Centre Of Mass Of A Solid Cone

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.

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

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

Centre of Mass of the Solid Cone

Have a look at the figure of a solid cone

Since it is symmetrical about the y-axis

$\text { So we can say that its } x_{c m}=0 \text { and } z_{c m}=0$

Now we will calculate its $y_{cm}$ which is given by

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

So Take a small elemental disc of mass dm of radius r at a vertical distance y from the bottom as shown in the figure.

So $d m=\rho d v=\rho\left(\pi r^2\right) d y$

Here
$
\rho=\frac{M}{V}=\frac{M}{\frac{1}{3} \pi R^2 H}
$

And from a similar triangle
$
\begin{aligned}
& \frac{r}{R}=\frac{H-y}{H} \\
& r=\left(\frac{H-y}{H}\right) R \\
& y_{c m}=\frac{\int y \cdot d m}{\int d m} \\
& y_{c m}=\frac{1}{M} \int_0^H y \cdot d m=\frac{1}{M} \int_0^H y \frac{3 M}{\pi R^2 H}\left(\pi r^2\right) d y=\frac{H}{4}
\end{aligned}
$

So, $\mathrm{y}_{\mathrm{cm}}=\frac{\mathrm{H}}{4}$ from the bottom o

Or, the Centre of Mass of a solid cone will lie at distance $\frac{3 h}{4}$ from the tip of the cone.

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Solved Examples Based on the Centre of Mass of Solid Cone

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 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 the option (2).

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 $\sqrt{\mathrm{x}} \mathrm{m}$. The value of $\mathrm{x}$ is_____________.

1) 2

2) 3

3) 4

4) 5

Key Concepts

Solution

$
\begin{aligned}
& \mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2+\mathrm{m}_3 \mathrm{x}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3} \\
& =\frac{\mathrm{M}(0)+\mathrm{M}(3)+\mathrm{M}(0)}{3 \mathrm{M}} \\
& \mathrm{x}_{\mathrm{cm}}=1 \\
& \mathrm{Y}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2+\mathrm{m}_3 \mathrm{y}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3} \\
& =\frac{\mathrm{M}(0)+\mathrm{M}(0)+\mathrm{M}(3)}{3 \mathrm{M}} \\
& \mathrm{Y}_{\mathrm{cm}}=1 \\
& \mathrm{Co} \text {-ordination of centre of mass }=\left(\mathrm{x}_{\mathrm{cm}}, \mathrm{y}_{\mathrm{cm}}\right)=(1,1) \\
& \mathrm{r}=\sqrt{1^2+1^2}=\sqrt{2}
\end{aligned}
$
The value of $\mathrm{x}=2$

Hence, the answer is the option (1).

Summary

The centre of mass is a critical concept in physics, representing the point where the entire mass of an object is considered to be concentrated. For a solid cone, the centre of mass lies along its central axis at a height of H/4 from the base. Understanding this concept is essential in fields like construction and sports, where balance and stability are key. The study extends to various shapes, like triangular laminas and systems of masses, emphasizing its broad applicability in real-world scenarios.