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A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5\; g$ are put one on top of the other at the $12.0 \;cm$ mark, the stick is found to be balanced at $45.0\; cm$. What is the mass of the metre stick?
Solution
Let $W$ and $W'$ be the respective weights of the metre stick and the coin.
The mass of the metre stick is concentrated at its mid-point, i.e., at the $50 cm$ mark.
Mass of the meter stick $=m'$
Mass of each coin, $m=5 g$
When the coins are placed $12 cm$ away from the end $P$, the centre of mass gets shifted by $5 cm$ from point $R$ toward the end $P$. The centre of mass is located at a distance of $45 cm$ from point $P$
The net torque will be conserved for rotational equilibrium about point $R$. $10 \times g (45-12)-m^{\prime} g (50-45)=0$
$\therefore m^{\prime}=\frac{10 \times 33}{5}=66 g$
Hence, the mass of the metre stick is $66 g$
