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