Question

The velocity of point of contact $R$, of the rod just before impact with the bumper is (in m/s):

• A. $1.2$
• B. $\sqrt{5}$
• C. $1.2$$\sqrt{5}$
• D. $2.4$

Answer 1

Answer: (C)

$1.2$$\sqrt{5}$

$PQ=l$
Let $v_1$ be the velocity of point of contact R of rod
just before impact with the bumper and $v_2$ be the velocity after
impact.
Loss in P.E. = gain in rotational energy
$mg\frac{l}{2}(1-cos\theta )=\frac{1}{2}{I}_P{\omega }_1^2$
$\Rightarrow mg\frac{l}{2}(1-cos{60}^0)=\frac{mgl}{4}=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }_1^2$
$\Rightarrow {\omega }_1=\sqrt{\frac{3g}{2l}}$
Velocity of point of contact R of rod before impact
$v_1=x{\omega }_1=x\sqrt{\frac{3g}{2l}}$
$=0.6\sqrt{\frac{3\times 10}{2\times 0.75}}$
$=1.2\sqrt{5}m/s$