Find Maximum Angular Velocity of Ball in Conical Pendulum

To find the maximum possible angular velocity of the ball, we analyze the forces acting on the ball of mass $m$ executing circular motion in a conical pendulum setup.

Let $\theta$ be the angle made by the string of length $L$ with the vertical axis.

The forces acting on the ball are:

1. The tension $T$ along the length of the string pointing towards the point of suspension.
2. The gravitational force $mg$ acting vertically downwards.

Resolving the tension $T$ into vertical and horizontal components:

• Vertical component: $T \cos\theta$
• Horizontal component: $T \sin\theta$

Since the ball moves in a horizontal circular path of radius $r = L \sin\theta$:

1. For vertical equilibrium:
   $T \cos\theta = mg \quad \text{--- (1)}$

2. The horizontal component provides the necessary centripetal force for circular motion:
   $T \sin\theta = m \omega^2 r$

Substituting $r = L \sin\theta$:
$T \sin\theta = m \omega^2 (L \sin\theta)$

Assuming $\sin\theta \neq 0$ (the pendulum is swung out into a circle):
$T = m \omega^2 L \quad \text{--- (2)}$

From Equation (2), the angular velocity $\omega$ is given by:
$\omega = \sqrt{\frac{T}{m L}}$

To obtain the maximum possible angular velocity $\omega_{\text{max}}$, the tension $T$ must reach its maximum allowable value $T_{\text{max}} = 324\text{ N}$.

Given values:

• Mass of the ball, $m = 0.5\text{ kg}$
• Length of the string, $L = 0.5\text{ m}$
• Maximum tension, $T_{\text{max}} = 324\text{ N}$

Substituting these values into the expression for $\omega_{\text{max}}$:
$\omega_{\text{max}} = \sqrt{\frac{324}{0.5 \times 0.5}}$

$\omega_{\text{max}} = \sqrt{\frac{324}{0.25}} = \sqrt{324 \times 4} = \sqrt{1296} = 36\text{ rad/s}$

(Note: The condition $\cos\theta = \frac{mg}{T_{\text{max}}} = \frac{0.5 \times 9.8}{324} < 1$ is satisfied, ensuring a valid physical configuration.)

Therefore, the maximum possible value of the angular velocity is $36\text{ rad/s}$.

Correct Option: (D) 36