Centre of Mass 1 Question 5
5. An $L$-shaped object made of thin rods of uniform mass density is suspended with a string as shown in figure. If $A B=B C$ and the angle is made by $A B$ with downward vertical is $\theta$, then
(2019 Main, 9 Jan)
(a) $\tan \theta=\frac{2}{\sqrt{3}}$
(b) $\tan \theta=\frac{1}{2 \sqrt{3}}$
(c) $\tan \theta=\frac{1}{2}$
(d) $\tan \theta=\frac{1}{3}$
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Answer:
Correct Answer: 5. (d)
Solution:
- Key Idea The centre of mass of a thin rod of uniform density lies at its centre.
The given system of rods can be drawn using geometry as,
where, $(COM) _1$ and $(COM) _2$ are the centre of mass of both rods $A B$ and $A C$, respectively.
So, in $\Delta A^{\prime} B B^{\prime}$,
$$ \tan \theta=\frac{A^{\prime} B^{\prime}}{A^{\prime} B}=\frac{\frac{a}{4}}{\frac{3 a}{4}}=\frac{1}{3} \text { or } \tan \theta=\frac{1}{3} $$
