Find the centre of mass of three particles, 100 gm, 150 gm and 200 gm placed at the vertices of an equilateral triangle of each side 0.5 m long. (Take 100 gm at origin and 150 gm along X-axis).

Question

Accepted Answer

>

> [!TLDR]
> Place the triangle's vertices using the given side length, then use the weighted-average formula for the centre of mass in each coordinate.
>
> **Setting up coordinates:** Equilateral triangle of side $a = 0.5$ m. Place:
> - $m_1 = 100$ g at the origin: $(x_1, y_1) = (0, 0)$
> - $m_2 = 150$ g along the X-axis at distance a: $(x_2, y_2) = (0.5, 0)$
> - $m_3 = 200$ g at the third vertex. For an equilateral triangle with the base along the X-axis from (0,0) to (0.5,0), the third vertex lies at the midpoint horizontally and at height $a\sin 60^\circ$:
> $(x_3, y_3) = \left(0.25,\ 0.5\sin 60^\circ\right) = (0.25,\ 0.433)$ m
>
> **Centre of mass formulas:**
> $X_{cm} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$, $\quad Y_{cm} = \dfrac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$
>
> **Substituting** (masses in grams, positions in metres; units cancel since they appear identically in numerator and denominator ratios):
> …

Q.Find the centre of mass of three particles, 100 gm, 150 gm and 200 gm placed at the vertices of an equilateral triangle of each side 0.5 m long. (Take 100 gm at origin and 150 gm along X-axis).

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.Find the centre of mass of three particles, 100 gm, 150 gm and 200 gm placed at the vertices of an equilateral triangle of each side 0.5 m long. (Take 100 gm at origin and 150 gm along X-axis).

🎓Unlock everything free for 14 days

Unlock everything free for 14 days