Question

A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is _______ (g = acceleration due to gravity).

• A. mg/3
• B. mg/2
• C. mg/4
• D. mg

Hint:

A common mistake is to assume the rod's center of mass just falls with acceleration g. However, because it's a rigid body rotating about a pivot, its acceleration is constrained. The acceleration of the center of mass is less than g, which means there must be an upward force (the tension) that is non-zero. The value is \(mg - ma_{cm}\).

Answer 1

The Correct Option is C

To find the tension in one string immediately after the other string is cut, we need to analyze the forces acting on the rod.

Initially, the rod is suspended by two strings, so the tension in each string supports half the weight of the rod. Hence, the initial tension in each string is:

T_1 = T_2 = \frac{mg}{2}

However, when one string is cut (say string B), the entire system dynamics change. The rod will begin to rotate about the point where the remaining string (A) is attached. Hence, we need to consider both the rotational and translational dynamics of the rod.

Translational Dynamics:

When string B is cut, the system's center of mass will experience a downward force. Immediately after the string is cut, string A will provide the necessary force (tension) to support the translational motion of the center of mass.

Rotational Dynamics:

Since the rod starts rotating about point A immediately after cutting B, we consider torque about point A. The torque due to the weight of the rod is:

\tau = mg \cdot \frac{l}{2} (where \frac{l}{2} is the distance from point A to the center of mass)

The tension in string A has to provide this torque as well as balance the weight of the rod.

Calculation:

Using the principle of rotational motion: \tau = I \cdot \alpha (where I is the moment of inertia and \alpha is the angular acceleration).

For a rod rotating about an end, the moment of inertia is: I = \frac{ml^2}{3}

Substituting in the values, we equate:

mg \cdot \frac{l}{2} = \frac{ml^2}{3} \cdot \alpha

Solving this gives us: \alpha = \frac{3g}{2l}

The tension in the string providing both translational and rotational balance immediately upon cutting can be derived by constructing the net vertical forces:

Tension, T = mg - ma_{\text{center of mass}}, where a_{\text{center of mass}} = \frac{\alpha l}{2} = \frac{3g}{4}.

Thus, T = mg - m \cdot \frac{3g}{4} = \frac{mg}{4}.

Hence, the tension in the remaining string immediately after the other string is cut is \frac{mg}{4}.