First slide

Location of centre on mass for multiple point masses

Question

Three particles, each of mass m, are placed at the corners of a right angled triangle as shown in figure. If OA = a and OB = b, The position vector of the centre of mass is

Easy

A

$\large \frac 13(a\hat i+b\hat j)$
B

$\large \frac 13(a\hat i-b\hat j)$
C

$\large \frac 23(a\hat i+b\hat j)$
D

$\large \frac 23(a\hat i-b\hat j)$

Solution

$\large \vec {x_c} = \frac{{m(0) + m(a) + m(0)}}{{3m}}\hat i + \frac{{m(0) + m(0) + m(b)}}{3}\hat j$
$\large \vec x_c = \frac{a}{3}\hat i + \frac{{b\hat j}}{3} = \frac{1}{3}\left( {a\hat i + b\hat j} \right)$

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