Centre Of Mass Of A Triangle

The concept of the centre of mass of a triangle is not just a fundamental principle in physics and mathematics but also has practical applications in our everyday lives. The centre of mass, or centroid, is the point where the entire mass of a triangle can be considered to be concentrated. In simpler terms, it’s the point where you can balance the triangle perfectly on the tip of a pencil. This concept is crucial in fields like engineering and architecture, where understanding the balance and stability of structures is essential. For instance, when designing bridges or buildings, engineers calculate the centroid to ensure that the structure can support its own weight evenly and remain stable under various conditions. Even in nature, the centre of mass plays a role in the way animals move and maintain balance. Understanding this principle helps us create safer, more efficient designs in both the man-made and natural worlds.

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This Story also Contains

1. Definition of Centre of Mass
2. Centre of Mass of a Triangle
3. Solved Examples Based on the Centre of Mass of a Triangle
4. Summary

Definition of Centre of Mass

The Centre of mass of a body is defined as a single point at which the whole mass of the body or system is imagined to be concentrated and all external forces are applied there. It is the point where if a force is applied it moves in the direction of the force without rotating.

Centre of Mass of a Continuous Distribution

The centre of mass of a continuous distribution is a key concept in physics that extends beyond simple, discrete systems to more complex, continuous ones. Unlike objects with distinct masses located at specific points, continuous distributions involve mass spread over a region, such as a rod, a plate, or even a fluid. To find the centre of mass in such cases, we consider each infinitesimally small mass element and calculate its contribution to the overall position.

$x_{c m}=\frac{\int x d m}{\int d m}, y_{c m}=\frac{\int y d m}{\int d m}, z_{c m}=\frac{\int z d m}{\int d m}$

Where dm is the mass of the small element. x, y, z are the coordinates of the dm part.

Centre of Mass of a Triangle

The centre of mass of a triangle, also known as the centroid, is a crucial concept in geometry and physics. It represents the point where the entire mass of the triangle is considered to be concentrated, effectively balancing the triangle perfectly. This point is found by intersecting the triangle's medians—the lines drawn from each vertex to the midpoint of the opposite side. The centroid divides each median into a 2:1 ratio, with the longer segment being closer to the vertex.

Where dm is the mass of the small element, x, y, z are the coordinates of the dm part.

Have a look at the figure of A triangular plate as shown in the figure.

Since it is symmetrical about the y-axis on both sides of the origin

So we can say that its $x_{c m}=0$

And it's $z_{c m}=0$ as the z-coordinate is zero for all particles of the semicircular ring.

Now we will calculate its which is given by

$y_{c m}=\frac{\int y \cdot d m}{\int d m}$

For this take an elemental strip of mass dm and thickness dy at a distance y from the origin on the y-axis

As shown in the figure

$\triangle A D E$ and $\triangle A B C$ will be similar

So,

$
\begin{aligned}
\frac{r}{R} & =\frac{H-y}{H} \\
r & =\left(\frac{H-y}{H}\right) R \\
\sigma & =\frac{\text { mass }}{\text { area }}=\frac{M}{\frac{1}{2} *(2 R) * H}
\end{aligned}
$
$
\sigma=\frac{M}{R H}
$
And, $\quad d m=\sigma d A=\sigma(2 r d y)$

$
\begin{aligned}
& y_{c m}=\frac{\int y \sigma d A}{M} \\
& y_{c m}=\frac{\int_H^0 y \cdot \sigma d y \cdot 2\left(\frac{H-y}{H}\right) \cdot R}{M}=\frac{H}{3}
\end{aligned}
$
So, $\mathrm{y}_{\mathrm{cm}}=\frac{\mathrm{H}}{3}$ from base

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Solved Examples Based on the Centre of Mass of a Triangle

Example 1: What is the centre of mass of a triangular lamina from the vertex of the triangular lamina if its height is H?

1) $\frac{H}{3}$
2) $\frac{4 H}{3}$
3) $\frac{2 H}{3}$
4) $\frac{H}{6}$

Solution:

We know that the centre of mass of the triangular plate has its centre of mass at a distance of $\frac{H}{3}$ from the base of the triangular plate. So from the vertex of the triangular plate, it is -

$\Rightarrow H-\frac{H}{3}=\frac{2 H}{3}$

Hence, the answer is the option (3).

Example 2: What is the centre of mass (in cm) of a triangular lamina from the vertex of the triangular lamina if its height is $20 \mathrm{~cm} ?$

1) 667

2) 13.33

3) 6.67

4) 8

Solution:

We know that the centre of mass of the triangular plate has its centre of mass at a distance of $\frac{H}{3}$ from the base of the triangular plate. So from the vertex of the triangular plate, it is -

$\Rightarrow H-\frac{H}{S}=\frac{2 H}{3}$

So, putting the value $H=20 \mathrm{~cm}$

From there we get the centre of mass of the triangular lamina $=13.33 \mathrm{~cm}$ from the vertex.

Hence, the answer is 13.33.

Example 3: Three identical spheres each of mass M are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to 3 m each. Taking the point of intersection of mutually perpendicular sides as the origin, the magnitude of the position vector of the centre of mass of the system will be $\sqrt{\mathrm{x}}$ M. The value of x is_____________.

1) 2

2) 3

3) 4

4) 5

Solution:

$
\begin{aligned}
& x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
& =\frac{M(0)+M(3)+M(0)}{3 M} \\
& x_{\mathrm{cm}}=1 \\
& Y_{\mathrm{cm}}=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3} \\
& =\frac{M(0)+M(0)+M(3)}{3 M} \\
& Y_{\mathrm{cm}}=1 \\
& \text { Co-ordination of centre of mass }=\left(x_{\mathrm{cm}}, y_{\mathrm{cm}}\right)=(1,1) \\
& r=\sqrt{1^2+1^2}=\sqrt{2}
\end{aligned}
$
The value of $\mathrm{x}=2$

Summary

The centre of mass of a triangle, or centroid, is the point where the entire mass of the triangle can be considered to be concentrated, balancing it perfectly. It is found at the intersection of the medians, dividing each median in a 2:1 ratio. Understanding this concept is essential in various fields, such as engineering, where it is used to design stable structures. The centroid of a triangular lamina is located at a distance of H/3 from the base, and in real-life applications, this principle aids in ensuring the balance and stability of objects and systems.