Question

Particles of mass 1 g, 2 g, 3 g, ...., 100 g are kept at the marks 1 cm, 2 cm, 3 cm, ..., 100 cm respectively on a metre scale, The MOI of the system of particles about perpendicular bisector of the metre scale is given as $\frac{40+x}{100}kg{m}^2$. Find $x$.

Answer 1

The MOI is calculated by adding MOI of individual particles.

For particle of $1cm-49cm$

MOI =$n{(50-n)}^2\Rightarrow total$

${\sum }_0^{49}(50-n)\times {10}^{-7}$

MOI of $51-100$ cm

MOI=$n{(n-50)}^2\Rightarrow {\sum }_{51}^{100}(n-50)\times {10}^{-7}$

$\therefore$total$=({\sum }_0^{49}(50-n{)}^2+{\sum }_{51}^{100}(n-50{)}^2)\times {10}^{-7}$

$={\sum }_0^{100}(n-50{)}^2\times {10}^{-7}$

$=0.43=\frac{43}{100}\therefore \frac{40+x}{100}=\frac{43}{100}\Rightarrow x=3$