Locate the centre of mass (C) of a system of three particles of masses 1 gram, 2 gram and 3 gram placed at the corners of an equilateral triangle of 1 metre side.

Question

Accepted Answer

>

> [!TLDR]
> Using coordinates for the equilateral triangle and the weighted-average centre-of-mass formula, the centre of mass lies at approximately (0.58 m, 0.43 m) measured from the 1 g vertex, about 0.73 m from it.
>
> Let the three masses be $m_1 = 1\ 	ext{g}$ at vertex A, $m_2 = 2\ 	ext{g}$ at vertex B, and $m_3 = 3\ 	ext{g}$ at vertex C (the apex), forming an equilateral triangle of side $a = 1\ 	ext{m}$, matching the figure's layout (A at the origin, B on the x-axis, C as the apex above the base).
>
> Take coordinates:
> $$A = (0, 0), \quad B = (1, 0), \quad C = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight) = (0.5,\ 0.866)$$
>
> The centre of mass coordinates are the mass-weighted average of the position vectors:
> $$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1+m_2+m_3} = \frac{1(0) + 2(1) + 3(0.5)}{1+2+3} = \frac{0+2+1.5}{6} = \frac{3.5}{6} \approx 0.58\ 	ext{m}$$
> $$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1+m_2+m_3} = \frac{1(0) + 2(0) + 3(0.866)}{6} = \frac{2.598}{6} \approx 0.43\ 	ext{m}$$
> …

Q.Locate the centre of mass (C) of a system of three particles of masses 1 gram, 2 gram and 3 gram placed at the corners of an equilateral triangle of 1 metre side.

Concept understanding — Center of Mass

What is the Center of Mass?

Why does this matter?

The precise definition

Breaking it down with an example

Key properties to remember

Q.Locate the centre of mass (C) of a system of three particles of masses 1 gram, 2 gram and 3 gram placed at the corners of an equilateral triangle of 1 metre side.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days