Question

A point mass $M=40kg$ is fixed at the very edge of a long plank of mass $80kg$ and length $1m$ which is pivoted such that it is in equilibrium. How far (approx.) from the pivot should a point mass of $100kg$ be attached so that the plank starts rotating with an angular acceleration of $1rad/s^2$ ?

• A. $\frac{1}{75}m$
• B. $\frac{1}{25}m$
• C. $\frac{1}{15}m$
• D. $\frac{1}{45}m$

Answer 1

Answer: (A)

$80gl=40g\left(\frac{1}{2}-l\right)$
$2l=\frac{1}{2}-l$
$3l=\frac{1}{2}$
$l=\frac{1}{6}m$

$=I\alpha =\left(\frac{80\times 1^2}{12}+80\times {\left(\frac{1}{6}\right)}^2+40\times {\left(\frac{1}{3}\right)}^2+100\times x^2\right)\times 1$
$\Rightarrow 1000x=\left(\frac{80}{12}+\frac{80}{36}+\frac{40}{9}+100x^2\right)$
$=\frac{\left(240+80+160+3600x^2\right)}{36}$
$300x^2-3000x+40=0$
$\Rightarrow 30x^2-300x+4=0$
$x=\frac{300\pm \sqrt{90000-480}}{60}$
$\approx \frac{300\pm 300\left(1-\frac{480}{90000}\times \frac{1}{2}\right)}{60}$
$\approx \frac{240}{300\times 60}$