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).
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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 \(mg - ma_{cm}\).
Step 1: Understanding the Question:
A horizontal rod is initially supported by two vertical strings at its ends. One string is cut. We need to find the tension in the remaining string at that instant. Immediately after being cut, the rod will start to rotate about the point where the remaining string is attached. Step 2: Key Formula or Approach:
We will use Newton's second law for both linear and rotational motion.
1. Linear Motion (for the center of mass): \( \Sigma F_y = m a_{cm} \)
2. Rotational Motion (about the pivot): \( \Sigma \tau = I \alpha \)
The key is to relate the linear acceleration of the center of mass (\(a_{cm}\)) to the angular acceleration (\(\alpha\)). For rotation about one end, \(a_{cm} = r \alpha = (l/2)\alpha\). Step 3: Detailed Explanation:
Let's assume the string at the right end is cut. The rod will pivot about the left end, where the remaining string with tension T is attached. 1. Torque Equation:
The forces acting on the rod are tension T (upwards at the pivot) and gravity mg (downwards at the center of mass, at distance l/2 from the pivot).
The torque about the pivot is produced only by gravity.
\[ \tau_{\text{pivot}} = mg \times \frac{l}{2} \]
The moment of inertia of a uniform rod about one end is \( I = \frac{1}{3}ml^2 \).
Using \( \tau = I\alpha \):
\[ mg \frac{l}{2} = \left(\frac{1}{3}ml^2\right) \alpha \]
Solving for \(\alpha\):
\[ \alpha = \frac{mg(l/2)}{ml^2/3} = \frac{3g}{2l} \]
2. Force Equation:
The net vertical force on the center of mass causes its linear acceleration \(a_{cm}\).
\[ F_{net} = mg - T = m a_{cm} \]
The linear acceleration of the center of mass is related to the angular acceleration by \(a_{cm} = r\alpha\), where \(r = l/2\).
\[ a_{cm} = \frac{l}{2} \alpha = \frac{l}{2} \left(\frac{3g}{2l}\right) = \frac{3g}{4} \]
Now substitute this \(a_{cm}\) back into the force equation:
\[ mg - T = m \left(\frac{3g}{4}\right) \]
Solving for T:
\[ T = mg - \frac{3mg}{4} = \frac{mg}{4} \]
Step 4: Final Answer:
The tension in the remaining string immediately after the other is cut is \(mg/4\). This corresponds to option (C).
