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Course by: Sunil Kumar Singh. E-mail the author

Center of mass and rigid bodies

Module by: Sunil Kumar Singh. E-mail the author

Summary: Under certain condition, COM of rigid bodies is same as geometric center.

Rigid bodies are composed of very small particles which interact with each other via electromagnetic force. They form a continuous distribution of mass. As such, expressions of COM in three coordinate directions involve evaluation of integrals as described in earlier module. This evaluation, however, is rendered difficult on two counts :

• Mass distribution may not be uniform.
• The body shape may be irregular.

The geometry of regularly shaped bodies are defined by mathematical equations. Such is not the case with irregular bodies. However, there is a good thing about center of mass (COM) that it represents the point where external force equivalently applies. This fact allows us to experimentally determine COM of even irregularly shaped bodies. We can balance a body on a pointed wedge. The COM of the body falls on the line of balance. In order to know the COM (a point), however, we need to balance the body with different orientation to get another line of balance. The point of intersection of the two lines of balance is the COM of the body.

COM of regular bodies with uniform density

We are saved from any mathematical calculation in cases of certain regularly shaped bodies with uniform density, which are symmetric to all the axes of the coordinate system involved. In all such cases, COM is same as geometric center. COMs of a sphere, spherical shell, ring, disc, cylinder, cone, rod, square plate etc. fall under this category where COM is simply the geometric center of the bodies.

However, there are cases of regular shaped bodies which are not symmetric to three (for three dimensional bodies) or two axes (for planar bodies). For example, consider the case of hemispherical body or the case of semi-circular wire. These bodies are not symmetric to all the axes involved. Here, geometric center and, thus, COM are not obvious. In this section, we shall evaluate COM of such regular shaped bodies, which are not symmetric about all axes of the coordinate system.

Regular bodies allow us to evaluate integrals as geometry is defined. Evaluation of integral is simplified if the mass is evenly distributed. Unless otherwise indicated, we shall consider rigid bodies of uniform density only. Further, some rigid bodies are combination of other regular bodies, whose COMs are known. In that case, we would employ the formula of COM for the system of particles.

L shaped rod or wire

In this case, we need not resort to integration as L-shaped rod is simply a combination of two rods with known COMs, which can be treated as combination of particles.

Let the origin of the planar coordinate system coincides with the corner of the L-shaped rod. COM of a uniform rod is the middle point of the rod. From the figure, the COM of rod along x-axis is ( l 1 2 , 0 l 1 2 , 0 ). On the other hand, COM of rod along y-axis is ( 0 , l 2 2 0 , l 2 2 ). These rods, therefore, can be considered as particles at these positions. Thus, the L-shaped rod system reduces to the case of two particles system separated by a distance. Let m 1 m 1 and m 2 m 2 be the mass of two rods respectively. Now, the linear mass density is :

γ = m 1 l 1 = m 2 l 2 γ = m 1 l 1 = m 2 l 2

The COM of the L-shaped rod is :

x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 x COM = γ l 1 x l 1 2 γ ( l 1 + l 2 ) x COM = l 1 2 2 ( l 1 + l 2 ) x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 x COM = γ l 1 x l 1 2 γ ( l 1 + l 2 ) x COM = l 1 2 2 ( l 1 + l 2 )

and similarly,

y COM = m 1 y 1 + m 2 y 2 m 1 + m 2 x COM = γ l 2 x l 2 2 γ ( l 1 + l 2 ) x COM = l 2 2 2 ( l 1 + l 2 ) y COM = m 1 y 1 + m 2 y 2 m 1 + m 2 x COM = γ l 2 x l 2 2 γ ( l 1 + l 2 ) x COM = l 2 2 2 ( l 1 + l 2 )

It is important to note that the expressions of COM are independent of mass term.

Example 1

Problem : A thin rod of length 3 m is bent at right angles at a distance 1 m from one end. If 1 m segment lies horizontally, determine the COM of the resulting L-shaped wire structure.

Solution : Here l 1 l 1 = 1m and l 2 l 2 = 2 m. The COM of the L - shaped rod is :

x COM = l 1 2 2 ( l 1 + l 2 ) x COM = 1 2 ( 1 + 2 ) = 1 6 m x COM = l 1 2 2 ( l 1 + l 2 ) x COM = 1 2 ( 1 + 2 ) = 1 6 m

and

y COM = l 2 2 2 ( l 1 + l 2 ) x COM = 4 2 ( 1 + 2 ) = 4 6 = 2 3 m y COM = l 2 2 2 ( l 1 + l 2 ) x COM = 4 2 ( 1 + 2 ) = 4 6 = 2 3 m

Semicircular uniform wire

A semicircular wire is not symmetric to all the axes of the coordinates. The wire is a planar body involving two dimensions (x and y). However, we note that the wire is symmetric about y-axis, but not about x-axis. This has two important implications :

• COM lies on y - axis (i.e on the axis of symmetry).
• We only need to evaluate "y" coordinate of COM as x COM x COM = 0.

Thus, we need to evalute the expression :

y COM = 1 M y m y COM = 1 M y m

Evaluation of an integral is based on identifying the differential element. The semicircular wire is composed of large numbers of particle like elements, each of mass "ⅆm". The idea of this particle like element of mass "dm" is that its center of mass i.e COM lies at "y" distance from the origin in y-direction. This element is linked to the mass "m" of the wire as :

m = γ l = ( M π R ) l m = γ l = ( M π R ) l

Now,

l = R θ y = R sin θ l = R θ y = R sin θ

m = ( M π R ) R θ = ( M π ) θ m = ( M π R ) R θ = ( M π ) θ

Putting in the integral,

y COM = 1 M R sin θ ( M π ) θ y COM = R π sin θ θ y COM = 1 M R sin θ ( M π ) θ y COM = R π sin θ θ

Integrating between θ = 0 and θ = π,

y COM = R π [ - cos θ ] 0 π y COM = R π [ - cos π + cos 0 ] = 2 R π y COM = R π [ - cos θ ] 0 π y COM = R π [ - cos π + cos 0 ] = 2 R π

The COM of the semicircular wire is ( 0 , 2 R π 0 , 2 R π ).

Semicircular unifrom thin disc

A semicircular disc is not symmetric to all the axes of the coordinates. The thin disc is a planar body involving two dimensions (x and y). However, we note that the disc is symmetric about y-axis, but not about x-axis. This has two important implications :

• COM lies on y - axis (i.e on the axis of symmetry).
• We only need to evaluate "y" coordinate of COM as x COM x COM = 0.

Thus, we need to evaluate the expression :

y COM = 1 M y m y COM = 1 M y m

Evaluation of an integral is based on identifying the differential element. The semicircular disc is composed of large numbers of semicircular thin wires. We must be careful in evaluating this integral. The elemental mass "dm" corresponds to a semicirular wire - not a particle as the case before. The term "y" in the integral denotes the distance of the COM of this elemental semicircular wire. The COM of this semi-circular thin wire element is at a distance 2r/Π (as determined earlier)in y-direction.

Now,

y = 2 r π m = σ A = ( π r r ) σ y = 2 r π m = σ A = ( π r r ) σ

where "σ" denoted the surface density. It is given by :

σ = M π R 2 2 = 2 M π R 2 σ = M π R 2 2 = 2 M π R 2

Putting in the integral,

y COM = 1 M ( 2 r π ) σ π r r y COM = 2 σ M r 2 r y COM = 1 M ( 2 r π ) σ π r r y COM = 2 σ M r 2 r

Integrating between r = 0 and r = R (radius of the semicircular disc),

y COM = 2 σ M [ r 3 3 ] 0 R y COM = 2 σ R 3 3 M y COM = 4 M R 3 3 M π R 2 = 4 R 3 π y COM = 2 σ M [ r 3 3 ] 0 R y COM = 2 σ R 3 3 M y COM = 4 M R 3 3 M π R 2 = 4 R 3 π

The COM of the semicircular disc is ( 0 , 4 R 3 π 0 , 4 R 3 π ).

COM rigid bodies with non-uniform density

Evaluation of integrals to determine COM of non-uniform rigid bodies may be cumbersome. There are two possibilities :

1. The variation in density is distinctly marked.
2. The variation in density is defined in mathematical form.

Variation in density is distinctly marked

If the variation in density is distinctly marked, then it is possible to determine COMs of individual uniform parts. Once COMs of the individual uniform parts are known, we can treat them as particle with their mass at their respective COMs. Thus, non-uniform rigid body is reduced to a system of particles, which renders easily for determination of its COM. This technique is illustrated in the example given here :

Example 2

Problem : A thin plate (axb) is made up of two equal portions of different surface density of σ 1 σ 1 and σ 1 σ 1 as shown in the figure. Determine COM of the composite plate in the coordinate system shown.

Solution : COM of the first half is at a 4 a 4 , whereas COM of the second half is at 3 a 4 3 a 4 from the origin of the coordinate system in x-direction. As x-axis lies at mid point of the height of the plate, y-component of COM is zero. Now two portions are like two particles having masses :

m 1 = a b 2 σ 1 m 1 = a b 2 σ 1

and

m 2 = a b 2 σ 2 m 2 = a b 2 σ 2

Also, the x-coordinates of the positions of these particles are x 1 = a 4 x 1 = a 4 and x 2 = 3 a 4 x 2 = 3 a 4 . The x-component of the COM of two particles system is :

x COM = m 1 x 1 + m 2 x 2 m 1 + m 2 x COM = m 1 x 1 + m 2 x 2 m 1 + m 2

x COM = a b 2 σ 1 a 4 + a b 2 σ 2 3 a 4 a b 2 σ 1 + a b 2 σ 2 x COM = a b 2 σ 1 a 4 + a b 2 σ 2 3 a 4 a b 2 σ 1 + a b 2 σ 2

x COM = ( σ 1 + 3 σ 2 ) a 4 ( σ 1 + σ 2 ) x COM = ( σ 1 + 3 σ 2 ) a 4 ( σ 1 + σ 2 )

Variation in density is defined in mathematical form

If the variation in density is well defined, then it is possible to evaluate integral to determine the COM for non-uniform rigid body. The technique is illustrated in the example given here :

Example 3

Problem : Linear density of a rod varies with length as γ = 2x + 3. If the length of the rod is "L", then find the COM of the rod.

Solution : The COM of the rod in one dimension is given as :

x COM = 1 M x m x COM = 1 M x m

Here,

m = γ x = ( 2 x + 3 ) x m = γ x = ( 2 x + 3 ) x

We can use this relation to determine COM as :

x COM = x m m x COM = x ( 2 x + 3 ) x ( 2 x + 3 ) x x COM = x m m x COM = x ( 2 x + 3 ) x ( 2 x + 3 ) x

x COM = 2 x 2 x + 3 x x 2 x x + 3 x x COM = 2 x 2 x + 3 x x 2 x x + 3 x

Evaluating between x = 0 and x = L, we have :

x COM = 2 L 3 3 + 3 L 2 2 2 L 2 2 + 3 L x COM = 2 L 3 3 + 3 L 2 2 2 L 2 2 + 3 L

x COM = 4 L 2 + 9 L 6 ( L + 3 ) x COM = 4 L 2 + 9 L 6 ( L + 3 )

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