Two identical rods are joined at one of their ends by a pin. Joint is smooth and rods are free to rotate about the joint. Rods are released in vertical plane on a smooth surface as shown in the figure

Question

Two identical rods are joined at one of their ends by a pin. Joint is smooth and rods are free to rotate about the joint. Rods are released in vertical plane on a smooth surface as shown in the figure. The displacement of the joint from its initial position to the final position is (i.e. when the rods lie straight on the ground) :

TestHub · Accepted Answer

Take the initial contact point of the vertical rod with the ground as origin. The joint starts at $(0, L)$ ; the rod centres are at $(0, L /2)$ and $(L /2, L)$ , so $x_{c m} = \frac{M \cdot 0 + M \cdot L /2}{2 M} = \frac{L}{4}$ .

The surface is smooth, so no external horizontal force acts and $x_{c m}$ stays $\frac{L}{4}$ . When the rods lie straight on the ground they form one line of length $2 L$ with the joint in the middle, so the system's centre of mass is at the joint itself: the joint ends at $x = \frac{L}{4}$ .

Displacement of the joint from $(0, L)$ to $(\frac{L}{4}, 0)$ : $Δ s = (\frac{L}{4})^{2} + L^{2} = \frac{17 L ^{2}}{16} = \frac{17 L}{4}$ - (B)

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