At the practice nets, a batsman finishes his turn and tosses his bat to the next player. The bat spins as it flies, handle over blade. The handle traces loops in the air, and so does the tip of the blade. Yet the bat as a whole clearly travels along an arc, rising and falling like a thrown ball. Somewhere inside the bat there must be a point that moves smoothly while everything else wheels around it.
A body is not a particle
Newton's second law, \(\vec F = m\vec a\), is a law for a particle. A particle has a single position and a single acceleration. Every earlier chapter used that law for blocks, cars and trains as if each were a point. A real body is not a point. A bat is made of countless particles, and while it spins they all move differently.
So a question has been waiting since the first chapter on forces. When we wrote \(F = ma\) for a block, which point of the block did the acceleration \(a\) describe? This chapter answers the question. The answer is a special point called the centre of mass. It is the point that moves as if the whole mass were there and all the outside forces acted on it.
Watch the two balls in Fig 1.1. Each one combines the flight with the spin, so each path loops and wobbles. Only the blue point C, on the rod and close to the heavier ball, draws a smooth curve. That curve is exactly the parabola of a single stone thrown with the same starting velocity. Gravity is the only outside force, and point C moves as if gravity acted on one particle of mass \(4\,\text{kg}\) placed at C.
The centre of mass of a body or a system is the point at the mass-weighted average of the positions of all its parts. It moves as a single particle carrying the total mass would move under the net outside force.
The second sentence of the definition is a promise, and the last two lessons of this chapter prove it. For now, notice what it offers. The motion of any body, however it spins, splits into two simpler motions. One is the motion of the centre of mass, which follows the particle laws of the earlier chapters. The other is a rotation about the centre of mass, which the unit on rotation treats.
A badminton racquet is thrown spinning across a court. Ignore air resistance. Which point of the racquet moves along a parabola?
Two particles on a line
Start with the simplest system: two particles on the x axis. A particle of mass \(m_1\) is at position \(x_1\), and a particle of mass \(m_2\) is at \(x_2\). The centre of mass is at the mass-weighted average of the two positions.
\[ x_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Each position is multiplied by its own mass, the products are added, and the sum is divided by the total mass \(M = m_1 + m_2\). A heavier particle counts for more in this average, just as a subject with more marks counts for more in a weighted average of results.
Now place the origin at the centre of mass itself, so that \(x_{cm} = 0\). The formula then says \(m_1 x_1 + m_2 x_2 = 0\). One mass is on each side of the origin, at distances \(d_1\) and \(d_2\), so \(m_1 d_1 = m_2 d_2\). The distances are in the inverse ratio of the masses: \(d_1 / d_2 = m_2 / m_1\).
Move the slider and three facts appear. With equal masses, C sits at the middle of the rod. As the right ball becomes heavier, C slides towards it, and with \(8\,\text{kg}\) against \(2\,\text{kg}\) it is only \(0.8\,\text{m}\) from the heavy ball. However far the slider goes, C never leaves the rod between the two balls, and the lower box always shows \(m_1 d_1 = m_2 d_2\).
The two balls on the rod behave like two children on a seesaw. A heavier child must sit closer to the pivot for the seesaw to balance. The balance point of a light rod with two masses is their centre of mass. Gravity pulls each mass in proportion to its mass. The chapter on Torque and Equilibrium proves this link between balance and the centre of mass.
Two children of masses \(30\,\text{kg}\) and \(45\,\text{kg}\) sit at the two ends of a light plank \(3\,\text{m}\) long in a park in Pune. Where should the pivot be for the plank to balance?
Put the origin at the lighter child, so \(x_1 = 0\) and \(x_2 = 3\,\text{m}\).
\(x_{cm} = \dfrac{30 \times 0 + 45 \times 3}{30 + 45} = \dfrac{135}{75} = 1.8\,\text{m}\) from the \(30\,\text{kg}\) child.
The pivot is therefore \(1.2\,\text{m}\) from the heavier child. Check: \(30 \times 1.8 = 54\) and \(45 \times 1.2 = 54\), so \(m_1 d_1 = m_2 d_2\) holds.
Particles of mass \(2\,\text{kg}\) and \(6\,\text{kg}\) are \(4\,\text{m}\) apart. How far is their centre of mass from the \(2\,\text{kg}\) particle, in metres?
Many particles in space
The same average works for any number of particles in space. Let particle \(i\) have mass \(m_i\) and position vector \(\vec r_i\) from a chosen origin, for \(i = 1, 2, \ldots, n\). The position vector of the centre of mass is
\[ \vec r_{cm} = \dfrac{m_1\vec r_1 + m_2\vec r_2 + \cdots + m_n\vec r_n}{m_1 + m_2 + \cdots + m_n} = \dfrac{1}{M}\sum_i m_i\vec r_i \]
where \(M = \sum_i m_i\) is the total mass. The two-particle formula is the case \(n = 2\) along one axis.
A vector equation is three equations, one along each axis. So each coordinate of the centre of mass is a separate weighted average of the same coordinate of the particles. In practice this is how every problem is solved: one axis at a time.
Find the centre of mass of the four point masses of Fig 1.2. They are \(1\,\text{kg}\) at \((2, 1)\), \(2\,\text{kg}\) at \((6, 1)\), \(3\,\text{kg}\) at \((6, 4)\) and \(2\,\text{kg}\) at \((2, 4)\), in metres.
The total mass is \(M = 1 + 2 + 3 + 2 = 8\,\text{kg}\). Work one axis at a time.
\(x_{cm} = \dfrac{1 \times 2 + 2 \times 6 + 3 \times 6 + 2 \times 2}{8}\) and \(y_{cm} = \dfrac{1 \times 1 + 2 \times 1 + 3 \times 4 + 2 \times 4}{8}\).
\(x_{cm} = 36 / 8 = 4.5\,\text{m}\) and \(y_{cm} = 23 / 8 = 2.875\,\text{m}\).
The centre of mass is at \((4.5, 2.875)\,\text{m}\). The middle of the rectangle is \((4, 2.5)\,\text{m}\); C lies right of it and above it, towards the \(3\,\text{kg}\) mass.
Masses of \(2\,\text{kg}\) at \((0, 0)\), \(1\,\text{kg}\) at \((3, 0)\) and \(1\,\text{kg}\) at \((0, 3)\), in metres, lie in a plane. Where is their centre of mass?
Where C can be, and where it cannot
The centre of mass need not coincide with any particle. Three equal masses at the corners of a triangle have their centre of mass inside the triangle. It is at the plain average of the three positions, where there is no mass at all. On the other hand, C can never lie outside the smallest region that encloses all the masses. A weighted average of positions always lies among those positions.
The coordinates of C depend on where the origin is placed, but the point itself does not. Shift the origin by a distance \(a\) along x. Every \(x_i\) falls by \(a\), and so \(x_{cm}\) falls by exactly \(a\) too, because the weights sum to one. The centre of mass stays at the same place in the system; only its label changes.
Two students find the centre of mass of the same four masses. One puts the origin at a corner and the other at the middle. They get different coordinates. Can both be correct? Explain.
Yes. The coordinates of the centre of mass depend on the origin, but the point does not. Moving the origin by a distance \(a\) reduces every particle's coordinate by \(a\). The weights \(m_i/M\) of the average sum to one, so \(x_{cm}\) also falls by exactly \(a\). Both students locate the same physical point and describe it from different origins.
Which of these statements about the centre of mass of a system of point masses are true?
The Earth has a mass of \(6.0 \times 10^{24}\,\text{kg}\) and the Moon \(7.4 \times 10^{22}\,\text{kg}\). Their centres are \(3.8 \times 10^{5}\,\text{km}\) apart. Estimate the distance of their common centre of mass from the centre of the Earth, in km.
A star and its planet also move about their common centre of mass. The planet is far lighter, so the star moves on a tiny circle while the planet swings around a huge one. Astronomers measure the small back-and-forth motion of the star from the shift in the colour of its light. In 1995 Michel Mayor and Didier Queloz found the first planet around a Sun-like star, 51 Pegasi b, from exactly this wobble.
Return now to the tossed bat of the opening. The blade is much heavier than the handle. So the centre of mass of the bat lies inside the blade, not at the middle of its length. That point follows the clean arc of a single thrown ball, whatever the spin. The handle and the tip of the blade loop around it, exactly as the two balls looped around C in Fig 1.1.
- Taking the centre of mass of two unequal masses at their midpoint.Weight each position by its mass. The centre of mass lies nearer the heavier mass, with \(m_1 d_1 = m_2 d_2\).
- Averaging the coordinates without the masses.The plain average is the centroid of the points. Only for equal masses is it the centre of mass.
- Mixing the axes, for example using y values in the x sum.Work one axis at a time: \(x_{cm}\) uses only the x coordinates, \(y_{cm}\) only the y coordinates.