Question

A weightless thread can bear tension up to $3.7 \text{ kg wt}$. A stone of mass $500 \text{ gram}$ is tied to it and revolved in circular path of radius $4 \text{ m}$ in vertical plane. Maximum angular velocity of the stone will be (acceleration due to gravity, $g = 10 \text{ m/s}^2$ )}

• A. 16 rad/s
• B. 4 rad/s
• C. 2 rad/s
• D. 8 rad/s

Hint:

String breaks at the bottom because tension must balance both weight and centripetal force.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a vertical circle, the tension in the string varies, reaching its absolute maximum at the lowest point.
To find the maximum possible angular velocity without breaking the thread, we evaluate the system at this lowest point and equate the tension to the breaking strength.
Step 2: Key Formula or Approach:
At the lowest point, tension must overcome gravity and provide centripetal force: \[ T_{\text{max}} = mg + F_{\text{centripetal}} \] \[ T_{\text{max}} = mg + m\omega^2 r \] Step 3: Detailed Explanation:
Convert breaking tension to Newtons: \[ T_{\text{max}} = 3.7 \text{ kg wt} = 3.7 \times g = 3.7 \times 10 = 37 \text{ N} \] Convert mass to kilograms: \[ m = 500 \text{ g} = 0.5 \text{ kg} \] Radius is given as $r = 4 \text{ m}$.
Substitute values into the maximum tension formula: \[ T_{\text{max}} = m(g + \omega^2 r) \] \[ 37 = 0.5 (10 + \omega^2 \cdot 4) \] Divide by 0.5: \[ 74 = 10 + 4\omega^2 \] Subtract 10: \[ 64 = 4\omega^2 \] Divide by 4: \[ \omega^2 = 16 \] Take the square root: \[ \omega = 4 \text{ rad/s} \] Step 4: Final Answer:
The maximum angular velocity is 4 rad/s.