Question

A rigid rod is placed against the wall as shown in the figure. When velocity of its lower end is 10 m/s and its base makes an angle \(\alpha\) =60° with horizontal, then the vertical velocity of its end B will be

Answer 1

Answer:

The velocity of its end B will be (10√3)/3 m/s

Explanation:

The figure is attached

Here α = 60°

Since the rod is rigid, the component of the velocities along the rod will be equal

Thus

$V_A\cos60^\circ=V_B\cos30^\circ$

Here $V_A=10 \text{ m/s}$

Thus

$10\cos60^\circ=V_B\cos30^\circ$

$\implies 10\times \frac{1}{2} =V_B\times \frac{\sqrt{3}}{2}$

$\implies V_B=\frac{10}{\sqrt{3}}$ m/s

or, $V_B=\frac{10\sqrt{3}}{3} \text{ m/s}$

Hope this helps.