First slide

Location of centre on mass for multiple point masses

Question

A thin uniform rod of length L is bent at its mid point as shown in the figure. The distance of centre of mass form the point ‘O’ is

Easy

A

$\large \frac{L}{2}\cos \frac{\theta }{2}$
B

$\large \frac{L}{4}\sin \frac{\theta }{2}$
C

$\large \frac{L}{4}\cos \frac{\theta }{2}$
D

$\large \frac{L}{8}sin\theta$

Solution

C₁ and C₂ are the COMs of the rods OA and OB respectively. Since the rods have equal masses, COM of the system must lie at the mid-point of the line C₁C₂. Now, OC₁=OC₂=L/4
$\large \therefore \;{C_1}C_2^2 = {\left( {\frac{L}{4}} \right)^2} + {\left( {\frac{L}{4}} \right)^2} - 2.\frac{1}{4}.\frac{1}{4}.COS\theta = {\left( {\frac{L}{2}\sin \frac{\theta }{2}} \right)^2}$
$\large \therefore \;{C_1}C = {C_2}C = \frac{L}{4}\sin \frac{\theta }{2}$
$\large \therefore OC = O{C_1}\sin \left( {\frac{{\pi - \theta }}{2}} \right) = \frac{L}{4}\sin \frac{\theta }{2}.\cos \frac{\theta }{2} = \frac{L}{8}\sin \theta$

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