Q5.2 Calculating the centre of mass II (S)
Equal masses are placed at three corners of a square. Where is the centre of mass of the system?Hint
Reveal
Solution
Reveal
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![\[ x = \frac{m_0 \times 0 + m_1 \times R + m_2 \times 0}{m_0 + m_1 + m_2} = R/3 \]](mastermathpng-0.png)

Solution
First we have to decide on a coordinate system: Lets draw a picture:
Choose the origin as the bottom left corner of the square; define the length of the sides of the square to be R and label the masses m0, m1 and m2. |
The vectors of the three masses are (0,0), (R,0) and (0,R). Therefore using Key Point 4.1, the centre of mass vector is (x,y) with
![\[ x = \frac{m_0 \times 0 + m_1 \times R + m_2 \times 0}{m_0 + m_1 + m_2} = R/3 \]](mastermathpng-0.png)
![\[ y = \frac{m_0 \times 0 + m_1 \times 0 + m_2 \times R}{m_0 + m_1 + m_2} = R/3 \]](mastermathpng-1.png)