CENTROIDS, CENTRE OF GRAVITY AND CENTRE OF MASS

Precession of a Spinning Top

Consider the spinning of top about an axis passing through point O. The weight Mg of the top acts through its C.G vertically downwards. If R be the distance of the C.G from O, the torque τ acting on the top is given by

τ =Mg Rsinα Also Δ =L Lsin α × Δθ

∴ sin

L L Δθ = Δ

α

Hence the rate of precession is given by

1 sin

sin sin sin

L Mg R Mg R Mg R

t L t L L L I

Δθ= Δ = τ = α = =

Δ α Δ α α ω

As the angular velocity decreases due to friction at the point O and air resistance, the rate of precession increases.

Uses of Gyroscopes

The effect of all this is that, once you spin a gyroscope, its axle wants to keep pointing in the same direction. If you mount the gyroscope in a set of gimbals so that it can continue pointing in the same direction, it will. This is the basis of the gyrocompass.

If you mount two gyroscopes with their axles at right angles to one another on a platform, and place the platform inside a set of gimbals, the platform will remain completely rigid as the gimbals rotate in any way they please. This is this basis of Inertial Navigation Systems (INS).

In an INS, sensors on the gimbals’ axles detect when the platform rotates. The INS uses those signals to understand the vehicle’s rotations relative to the platform. If you add to the platform a set of three sensitive accelerometers, you can tell exactly where the vehicle is heading and how its motion is changing in all three directions.

With this information, an airplane’s autopilot can keep the plane on course, and a rocket’s guidance system can insert the rocket into a desired orbit!

SAQ 9

(a) What is a Gyroscope?

(b) What is Precession?

1.10 CENTROIDS, CENTRE OF GRAVITY AND CENTRE OF MASS

It is often required to define a body such that the length of wire, area of a rectangular plate; the volume, the mass or the gravitational forces acting on a body may be assumed to be concentrated at that point. Such points are often called as central points. Some commonly used central points are :

Physics

n

• Centroid of the length of a curve, line.

• Centroid of the volume of a body.

• Mass centre of the mass of a body.

• Centre of gravity of the gravitational forces acting on a body.

Centre of Gravity

Centre of gravity of a body is a point through which the resultant of the distributed gravity forces act irrespective of the orientation of the body.

ΔM1

Consider a body of mass M. Let this body be composed of ‘n’ number of masses ΔΜ₁, ΔΜ₂, ΔΜ₃, . . . , ΔΜn, distributed within the body such that

Μ = ΔΜ1 + ΔΜ2 + . . . + ΔΜn

The distance of these masses with respect to the axes be, (x₁, y₁), (x₂, y₂), . . . , (xn, yn).

Let the centre of gravity of the whole mass M lie at a distance (xc , yc) with respect to the reference axes. Let us assume that the gravitational field is uniform and parallel. Gravitational force acting on the mass ΔΜ1 = ΔΜ1 g. Similarly, we can find the gravitational forces acting the masses ΔΜ2, ΔΜ3, . . . ΔΜn. Τo find the resultant of parallel forces ΔΜ1 g, ΔΜ2 g, ΔΜ3 g, . . . , ΔΜn g, we apply the principle of moments.

The moment of the resultant of all forces about the y-axis = the sum of the moments of all the forces about the y-axis :

Mg x_c = Δ( M g x₁ ) ₁ + Δ( M g x₂ ) ₂ +. . . (+ ΔM g x_n )

Mechanics

It is the point where the entire mass of the body may be assumed to be concentrated.

The centre of mass and centre of gravity of a body are different only when the gravitational field is not uniform and parallel.

The earth exerts a gravitational force on each of the particles forming a body.

These forces can be replaced by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body.

The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid.

Concept of Centroid

One Dimensional Body

Consider a body in the shape of a curved homogeneous wire of uniform cross-section and of a length L.

Figure 1.22

Divide the length of the wire into small elements of lengths ΔL1, ΔL2, ΔL3, . . . , ΔLn.

Let the uniform area of cross section = A, density of the wire is ρ.

The mass M of the wire of length L = A ρ L The mass of an element of length ΔL1 = ΔM1

ΔM1 = A ρ ΔL1

Similarly, the masses ΔM2, ΔM3, . . . , ΔMn of other elements can be determined.

Let the distance of the centers of these lengths with respect to the axes be (x₁, y₁), (x₂, y₂), . . . , (xn, yn).

Applying the principle of moments,

( )

1

Because the density, ρ, and the area of the cross section, A, are constant over the entire length of the wire, the coordinates of CG of the wire become the co-ordinates of the centroid of the wire : generally called the centroid of a line segment.

Centroid of Two-dimensional Body

We can derive the centroid of an area as we have derived for the line segment.

( )

xc, yc are the centroids of the plate, generally called as the coordinates of the centroid of an area. Again, as the density and thickness of the plate are constant over the entire area so, the coordinates of the CG become the coordinates of the centroid of the area.

Figure 1.23

Generally the term centroid is used for the centre of the gravity of a geometrical figure and the term centre of gravity is used when referring to actual physical bodies.

Determination of Centroid and Centre of Gravity

By considering a plane figure to be made up of a number of small elements of length or area, we cannot generate the true shape of the figure. To generate the shape of a figure, we have to make the size of these elements very small and their number very large.

Mathematically, we replace the terms ΔL and ΔA in the early expressions of centroid and CG by the integral term. Since ΔL and ΔA is considered infinitesimally small, expressions of centroid and CG are rewritten as

Mechanics

_c x dL

x

dL

=

∫

∫

_c x dA x

dA

=

∫

∫

_c x dm x

dm

=

∫

∫

_c y dL y

dL

=

∫

∫

_c y dA y

dA

=

∫

∫

_c y dm y

dm

=

∫

∫

where dL, dA and dm denote the length, area and mass respectively of a differential element chosen and the (xc, yc) the coordinates of its centroid. The integral

∫

_{x dA} ^is

known as the first moment of area w.r.t. the y-axis. Similarly, the integral

∫

y dA denotes the first moment of area w.r.t. the x-axis.

Figure 1.24

x A=

∫

x dA=Qy

= first moment with respect to y.

y A=

∫

y dA=Qx

= first moment with respect to x.

x L=

∫

x dL y L=

∫

y dL

Figure 1.25

Example 1.9

Compute the centroid of a triangular lamina.

Solution

Consider the right triangle depicted below. Suppose that the right triangle is constructed from a material with a uniform density of ρ units of mass per unit of area.

Physics ^y

We are going to compute the moment of the triangle about the y-axis.

• To simplify the calculation, we cut the triangle up into a set of thin vertical strips.

• We will compute the moment of each of the strips about the y-axis and then integrate to add up all of the moments.

The moment of a strip centered at x is given by

Moment = Mass Distance× =m x x( )

The mass of the strip, m (x), is given by the density ρ times the area of the strip.

The area is the product of the little bit of thickness dx and the height of the strip, f (x).

m x( )λ = ρA x( )= ρ f x dx( ) A simple argument shows that

( ) h

f x x

= b

Hence, Moment of the small strip w.r.t. y-axis h h ² x dx x x dx

b b

⎛ ⎞

= ρ⎜ ⎟ = ρ

⎝ ⎠

and Total moment of all such strips w.r.t. y-axis ^{2 3}

0 3 0

b b

h h

b x dx b

=

∫

ρ = ρ ^x

To compute the distance from the reference line to the center of mass, we use the formula

Total moment=(Total mass)d_c

or Total moment

= Total mass dc

The mass is easy to compute.

1

Mechanics

This says that if we lay a line parallel to the y-axis at 2

x=3 b, the triangle will balance exactly on the line. If you look back up at the picture of the triangle above, you will see that this answer appears to be perfectly reasonable.

Example 1.10

Determine the centroid of an area of a semi-circle as shown in Figure 1.27.

C

You will note that it is not possible to divide a semi-circle into a few suitable standard areas for each of which magnitude as well as location of its centroid is known. Hence in such a situation, we should consider a very large number of very thin strips of width dx and assume that its centroid is at the centre of the strip.

Consider the semi-circle with centre C and its straight boundary ECD along y-axis.

The axis CX is therefore the axis of symmetry of the area. This means that portion of quadrant CFE is the mirror image of quadrant CFD.

Let us consider of an area as a strip AB at a distance x from C with width dx tending to zero. If a is the radius of semi-circle,

then AB = 2a sin θ

Since value of Considering the first moment of elemental areas about y-axis, we have

A x = ∑ A x_{i i}

Determine the centroid of the shaded area shown in Figure 1.28.

C

as shown in Figure 1.28.

Now A= A₁ + A₂ − A₃

Mechanics

Determine the CG of the thin uniform wire bent into shape ABC as shown in Figure 1.29.

Taking moment about x-axis,

L y = ×5 2.5 5 6.5+ ×

or, y =4.5 m

Example 1.13

Determine the CG of a wire of uniform cross-section bent into shape of a semi-circle as shown in Figure 1.30.

Take radius of wire = a.

Refer Figure 1.30 showing length of wire π a, where a = radius of semi-circle with centre at 0.

(a) Determine the centroid of the area as shown in Figure 1.31 where D is the contact point of the circle to the edge GF.

Figure 1.31

Mechanics

(b) Determine the position of centroid of a quadrant OAB of a circle, where arc AB subtends an angle of 90^o at the centre O and radius OA = a.

(c) Determine the centroid of the area shown in Figure 1.32, where portion BCE is a quadrant of circle of radius = 10 cm.

10 cm 10 cm

y

D E

C

A B x

10 cm

Figure 1.32

(d) Determine the centroid of length in form of a channel ABCD as shown in Figure 1.33.

20 cm 10 cm

B A

C D

x M

Figure 1.33

1.11 SUMMARY

Let us summarise what we have learnt in this unit.

Motion is a common phenomenon but by mechanics point of view, it is an extremely important phenomenon. In this unit, we have understood the cause of motion and found that a body at rest may be set into motion, on being acted upon by external forces. We have also discussed Newton’s law of gravitation. A particle may move along a circular path with uniform or variable velocity. We have also discussed angular velocity, as well as, angular momentum. The concept of radius of gyration has also been introduced while discussing moment of inertia.

In this unit, we also learnt that whenever a constant force F acts on an object while it experiences a displacement, ‘d’, we say that the force does work W on the object. The amount of work done is a scalar quantity, and is calculated from W = F . d. The concepts of work, power, and energy have been explained in detail in this unit along with different forms of energy. The concept of gyroscope has also been introduced.

Centre of gravity is a point where the entire mass or weight is assumed to be

concentrated. Centre of mass, for a body with a system of particles is the point where the entire mass of the system is assumed to be concentrated. Generally, a given plane area can be divided into number of smaller areas Ai for which the centroid Gi (xi, yi) are known. The location of the centroid for a composite area can be computed with the help of following expressions :

, ^{i i}, ^{i i}

i

A x A x

A A x y

A A

∑ ∑

= ∑ = =

In document Unit 1.Newton's Laws (Page 35-46)