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Locating the center of mass

Main article: Locating the center of mass

Plumb line method

In two dimensions

An experimental method for locating the center of mass in a particular plane 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 located in the plane formed by the two points and the plumb lines.

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 centre of buoyancy of a ship, and ensure it would not capsize.

In three dimensions

An experimental method to determine the location of the center of mass in three dimensions supports the object at three points and measures the forces, F₁, F₂, and F₃ that resist the weight of the object, W= -Wk---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,

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

or

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

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

$\mathbf {R} ^{*}={\vec {k}}\times (\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\mathbf {r} _{3}\times \mathbf {F} _{3}).$

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

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

The three dimensional coordinates of the center of mass can be determined by performing this experiment with the object positioned so that two different horizontal planes passing through it. The center of mass will be the intersection of the two lines L₁ and L₂ obtained from the two experiments.

Center of gravity

Diagram of an educational toy that balances on a point: the CM (C) settles below its support (P)

The suspending chair trick makes use of the human body's surprisingly high center of mass

The benefits of using the "center of mass" to model a mass distribution can be seen by considering the resultant of the gravity forces on a continuous body. Consider a body of volume V with density ρ(r) at each point r in the volume. In a parallel gravity field the force f at each point r in the volume V is given by,

$\mathbf {f} (\mathbf {r} )=-g\rho (\mathbf {r} )dV{\vec {k}}.$

Choose a reference point R in the volume and compute the resultant force and torque at this point,

$\mathbf {F} =\int _{V}\mathbf {f} (\mathbf {r} )=\int _{V}\rho (\mathbf {r} )dV(-g{\vec {k}})=-Mg{\vec {k}},$

and

$\mathbf {T} =\int _{V}(\mathbf {r} -\mathbf {R} )\times \mathbf {f} (\mathbf {r} )=\int _{V}(\mathbf {r} -\mathbf {R} )\times (-g\rho (\mathbf {r} )dV{\vec {k}})=(\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV)\times (-g{\vec {k}})$

If the reference point R is chosen so that it is the center of mass, then

$\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV=0,$

which means the resultant force F has zero associated torque T.

is often used interchangeably with the term center of gravity because any uniform gravitational field g acts on a system as if the mass M of the system were concentrated at the center of mass R. The center of gravity is defined as the average position of weight distribution, and mass and weight are technically different properties. However, because weight and mass are proportional, the center of gravity and center of mass refer to the same point of an object for almost all objects on and near Earth's surface. Generally, physicists prefer to use the term center of mass, as an object has a center of mass whether or not it is under the influence of gravity. In addition, the term "center of gravity" refers to the single point associated with an object where the force of gravity can be considered to act,

Specifically, the gravitational potential energy is equal to the potential energy of a point mass M at R, and the gravitational torque is equal to the torque of a force Mg acting at R. In a uniform gravitational field, the center of mass is a center of gravity, and in common usage, the two phrases are used as synonyms.

In a non-uniform field, gravitational effects such as potential energy, force, and torque can no longer be calculated using the center of mass alone. In particular, a non-uniform gravitational field can produce a torque on an object, causing it to rotate. The center of gravity, an application point of the resultant gravitational force, may not exist or not be unique; see centers of gravity in non-uniform fields.

Momentum

Further information: Derivation of the center of mass, Center of mass frame

Consider a system of body P_i, i=1,...,n located at the coordinates r_{i</sum> and velocities vi</sum>. 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 linear and angular momentum relative to the reference point R is

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

and

$\mathbf {H} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times {\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+(\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} ))\times \mathbf {V} .$

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

$\mathbf {p} =M\mathbf {V} ,\quad \mathbf {H} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times {\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} ).$

$\mathbf {F} _{i}+\sum _{j=1}^{n}\mathbf {F} _{ij}=m_{i}{\dot {\mathbf {a} }}_{i},\quad i=1,\ldots ,n.$

Introduce the reference point R so the position, velocity and acceleration of the bodies are given by

$\mathbf {r} _{i}=(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {R} ,\quad \mathbf {v} _{i}={\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} ,\quad \mathbf {a} _{i}={\frac {d^{2}}{dt^{2}}}(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {A} .$

The resultant force and torque of the system of forces at the reference point R is

$\mathbf {F} ={\frac {d^{2}}{dt^{2}}}(\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} ))+(\sum _{i=1}^{n}m_{i})\mathbf {A} ,$

and

$\mathbf {T} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times {\frac {d^{2}}{dt^{2}}}(\mathbf {r} _{i}-\mathbf {R} )+(\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} ))\times \mathbf {A} .$

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

$\mathbf {F} =M\mathbf {A} ,\quad \mathbf {T} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times {\frac {d^{2}}{dt^{2}}}(\mathbf {r} _{i}-\mathbf {R} ).$

$\mathbf {F} =\sum _{i=1}^{n}(\sum _{j=1}^{n}\mathbf {F} _{ij})=0,\quad \mathbf {T} =\sum _{i=1}^{n}(\sum _{j=1}^{n}(\mathbf {r} _{j}-\mathbf {R} )\times \mathbf {F} _{ij})=\sum _{i=1}^{n}\sum _{j=1}^{n}(\mathbf {r} _{j}-\mathbf {r} _{i})\times \mathbf {F} _{ij}=0.$

The force terms cancel because F_ij=-F_ji, and the torque terms cancel because the relative vectors r_j-r_i are parallel to the forces F_ij.

Let the velocities of the particles be represented by the velocity V of the reference point R plus the relative velocity v_i, such that

$\mathbf {V} _{i}=(\mathbf {V} _{i}-\mathbf {V} )+\mathbf {V} =\mathbf {v} _{i}+\mathbf {V} ,$

then Newton's second law for the system of particles becomes

$0=\sum _{i=1}^{n}(m_{i}{\dot {\mathbf {V} }}_{i})=\sum _{i=1}^{n}m_{i}({\dot {\mathbf {v} }}_{i}+{\dot {\mathbf {V} }})=M{\dot {\mathbf {V} }}+\sum _{i=1}^{n}m_{i}{\dot {\mathbf {v} }}_{i}.$

If the reference point R is chosen so that it is the center of mass, then

$\sum _{i=1}^{n}m_{i}{\dot {\mathbf {v} }}_{i}={\frac {d^{2}}{dt^{2}}}(\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} ))=0.$

This shows that the velocity of the center of mass is constant, because its acceleration is zero,

$0=M{\dot {\mathbf {V} }}.$

For any system with no external forces, the center of mass moves 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 satisfy Newton's Third Law.

Consider the set of particles, P_i, i=1,...,n of mass m_i that are located at the coordinates r_i and moving with velocity V_i. Assume there are not external forces and each of the particles applies a force on all of the rest of the particles. Netwon's second law yields the equations,

$\sum _{j=1}^{n}\mathbf {F} _{ij}=m_{i}{\dot {\mathbf {V} }}_{i},\quad i=1,\ldots ,n.$

Introduce the reference point R and compute the resultant force and torque of the system of forces at this reference point, that is

$\mathbf {F} =\sum _{i=1}^{n}(\sum _{j=1}^{n}\mathbf {F} _{ij})=0,\quad \mathbf {T} =\sum _{i=1}^{n}(\sum _{j=1}^{n}(\mathbf {r} _{j}-\mathbf {R} )\times \mathbf {F} _{ij})=\sum _{i=1}^{n}\sum _{j=1}^{n}(\mathbf {r} _{j}-\mathbf {r} _{i})\times \mathbf {F} _{ij}=0.$

The force terms cancel because F_ij=-F_ji, and the torque terms cancel because the relative vectors r_j-r_i are parallel to the forces F_ij.

Let the velocities of the particles be represented by the velocity V of the reference point R plus the relative velocity v_i, such that

$\mathbf {V} _{i}=(\mathbf {V} _{i}-\mathbf {V} )+\mathbf {V} =\mathbf {v} _{i}+\mathbf {V} ,$

then Newton's second law for the system of particles becomes

$0=\sum _{i=1}^{n}(m_{i}{\dot {\mathbf {V} }}_{i})=\sum _{i=1}^{n}m_{i}({\dot {\mathbf {v} }}_{i}+{\dot {\mathbf {V} }})=M{\dot {\mathbf {V} }}+\sum _{i=1}^{n}m_{i}{\dot {\mathbf {v} }}_{i}.$

If the reference point R is chosen so that it is the center of mass, then

$\sum _{i=1}^{n}m_{i}{\dot {\mathbf {v} }}_{i}={\frac {d^{2}}{dt^{2}}}(\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} ))=0.$

This shows that the velocity of the center of mass is constant, because its acceleration is zero,

$0=M{\dot {\mathbf {V} }}.$

$\mathbf {p} =M\mathbf {v} _{\mathrm {cm} },$

where M indicates the total mass, and v_cm is the velocity of the center of mass. This velocity can be computed by taking the time derivative of the position of the center of mass. An analogue to Newton's Second Law is

$\mathbf {F} =M\mathbf {a} _{\mathrm {cm} },$

where F indicates the sum of all external forces on the system, and a_cm indicates the acceleration of the center of mass. It is this principle that gives precise expression to the intuitive notion that the system as a whole behaves like a mass of M placed at R.

The angular momentum vector for a system is equal to the angular momentum of all the particles around the center of mass, plus the angular momentum of the center of mass, as if it were a single particle of mass $M$:

$\mathbf {L} _{\mathrm {sys} }=\mathbf {L} _{\mathrm {cm} }+\mathbf {L} _{\mathrm {around\,cm} }.$

This is a corollary of the parallel axis theorem.