Question

For the system shown in the figure, assume that the cylinder remains in contact with the two wedges moving with constant velocity as indicated. What will be the velocity of cylinder?

• A. $\sqrt{19-4\sqrt{3}}\frac{u}{2}\text{m/s}$
• B. $\frac{\sqrt{13}u}{2}\text{m/s}$
• C. $\sqrt{3}u\text{m/s}$
• D. $\sqrt{7}u\text{m/s}$

Answer 1

The correct option is D $\sqrt{7}u\text{m/s}$

Method - 1

Let the downward velocity of the cylinder be $V$, whose horizontal and vertical components are $V_x$ and $V_y$ as shown in the figure below.
As cylinder will remains in contact with wedge $A$,
$\Rightarrow V_x=2u$


From the wedge constraint, the components of the velocities of cylinder and wedge along the common normal should be equal.

So, $u\sin {30}^{\circ }=V_y\cos {30}^{\circ }-V_x\sin {30}^{\circ }$
Or, $V_y=V_x\frac{\sin {30}^{\circ }}{\cos {30}^{\circ }}+\frac{u\sin {30}^{\circ }}{\cos {30}^{\circ }}$

Or, $V_y=V_x\tan {30}^{\circ }+u\tan {30}^{\circ }$
$\because V_x=2u$
$\Rightarrow V_y=3u\tan {30}^{\circ }=\sqrt{3}u$
Thus, $V=\sqrt{V_x^2+V_y^2}=\sqrt{(2u{)}^2+(\sqrt{3}u{)}^2}=\sqrt{7}u$

Method - II

In the frame of $A$,
Velocity of B w.r.t A, $V_{BA}=V_B-V_A=u-(-2u)=3u$

From the wedge constraint, the components of the velocities of cylinder and wedge along the common normal should be equal.

$3u\sin {30}^{\circ }=V_y\cos {30}^{\circ }$
$\Rightarrow V_y=3u\tan {30}^{\circ }=\sqrt{3}u$
$\text{and}{V}_x=2u\Rightarrow V=\sqrt{V_x^2+V_y^2}=\sqrt{7}u$