Topic 2B-02: Center of mass

Two point-like particles of masses m1 and m2 are attached to end points Q1 and Q2 of a massless rod of length L respectively. We try to balance the rod at a point O as shown in Fig. 1. The rod forms an angle θ with a horizontal line. The length OQ1 is denoted as x. We need to vary x and θ until the point O is found. Such a point O is called the center of mass, or the center of gravity of the two particle system.

The situation can be abstracted as shown in Fig. 2, where position vector r1 originates from point O and ends at point Q1, and position vector r2 originates from point O and ends at point Q2. Force F1 originates from point Q1, points vertically downward, and has a magnitude of m1g. Force F2 originates from point Q2, points vertically downward, and has a magnitude of m2g. F1 and F2 are gravitational pulls acting on particles m1 and m2 respectively. When the rod is balanced at point O, the rod will not rotate around point O. The torque τ1 = r1 x F1 tries to rotate the rod counter-clockwise around O, and the torque τ2 = r2 x F2 tries to rotate the rod clockwise around point O. In order to keep the rod stationary, we must have

τ1 + τ2 = 0.

Since τ1 originates from point O and points vertically out of the paper, and τ2 originates from point O and points vertically into the paper, we must have |τ1 = |τ2| in order to have τ1 + τ2 = 0. From Eq. (2) of Topic 2B-01, we have

|τ1| = |r1|·|F1|·sinθ
= x·m1g·sinθ ,

and

|τ2| = |r2|·|F2|·sinθ
= (L - x)·m2g·sinθ ,

Thus |τ1| = |τ2| leads to

m1x = m2(L - x),

or

x = m2L/(m1 + m2) .                                                 (1)

This concept of the center of mass can be refined further so that it does not need to depend on the gravitational pulls. From Eq. (1) we have

|m1r1| = m1x
=m1m2L / (m1 + m2),

and

|m2r2| = m2(L - x)
=m1m2L / (m1 + m2).

Thus |m1r1| = |m2r2|. Since m1r1 and m2r2 are pointing at the opposite directions, we have

m1r1 + m2r2 = 0.                                                 (2)

Eq. (2) can be considered as the definition of the center of mass. It says that first find a point O that satisfies Eq. (2), then point O is the center of mass. To prove that such a point O is indeed the center of mass as discussed in Eq. (1), we assume a force m1f1 applies on point Q1, and a force m2f2 applies on point Q2 as shown in Fig. 3. We also assume that
|f1| = |f2| = |f| and f1 // f2 // f, where f is a mathematical vector without a specific originating point. Torques τ1 and τ2 are defined as

τ1 = r1 x (m1f1)

and

τ2 = r2 x (m2f2)

respectively. Following the second kind of addition rule for torques as discussed in Topic 2B-02, we have

τ = τ1 + τ2
= r1 x (m1f1) + r2 x (m2f2)
= (m1r1 + m2r2) x f.

Substituting Eq. (2) into the above equation, we get

τ = 0 .

If |f1| = |f2| = g, and both f1 and f2 point vertically downward, τ becomes the torque around point O generated by the gravitational pulls. τ = 0 means that the two particle system under consideration can be balanced at point O. Thus point O is the center of mass as discussed in Fig. 1.

Though Eq. (2) is an elegant way to define the center of mass, it is not easy to use. There is a more practical way to compute the center of mass. In Fig. 4 P is a reference point, like the origin of a coordinate system. Position vectors R1 and R2, originating from point P, define points Q1 and Q2 respectively. Position vector R defines the center of mass point O. Position vectors r1 and r2 are drawn from O to Q1 and Q2 respectively. We have

R1 = R + r1,

and

R2 = R + r2.

Thus

m1R1 + m2R2 = m1(R + r1) + m2(R + r2)
= (m1 + m2)R + (m1r1 + m2r2).

Since point O is the center of mass and Eq. (2) is satisfied, the above equation leads to

R = (m1R1 + m2R2) / (m1 + m2).                                                 (3)

Eq. (3) allows us to compute the center of mass easily when R1 and R2 are known.

It is not difficult to extend the discussions about a two particle system to a N-particle system. Let particles of masses m1, m2, · · ·, mN located at points Q1, Q2, · · ·, QN respectively. Let position vectors r1, r2, · · ·, rN originate from point O and ends at points
Q1, Q2, · · ·, QN respectively. If the equation

m1r1 + m2r2 + · · · + mNrN = 0                                                 (4)

is satisfied, then O is the center of mass of this N-particle system. To demonstrate the validity of this statement, we imagine that each particle in the system is connected to point O via a massless rod as shown in Fig. 5. Let forces m1f1, m2f2, · · ·, mNfN applied to points Q1, Q2, ·, ·, ·, QN respectively. We assume that |f1| = |f2| = · · · = |fN| = |f| and f1 // f2 // · · · // fN // f, where f is a mathematical vector without a specific originating point. The torque of this system around point O is

τ = r1 x (m1f1) + r2 x (m2f2) + · · · + rN x (mNfN)
= (m1r1 + m2r2 + · · · + mNrN) x f.

Eq. (4) then says that τ = 0. If f is chosen to point vertically downward, and |f| is set to g, then τ is the torque around point O generated by the gravitational pulls of this N-particle system. τ = 0 means that the system as depicted in Fig. 5 can indeed be balanced at point O. Thus point O can be properly called the center of mass (or center of gravity) of this system.

The extension of Eq. (3) to the N-particle system is also straightforward. We have

R = (m1R1 + m2R2 + · · · + mNRN) / (m1 + m2 + · · · + mN) ,                                                 (5)

where R1, R2, · · ·, RN are position vectors that originate from a reference point P and define points Q1, Q2, · · ·, QN respectively, and R is the position vector that defines point O.

Eq. (5) is actually more powerful than it appears. We can group any number of particles under consideration as a subsystem. Let us consider the center of mass point of particles 1 and 2, and denote the postion vector for this center of mass point as R12. By applying Eq.(5) to the subsystem of particles 1 and 2, we have

(m1 + m2)R 12 = (m1R1 + m2R2).

Substituting this equation into Eq. (5), we get

R = {(m1 + m2)R 12 + · · · + mNRN} / {(m1 + m2) + · · · + mN} .

This means that particles 1 and 2 can be replaced by a particle of mass (m1 + m2) located at the position R 12. This simplifying process can be applied to any group of particles in the system and as many groups as we want. We will be using this property later on when it comes to find center of mass points in the exercises.

The number N can be extended to infinity, and the distance among neighboring particles can be reduced to infinitesimally small. Thus the discrete particle system will approach a continuous and extended object. Equations (4) and (5) will also approach a continuous limit. We will not get into details here but rather differ the discussion to upcoming exercises.

There is an important property related to the center of mass and the gravitational pulls. Let us consider the N-particle system as depicted in Fig. 6. Let forces F1, F2, · · · FN apply on points Q1, Q2, · · ·, QN respectively. R1, R2, · · ·, RN are position vectors that originate from a reference point P and define points Q1, Q2, · · ·, QN respectively. Position vector R defines the center of mass point O as shown in Fig. 6. Position vectors OQ1, OQ2, · · ·, OQN are denoted as r1, r2, · · ·, rN respectively. The torque vector around the reference point P is

τ = R1 x F1 + R2 x F2 + · · · + RN x FN .

Since R1 = r1 + R, R2 = r2 + R, · · ·, RN = rN + R, we have, in the mathematical sense,

τ = (r1 x F1 + r2 x F2 + · · · + rN x FN) + R x (F1 + F2 + · · · + FN).

The first term at the right-hand side of the above equation is the torque around the center of mass point O so is denoted as τO. The sum (F1 + F2 + · · · + FN) is a mathematical sum. It means that F1, F2, · · ·, FN are all displaced without changing their magnitudes and the orientation such that they all originate from point O and then summed. Let F1, F2, · · ·, FN be gravitational pulls, that is,

F1 = m1g, F2 = m2g, · · , FN = mNg,

where g is the gravitational acceleration vector that has the magnitude g and points vertically downward. Then

τO = (m1r1 + m2r2 + · · · + mNrN) x g.

Using Eq. (4), we get τO = 0. The mathematical sum F1 + F2 + · · · + FN can be written as

F1 + F2 + · · · + FN = (m1 + m2 + · · · + mNg.

Thus

τ = R x (Mg),                                                 (6)

where M = m1 + m2 + · · · + mN is the total mass of the system. This means that the torque of a N-particle system due to the gravitational pulls around a point P is equal to the torque of a single particle of mass M located at the center of mass point O. This property is preserved at the limit of a continuous object.