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).
Show Hint
Locate the instantaneous center of the bar using the perpendiculars to the velocities at A and B.
Step 1: Write coordinates as functions of the bar angle.
Let $\theta$ be the angle the bar makes with the floor. With A on the wall and B on the floor, $A = (0, L\sin\theta)$ and $B = (L\cos\theta, 0)$, so the midpoint is $G = (L\cos\theta/2, L\sin\theta/2)$.
Step 2: Differentiate with respect to time.
Since $L$ is fixed, differentiating gives $\dot{y}_A = L\cos\theta\, \dot\theta$ and the velocity components of $G$ as $\dot{x}_G = -\tfrac{1}{2}L\sin\theta\,\dot\theta$ and $\dot{y}_G = \tfrac{1}{2}L\cos\theta\,\dot\theta$.
Step 3: Combine the components.
The speed of G is $v_G = \sqrt{\dot{x}_G^2 + \dot{y}_G^2} = \tfrac{1}{2}L\dot\theta\sqrt{\sin^2\theta + \cos^2\theta} = \tfrac{1}{2}L\dot\theta$. Since $v_A = |\dot{y}_A| = L\cos\theta\,\dot\theta$, we get $L\dot\theta = v_A/\cos\theta$.
Step 4: Substitute the numbers.
$v_G = \dfrac{v_A}{2\cos\theta} = \dfrac{6}{2\cos45^{\circ}} = \dfrac{6}{1.4142} = 4.24$ m/s, at $\theta = 45^{\circ}$.
Final Answer:
Both the geometric route and the calculus route land on the same result once the constraint equations are set up correctly.
\[ \boxed{v_G = 4.24 \ \text{m/s}} \]
