Integrals: Centre of Mass and Centre of Gravity

CENTRE OF MASS AND CENTRE OF GRAVITY

As a physical application of multiple integral, we shall see centre of mass and cen- tre of gravity of two dimensional and three dimensional objects.

Many Mechanical systems and structures behave as if their masses were con- centrated at a single point. We call this point the system's centre of mass.

A rigid body is composed of many particles, each of which is affected by gravity. These individual forces can be replaced by a single force acting at a point, called centre of mass or centre of gravity.

A two dimensional or three dimensional object or body is homogeneous if its composition is uniform in the entire body. Otherwise we call it inhomogeneous body.

1. Centre of Mass of a Two Dimensional Object – A Lamina 

In practical situations we have to find the centre of mass of a flat thin plate, a disk of aluminum, a triangular sheet of steel etc. Such a flat object is called a lamina, which can be considered as a two dimensional plane region.

The density of a homogeneous lamina is defined as mass per unit area.

For a homogeneous lamina, if total mass is M and total area is A, then the density ρ = M/A

For an inhomogeneous lamina density varies from point to point.

Let (x, y) be a point of the lamina and ρ (x, y) is the density at the point with element area ΔA and element mass ΔM at the point

2. Centroid

In the special case, when the lamina is homogeneous, density ρ is a constant. Then the centre of mass or centre of gravity is called the centroid of the lamina.

3. Centre of Mass or Centre of Gravity of Three Dimensional Solid

Let (x, y, z) be a point inside a volume V. Let ρ (x, y, z) be the volume density at the point (x, y, z) and ΔV = dxdydz be the element volume at the point and ΔM is the element mass of the point,

Then the mass of the solid is given by

Here also when the density ρ (mass/ unit Volume) is a constant, then the centre of mass or centre of gravity is called the centroid of the solid.

WORKED EXAMPLES

Example 1 

Find the Centre of gravity of a quarter of the ellipse assuming surface density 1 at all points.

Solution

where D is the shaded region in the Fig 5.84.

Example 2 

Find the centroid of the area enclosed by the parabola y2 = 4ax, the x-axis and its latus rectum.

Solution

The given region is the area bounded by y2 = 4ax, the x-axis and its latus rectum x = a. We have to find the centroid.

⸫ Density is a constant, say 1.

⸫ ρ(x, y) = 1          (1)

Let M be the mass of the bounded area D as in Fig 5.85.

In D, the limits of x are x = 0 and x = a

To find the limits of y, take a strip parallel to the y-axis.

Example 3 

Find the mass and centre of mass of the rectangle 0 ≤ x ≤ 4, 0 ≤ y ≤ 2 having density ρ(x, y) = 2xy.

Solution

Given the rectangular region 0 ≤ x ≤ 4, 0 ≤ y ≤ 2, with variable density ρ = 2xy. 

Let M be the mass of the bounded area D as in Fig 5.86.

Example 4 

Find the centre of gravity of the triangular lamina with vertices (0, 0), (0, 1), (1, 0) and the surface density ρ(x, y) = xy.

Solution

The area of the region is bounded by

x + y = 1, x = 0, y = 0

Given variable surface density is xy

⸫     ρ(x, y) = xy     (1)

Let M be the mass of the bounded area D as in Fig 5.87.

Example 5 

Find the centroid of the upper half of the circle x2 + y2 = a2.

Solution

Required the centroid of the plane region, the semi circle x2 + y2 = a2.

Example 6 

Find the mass and centroid of the volume bounded by the coordinates planes and the plane with volume density ρ.

Solution

Given the volume bounded by the coordinates and the plane

To find z limits, take a strip parallel to the z-axis, lower end z = 0 and upper end is on the plane

Similarly, by symmetry

Example 7 

Find the centre gravity of the solid octant of the ellipsoid if the density at any point varies as xyz. 

Solution

We shall find the centre gravity of the solid in the first octant of the ellipsoid.

In the first octant,

x varies from 0 to a

Let M be the volume of the bounded region.

Similarly, by symmetry, we get

Example 8 

Find the centre of mass if a solid of constant density ρ bounded below by the disk x2 + y2 ≤ 4 in the z = 0 plane and above by the paraboloid z = 4 − x2 - y2.

Solution

The solid is bounded by the paraboloid z = 4 − x2 - y2 and the disk x2 + y2 ≤ 4  given volume density is a constant and equal to ρ.

Put x = 2 sin θ  ⸫ dx = 2 cos θ dθ

EXERCISE 

1. Find the mass and centre of gravity of the rectangular region in the xy plane 0 ≤ x ≤ 4, 0 ≤ y ≤ 2 having variable density xy.

2. Find the mass of the lamina bounded by the curve y = x2 - 2x and y = 2x if the density at any point is xy. Also find the centre of mass.

3. Find the centroid of the plane region bounded by y2 = -x and y = x + 2 with constant density P.

4. Find the centre of mass of a solid of mass a solid of constant density bound below by the paraboloid z = x2 + y2 above the plane z = 4.

5. A solid cube in the first octant is bounded by the coordinate planes and by the planes x = 1, y = 1, z = 1 and the density if the cube is P(x, y, z) = x + y + z + 1. Find the mass of the solid and the centre of mass.

6. Find the mass and centre of mass of a triangular lamina with vertices (0, 0), (1, 0) and (0, 2) if the density is δ (x, y) = 1 + 3x + y

7. Find the centre of mass of a semicircular lamina if the density at any point is proportional to the distance from the centre of the circle.

[Hint: ρ(x, y) = use polar coordinates]

8. Find the mass and centre of mass of the triangular region with vertices (0, 0), (2, 1) (0, 3) and density ρ(x, y) = x + y

ANSWERS TO EXERCISE