Question

A rod of mass \(m\) and length \(l\) is attached to two ideal strings. Find tension in the left string just after the right string is cut.

• A. \( \frac{mg}{2} \)
• B. \( \frac{mg}{4} \)
• C. \( \frac{2}{3} mg \)
• D. \( \frac{mg}{5} \)

Hint:

When solving for tension in dynamic systems, remember to consider both the gravitational and the rotational forces that act on the object.

Answer 1

The Correct Option is B

To determine the tension in the left string just after the right string is cut, we follow these steps:

1. Initially, the rod is in equilibrium with both strings supporting its weight. The tension in each string is: \(\frac{mg}{2}\).
2. When the right string is cut, the left string must support the entire weight of the rod and provide torque to make the rod rotate about the pivot point at the left string.
3. The rod will rotate about the left string. Just after the cut, the rod is in free fall, except for the point at which the left string is attached.
4. To find the tension in the left string, we apply the torque equation about the pivot (left end):
   • Torque due to gravity about the pivot is: \(\text{Torque} = mg \times \frac{l}{2}\).
   • To balance this torque, the tension \(T\) creates an opposite torque about the pivot.
     • Considering rotational equilibrium (angular acceleration \(\alpha\)), using the equation: \(T \cdot l = I \cdot \alpha\)
     • For a rod about one end, \(I = \frac{1}{3}ml^2\).
5. Equating and solving the above expressions with \(\alpha = \frac{3g}{2l}\) (linear acceleration at the center), \(T \cdot l = \frac{1}{3}ml^2 \cdot \left( \frac{3g}{2l} \right)\).
6. After simplification, we find: \(T = \frac{mg}{4}\).

Therefore, the tension in the left string just after the right string is cut is \(\frac{mg}{4}\). Thus, the correct answer is \(\frac{mg}{4}\).