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

The Correct Option is B

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.

Answer 2

A ball of mass 0.5 kg is attached to a string of length 50 cm and rotated in a horizontal circular path. The maximum tension the string can bear is 400 N. We need to find the maximum possible angular velocity in rad/s.

Concept Used:

For a conical pendulum (mass moving in horizontal circle), the tension in the string provides the centripetal force. The relationship is given by:

\[ T \cos \theta = mg \] \[ T \sin \theta = m \omega^2 r \]

where \( r = l \sin \theta \) is the radius of the circular path, \( l \) is the string length, \( \omega \) is the angular velocity, and \( \theta \) is the angle the string makes with the vertical.

Step-by-Step Solution:

Step 1: Identify the given values and convert to SI units.

\[ m = 0.5 \text{ kg}, \quad l = 50 \text{ cm} = 0.5 \text{ m}, \quad T_{\text{max}} = 400 \text{ N} \]

Step 2: For the maximum angular velocity, the tension will be maximum. We use the vertical force balance:

\[ T \cos \theta = mg \] \[ \cos \theta = \frac{mg}{T} = \frac{0.5 \times 9.8}{400} = \frac{4.9}{400} = 0.01225 \]

Step 3: Find \( \sin \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (0.01225)^2} \approx \sqrt{1 - 0.000150} \approx \sqrt{0.99985} \approx 0.999925 \]

Since \( \cos \theta \) is very small, \( \theta \approx 90^\circ \), so \( \sin \theta \approx 1 \).

Step 4: Use the horizontal force equation to find the angular velocity:

\[ T \sin \theta = m \omega^2 (l \sin \theta) \]

Canceling \( \sin \theta \) from both sides (since \( \sin \theta \neq 0 \)):

\[ T = m \omega^2 l \]

Step 5: Substitute the maximum tension and solve for \( \omega \):

\[ 400 = 0.5 \times \omega^2 \times 0.5 \] \[ 400 = 0.25 \omega^2 \] \[ \omega^2 = \frac{400}{0.25} = 1600 \] \[ \omega = \sqrt{1600} = 40 \text{ rad/s} \]

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