Moments and Center of Mass

Definition of center of mass

The center of mass is the unique point at the center of a distribution of mass in space that has the property that the weighted position vectors relative to this point sum to zero. In analogy to statistics, the center of mass is the mean location of a distribution of mass in space.

A system of particles

In the case of a system of particles $P i, i = 1, \dots, n$ , each with mass $m i$ that are located in space with coordinates $r i, i = 1, \dots, n$ , the coordinates R of the center of mass satisfy the condition

\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )=\mathbf {0} .

Solving this equation for R yields the formula

\mathbf {R} ={\frac {1}{M}}\sum _{i=1}^{n}m_{i}\mathbf {r} _{i},

where $M=\sum _{i=1}^{n}m_{i}$ is the total mass of all of the particles.

A continuous volume

If the mass distribution is continuous with the density ρ(r) within a solid Q, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass R over the volume V is zero, that is

\iiint \limits _{Q}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV=0.

Solve this equation for the coordinates R to obtain

\mathbf {R} ={\frac {1}{M}}\iiint \limits _{Q}\rho (\mathbf {r} )\mathbf {r} dV,

where M is the total mass in the volume.

If a continuous mass distribution has uniform density, which means ρ is constant, then the center of mass is the same as the centroid of the volume.

Locating the center of mass

Plumb line method

The experimental determination of a body's centre of mass makes use of gravity forces on the body and is based on the fact that the centre of mass is the same as the centre of gravity in the parallel gravity field near the earth's surface.

The center of mass of a body with an axis of symmetry and constant density must lie on this axis. Thus, the center of mass of a circular cylinder of constant density has its center of mass on the axis of the cylinder. In the same way, the center of mass of a spherically symmetric body of constant density is at the center of the sphere. In general, for any symmetry of a body, its center of mass will be a fixed point of that symmetry.

In two dimensions

An experimental method for locating the center of mass is to suspend the object from two locations and to drop plumb lines from the suspension points. The intersection of the two lines is the center of mass.

The shape of an object might already be mathematically determined, but it may be too complex to use a known formula. In this case, one can subdivide the complex shape into simpler, more elementary shapes, whose centers of mass are easy to find. If the total mass and center of mass can be determined for each area, then the center of mass of the whole is the weighted average of the centers. This method can even work for objects with holes, which can be accounted for as negative masses.

A direct development of the planimeter known as an integraph, or integerometer, can be used to establish the position of the centroid or center of mass of an irregular two-dimensional shape. This method can be applied to a shape with an irregular, smooth or complex boundary where other methods are too difficult. It was regularly used by ship builders to compare with the required displacement and center of buoyancy of a ship, and ensure it would not capsize.

In three dimensions

An experimental method to locate the three-dimensional coordinates of the center of mass begins by supporting the object at three points and measuring the forces, F₁, F₂, and F₃ that resist the weight of the object, $\mathbf {W} =-W\mathbf {\hat {k}}$ ( $\mathbf {\hat {k}}$ is the unit vector in the vertical direction). Let r₁, r₂, and r₃ be the position coordinates of the support points, then the coordinates R of the center of mass satisfy the condition that the resultant torque is zero,

\mathbf {T} =(\mathbf {r} _{1}-\mathbf {R} )\times \mathbf {F} _{1}+(\mathbf {r} _{2}-\mathbf {R} )\times \mathbf {F} _{2}+(\mathbf {r} _{3}-\mathbf {R} )\times \mathbf {F} _{3}=0,

or

\mathbf {R} \times \left(-W\mathbf {\hat {k}} \right)=\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\mathbf {r} _{3}\times \mathbf {F} _{3}.

This equation yields the coordinates of the center of mass R* in the horizontal plane as,

\mathbf {R} ^{*}=-{\frac {1}{W}}\mathbf {\hat {k}} \times (\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\mathbf {r} _{3}\times \mathbf {F} _{3}).

The center of mass lies on the vertical line L, given by

\mathbf {L} (t)=\mathbf {R} ^{*}+t\mathbf {\hat {k}} .

The three-dimensional coordinates of the center of mass are determined by performing this experiment twice with the object positioned so that these forces are measured for two different horizontal planes through the object. The center of mass will be the intersection of the two lines L₁ and L₂ obtained from the two experiments.

The moment of inertia can be defined as the second moment about an axis and is usually designated the symbol I. The moment of inertia is very useful in solving a number of problems in mechanics. For example, the moment of inertia can be used to calculate angular momentum, and angular energy. Moment of inertia is also important in beam design.

Linear and angular momentum

The linear and angular momentum of a collection of particles can be simplified by measuring the position and velocity of the particles relative to the center of mass. Let the system of particles P_i, i = 1, ..., n of masses m_i be located at the coordinates r_i with velocities v_i. Select a reference point R and compute the relative position and velocity vectors,

\mathbf {r} _{i}=(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {R} ,\quad \mathbf {v} _{i}={\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {v} .

The total linear momentum and angular momentum of the system are

\mathbf {p} ={\frac {d}{dt}}\left(\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\right)+\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {v} ,

and

\mathbf {L} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times {\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+\left(\sum _{i=1}^{n}m_{i}\right)\left[\mathbf {R} \times {\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+(\mathbf {r} _{i}-\mathbf {R} )\times \mathbf {v} \right]+\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {R} \times \mathbf {v}

If R is chosen as the center of mass these equations simplify to

\mathbf {p} =m\mathbf {v} ,\quad \mathbf {L} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times {\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+\sum _{i=1}^{n}m_{i}\mathbf {R} \times \mathbf {v}

where m is the total mass of all the particles, p is the linear momentum, and L is the angular momentum.

The law of conservation of momentum predicts that for any system not subjected to external forces the momentum of the system will remain constant, which means the center of mass will move with constant velocity. This applies for all systems with classical internal forces, including magnetic fields, electric fields, chemical reactions, and so on. More formally, this is true for any internal forces that cancel in accordance with Newton's Third Law.

Moment of Inertia

Shape moment of inertia for flat shapes

The area moment of inertia takes only shape into account, not mass.

It can be used to calculate the moment of inertia of a flat shape about the x or y axis when I is only important at one cross-section.

$I_{x}=\int y^{2}\,dA$

$I_{y}=\int x^{2}\,dA$

$I_{z}=\int r^{2}\,dA$

Because, for flat shapes, $r^{2}=x^{2}+y^{2}$ the following is true $I_{z}=I_{x}+I_{y}$

The shape moment of inertia of the cross-section of a beam is used in Structural Engineering in order to find the stress and deflection of the beam.

Shape moment of inertia for 3D shapes

The moment of inertia I=∫r²dm for a hoop, disk, cylinder, box, plate, rod, and spherical shell or solid can be found from this figure.

Mass moment of inertia

The mass moment of inertia takes mass into account. The mass moment of inertia of a point mass about a reference axis is equal to mass multiplied by the square of the distance from that point mass to the reference axis:

$I_{pointmass}=r^{2}m\,$

The metric units are kg*m^2.

The mass moment of inertia of any body of mass rotating around any axis is equal to the sum of the mass moment of inertia of each of the particles of that body:

$I_{m}=\sum r_{i}^{2}m_{i}$

Rather than adding up each particle individually, sometimes we can take a mathematical shortcut by integrating over all the particles:

$I_{m}=\int r^{2}\,dm$

Radius of gyration

The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. If the mass of an object was 2kg, and the moment of inertia was $18kg*m^{2}$ , then the radius of gyration would be 3m. In other words, if all of the mass was concentrated at a distance of 2m from the axis, then the moment of inertia would still be $18kg*m^{2}$ . Radius of gyration is represented with a k.