GATE 2026 ME – Question 50
A rigid slender bar, AB, is sliding against two mutually perpendicular frictionless walls, as shown in the figure below. The velocity of A in the downward direction at a given instant is 6 m/s. At that instant, the magnitude of absolute velocity of the midpoint G is ________ m/s (*rounded off to 2 decimal places*).

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Correct answer: 4.22 to 4.26
Explanation
The instantaneous centre of rotation is at the corner of the rectangle formed by the two walls, a distance $L\cos 45°$ horizontally from A. So $v_A = \omega L\cos 45°$, which gives $\omega L = \frac{6}{0.7071} = 8.485$. The midpoint is at a distance $\frac{L}{2}$ from the instantaneous centre, so $v_G = \frac{\omega L}{2} = 4.24$ m/s.