Question

A uniform thin rod of weight W is supported horizontally by two vertical props at its ends. At $t=0,$ one of these supports is kicked out. The force on the other support immediately thereafter is

• A. $4W$
• B. $\frac{W}{2}$
• C. $2W$
• D. $\frac{W}{4}$

Answer 1

The correct option is D $\frac{W}{4}$

$W-F=ma$
angular motion
$\frac{WL}{2}=I\theta$
$I$ at the end of rod $=\frac{1}{3}m{L}^2$
$\theta =\frac{a}{\left(\frac{L}{2}\right)}$
$\frac{WL}{2}=\frac{I}{3}m{L}^2\times \frac{a}{4}$
$\frac{W}{2}=\frac{2ma}{3}$
$W=\left(\frac{4ma}{3}\right)$
$ma=\left(\frac{3}{4}W\right)$
$W-F=\frac{3}{4}W$
$\frac{W}{4}=F$