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A mass $M= 40\ kg$ is fixed at the very edge of a long plank of mass $80\ kg$ and length $1\ m$ which is pivoted such that it is in equilibrium. How far (approx.) from the pivot should a mass of $100\ kg$ be attached so that the plank starts rotating with an angular acceleration of $1\ rad/s^2$?
$\frac{1}{{75}}\,m$
$\frac{1}{{25}}\,m$
$\frac{1}{{15}}\,m$
$\frac{1}{{45}}\,m$
Solution
$80 \mathrm{gl}=40 \mathrm{g}\left(\frac{1}{2}-l\right)$
$2 l=\frac{1}{2}-l$
$3 l=\frac{1}{2}$
$l=\frac{1}{6} \mathrm{m}$
$100 \mathrm{g} \mathrm{x}=1 \alpha \Rightarrow 1000 \mathrm{x}$
$=\left(\frac{80 \times 1^{2}}{12}+80 \times\left(\frac{1}{6}\right)^{2}+40 \times\left(\frac{1}{3}\right)^{2}+100 \times x^{2}\right) \times 1$
$=\frac{\left(240+80+160+3600 x^{2}\right)}{36}$
$\Rightarrow 300 x^{2}-3000 x+40=0$
$\Rightarrow 30 \mathrm{x}^{2}-300 \mathrm{x}+4=0$
$x=\frac{300 \pm \sqrt{90000-480}}{60}$
$\approx \frac{240}{300 \times 60}$
