Question

A uniform rod $AB$ of mass $2kg$ and length $1m$ is placed on a sharp support $O$ , such that $AO=a=25cm$ and $OB=b=75cm$ . A spring of force constant $600N{m}^{-1}$ is attached to the end $B$ , as shown. Initially, the spring is extended by $2cm$ at equilibrium. As soon as the thread is burnt, find the normal force (in $N$ ) exerted by the support at $O$ .

Answer 1

Answer: 8

$I$ about $O$ is $\frac{m{L}^2}{12}+\frac{m{L}^2}{16}=\frac{7m{L}^2}{48}$
Spring force $\Rightarrow K\Delta x=600\times \frac{2}{100}=12N$
Torque, just after burning the string,
$\tau =\left(2\times 10\times \frac{1}{4}\right)+12\times \frac{3}{4}$
$\Rightarrow \tau =5+9=14Nm$
$\therefore \alpha =\frac{14\times 48}{7\times 2\times 1}=48rad{s}^{-2}$
$\therefore$ Acceleration of COM,
$a=48\times \frac{1}{4}=12m{s}^{-2}$
Now, total vertical force
$12+20-N=2\times 12$
$\Rightarrow N=32-24=8N$