Question

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).

Hint:

Locate the instantaneous center of the bar using the perpendiculars to the velocities at A and B.

Answer 1

Correct Answer: 4.24

Step 1: Set up the geometry.
End A slides on the vertical wall, so its velocity \(v_A\) stays vertical. End B slides on the floor, so its velocity \(v_B\) stays horizontal. The bar makes \(45^{\circ}\) with the floor, and G is the midpoint of AB.

Step 2: Find the instantaneous center (IC) of the bar.
The IC lies where the perpendicular to \(v_A\) (a horizontal line through A) meets the perpendicular to \(v_B\) (a vertical line through B). Taking A on the wall at height \(L\sin45^{\circ}\) and B on the floor at distance \(L\cos45^{\circ}\) from the wall, the IC sits directly above B at the same height as A.

Step 3: Get the distance from IC to A and to G.
The distance from IC to A equals \(L\cos45^{\circ}\), so the angular velocity is \(\omega = v_A/(L\cos45^{\circ})\). Since G is the midpoint of AB, the distance from IC to G works out to \(L/2\), because IC, A and B form a right angle at IC with legs \(L\cos45^{\circ}\) and \(L\sin45^{\circ}\), and the midpoint of the hypotenuse AB sits at \(L/2\) from that right-angle vertex.

Step 4: Compute \(v_G\).
\(v_G = \omega \times (L/2) = \dfrac{v_A}{L\cos45^{\circ}} \times \dfrac{L}{2} = \dfrac{v_A}{2\cos45^{\circ}}\).
Putting \(v_A = 6\) m/s and \(\cos45^{\circ} = 0.7071\):
\(v_G = \dfrac{6}{2 \times 0.7071} = \dfrac{6}{1.4142} = 4.24\) m/s.

Final Answer:
The midpoint of a bar sliding on two perpendicular walls always sits at distance \(L/2\) from the instantaneous center, giving a clean closed form for its speed. \[ \boxed{v_G = 4.24 \ \text{m/s}} \]