Question

A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad/s is,:

• A. 1600
• B. 40
• C. 1000
• D. 20

Answer 1

Answer: (B)

40

Answer 2

The tension in the string is related to the centripetal force required for circular motion:

$T = mr\omega^2$

where:

• $T = 400 \, \text{N}$ is the maximum tension.
• $m = 0.5 \, \text{kg}$ is the mass of the ball.
• $r = 0.5 \, \text{m}$ is the radius (length of the string).
• $\omega$ is the angular velocity.

Rearranging the formula to solve for $\omega$:

$\omega = \sqrt{\frac{T}{mr}} = \sqrt{\frac{400}{0.5 \times 0.5}} = \sqrt{\frac{400}{0.25}} = \sqrt{1600} = 40 \, \text{rad/s}$

Thus, the maximum possible angular velocity of the ball is 40 rad/s.