) \\ T - mgcos θ = \frac{m v^{2}}{l} . . . . . . . . (ii) \\ V^{2} = g l (2 - 3 \cos θ) \end{matrix}$
$(a) θ = π, V^{2} = 5 gl, hence true$
$(b) 3 .5 = 2 - 3 \cos θ, \cos θ = - \frac{1}{2}$
$\Rightarrow A = 120^{\circ} hence true$
$(c) 2 = 2 - 3 \cos θ, \cos θ = 0, θ \Rightarrow \frac{π}{2}, hence true$
$(d) 1 = 2 - 3 \cos θ, \cos θ = \frac{1}{3}, θ > 60^{\circ}, hence true$

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