Rotational Motion Question 157

Question: The centre of mass of three bodies each of mass 1 kg located at the points (0,0), (3,0) and (0,4) in the XY plane is

Options:

A) $\left( \frac{4}{3},1 \right)$

B) $\left( \frac{1}{3},\frac{2}{3} \right)$

C) $\left( \frac{1}{2},\frac{1}{2} \right)$

D) $\left( 1,\frac{4}{3} \right)$

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Answer:

Correct Answer: D

Solution:

[d]

$X _{CM}=\frac{m _{1}x _{1}+m _{2}x _{2}+m _{3}x _{3}}{m _{1}+m _{2}+m _{3}}$

$=\frac{1\times 0+1\times 3+1\times 0}{1+1+1}$

$Y _{CM}=\frac{m _{1}y _{1}+m _{2}y _{2}+m _{3}y _{3}}{m _{1}+m _{2}+m _{3}}$

$=\frac{1\times 0+1\times 0+1\times 4}{1+1+1}=\frac{4}{3}$

Therefore the coordinates of centre of mass are

$\left( 1,\frac{4}{3} \right)$



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