Centre of Mass of Continuous Bodies

The concept of the centre of mass (COM) is crucial in understanding the motion of a system of particles or a continuous body. For a system of discrete particles, the centre of mass is the point where the total mass of the system can be considered to be concentrated. For continuous bodies, the definition is similar, but the calculation involves an integration over the volume of the object.

Definition

The centre of mass of a continuous body is the point at which the distribution of mass is balanced in all directions. It is the point that moves as if all the mass of the body were concentrated there and all external forces were applied there.

Calculation

The position of the centre of mass $\vec{R}_{\text{COM}}$ of a continuous body is given by the integral:

$$ \vec{R}_{\text{COM}} = \frac{1}{M} \int \vec{r} \, dm $$

where:

• $\vec{r}$ is the position vector of a mass element $dm$
• $M$ is the total mass of the body, given by $M = \int dm$

For bodies with uniform density $\rho$, the integral can be expressed in terms of volume $dV$:

$$ \vec{R}_{\text{COM}} = \frac{1}{M} \int \vec{r} \, \rho \, dV $$

Differences and Important Points

Aspect	Discrete System	Continuous Body
Basic Definition	Weighted average of positions of individual masses	Point where mass distribution is balanced
Calculation	Summation: $\vec{R}_{\text{COM}} = \frac{1}{M} \sum m_i \vec{r}_i$	Integration: $\vec{R}_{\text{COM}} = \frac{1}{M} \int \vec{r} \, dm$
Density	Not applicable (individual point masses)	Uniform or variable density considered
Mass Element	Individual mass $m_i$	Infinitesimal mass element $dm$
Total Mass	Sum of individual masses: $M = \sum m_i$	Integral of mass elements: $M = \int dm$

Examples

Example 1: Centre of Mass of a Rod

Consider a uniform rod of length $L$ and mass $M$. To find its centre of mass, we can assume the rod is lying along the x-axis with one end at the origin.

Since the rod has uniform density, its centre of mass will be at the midpoint. Mathematically, we can express the position of the centre of mass as:

$$ x_{\text{COM}} = \frac{1}{M} \int_0^L x \, dm $$

Since the density is uniform, $dm = \rho \, dx = \frac{M}{L} dx$. Substituting this into the integral, we get:

$$ x_{\text{COM}} = \frac{1}{M} \int_0^L x \frac{M}{L} dx = \frac{1}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{L}{2} $$

So the centre of mass is at $L/2$, which is the midpoint of the rod.

Example 2: Centre of Mass of a Hemisphere

Consider a solid hemisphere of radius $R$ and mass $M$. To find its centre of mass, we can use spherical coordinates and symmetry arguments.

The centre of mass will lie on the axis of symmetry (the z-axis) and can be found by integrating over the volume of the hemisphere:

$$ z_{\text{COM}} = \frac{1}{M} \int z \, dm $$

Using spherical coordinates, $dm = \rho \, dV = \rho \, R^2 \sin(\theta) \, d\theta \, d\phi \, dr$. The limits of integration for $\theta$ are $0$ to $\pi/2$, for $\phi$ are $0$ to $2\pi$, and for $r$ are $0$ to $R$. The integral becomes:

$$ z_{\text{COM}} = \frac{1}{M} \int_0^R \int_0^{\pi/2} \int_0^{2\pi} (r \cos(\theta)) \rho R^2 \sin(\theta) \, d\phi \, d\theta \, dr $$

Solving this integral will give the position of the centre of mass along the z-axis. The result is:

$$ z_{\text{COM}} = \frac{3R}{8} $$

This indicates that the centre of mass of a solid hemisphere is located at a height of $3R/8$ above the flat face.

Understanding the centre of mass of continuous bodies is essential in many areas of physics, including mechanics, dynamics, and even astrophysics. It simplifies the analysis of motion and provides insights into the behavior of complex systems.