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Q. 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 $12 \,cm$ mark, the stick is found to be balanced at $45 \,cm$. The mass of the metre stick is

System of Particles and Rotational Motion

Solution:

Let $m$ be the mass of the metre stick concentrated at $C$, the $50 \,cm$ mark as shown in the figure.
image
In equilibrium, taking moments of forces about $C'$, we get
$10 g(45-12) = mg (50-45);$
$10g \times 33 = mg \times 5$
$ m = \frac{10 \times 33}{5} = 66\,g$