Center of mass, systems of particles and rocket propulsion

In collisions you followed the momentum of the two objects separately. When the number of objects grows (an explosion, a person and a boat, a rocket and its exhaust), it is very useful to summarize the whole system by the motion of a single point. This point is the center of mass. Whatever happens inside the system, the center of mass moves obeying only the external forces.

Golden rules:

1. The center of mass is a weighted average with the masses, not the midpoint.
2. $\vec v_{cm} = \vec p_{\text{total}}/M$.
3. $\sum\vec F_{\text{ext}} = M\vec a_{cm}$. Internal forces (explosion, pushing apart, walking) cannot move the center of mass.
4. If there is no external force and the system is initially at rest, the center of mass stays in place.

Position of the center of mass

In a system of particles:

$x_{cm} = \frac{\sum m_ix_i}{M} \qquad y_{cm} = \frac{\sum m_iy_i}{M} \qquad M = \sum m_i$

In vector form $\vec r_{cm} = \frac{1}{M}\sum m_i\vec r_i$. Calculate each component separately.

Worked example 1. On the x axis, 1.00 kg ($x = 0$), 2.00 kg ($x = 2.00$ m) and 5.00 kg ($x = 4.00$ m).

$x_{cm} = \frac{(1.00)(0) + (2.00)(2.00) + (5.00)(4.00)}{8.00} = \frac{24.0}{8.00} = 3.00\ \text{m}$

It is close to the heavy 5 kg. The center of mass always stays between the outermost particles.

Worked example 2. 2.00 kg at $(0, 0)$, 3.00 kg at $(4.00, 0)$, 1.00 kg at $(2.00, 3.00)$ (m).

x_cm = (2·0 + 3·4.00 + 1·2.00) / 6.00 = 14.0 / 6.00 = 2.33 m
y_cm = (2·0 + 3·0    + 1·3.00) / 6.00 =  3.00 / 6.00 = 0.500 m

PITFALL · not the geometric center. The centroid of the triangle formed by the three points is the center of mass if the masses are equal. If the masses are different, you must calculate the weighted average.

Continuous objects

The sum becomes an integral: $x_{cm} = \frac{1}{M}\int x\,dm$. For a rod $dm = \lambda\,dx$, where $\lambda$ is the linear mass density (kg/m). In uniform and symmetric objects the center of mass is at the geometric center: in the middle of a uniform rod, at the center of a uniform disc.

Worked example 3. A 1.50 m rod has density $\lambda = 2x$ kg/m (increasing from one end to the other).

M = ∫0^1.5 2x dx = [x²] = 2.25 kg
∫ x dm = ∫0^1.5 2x² dx = [2x³/3] = 2.25 kg·m
x_cm = 2.25 / 2.25 = 1.00 m   (shifted from the middle, 0.75 m, toward the heavy end)

Motion of the system

Take the derivative of the position formula with respect to time:

$\vec v_{cm} = \frac{\sum m_i\vec v_i}{M} = \frac{\vec p_{\text{total}}}{M} \qquad \sum\vec F_{\text{ext}} = M\vec a_{cm}$

The total momentum of the system is as if all the mass were gathered at the center of mass and moving with $v_{cm}$. By Newton’s 3rd law, internal forces cancel each other in pairs, so only external forces remain in the equation.

The center of mass in a collision

Recall the collision on the previous page: a 1500 kg car hit a stationary 2500 kg truck at 20.0 m/s and stuck to it. The velocity of the center of mass before the collision:

$v_{cm} = \frac{(1500)(20.0) + (2500)(0)}{4000} = 7.50\ \text{m/s}$

After the collision the velocity of the stuck wreckage was also 7.50 m/s. This is not a coincidence. Collision forces are internal forces and cannot change the velocity of the center of mass. Objects that stick together already move with the center of mass. This is a quick way to check in perfectly inelastic collision questions.

Which force is internal and which is external depends on how you choose the system. If you choose the car and the truck together as the system, the collision force is internal. If you choose only the car, the force exerted by the truck is an external force and the momentum of the car changes. Write the system clearly at the start of the question.

Consequences:

• Explosion: the fragments of a shell exploding in the air spread out, but the center of mass continues on the old parabola.
• Recoil: in a system initially at rest, the total momentum stays zero. The rifle recoils in the direction opposite to the bullet: $mv + MV = 0$.
• Walking in a boat: on frictionless water, the center of mass of the boat and person system does not move. If the person goes forward, the boat goes back.

PITFALL · the distance when walking in a boat. The person walks $L$ relative to the boat, and the boat goes back $d$ relative to the shore. The displacement of the person relative to the shore is $L - d$. The equation is $m(L - d) = Md$, and from it $d = \dfrac{mL}{m + M}$. Writing only $M$ in the denominator is the most common mistake.

Rocket propulsion (short note)

A rocket goes forward by throwing exhaust gas backward. This is continuous recoil; no external force is needed, and it works in a vacuum too. If the exhaust leaves with speed $v_e$ relative to the rocket (exhaust speed):

$\text{Thrust} = v_e\left|\frac{dM}{dt}\right| \qquad v_f - v_i = v_e\ln\frac{M_i}{M_f}$

Worked example 4. $v_e = 2800\ \text{m/s}$, the initial mass is 3 times the final mass, and fuel burns at 15.0 kg/s.

Δv = 2800 · ln 3 = 2800 · 1.0986 = 3.08 × 10³ m/s
Thrust = 2800 · 15.0 = 4.20 × 10⁴ N

The speed increases with the logarithm of the mass ratio. Doubling the ratio does not double the speed; that is why rockets have many stages.

Mixed exam examples

Worked example 5 (explosion in the air). A 10.0 kg shell is moving horizontally at 30.0 m/s at the top of its trajectory when it splits into two pieces. The 4.00 kg piece is at rest immediately after the explosion (and then falls straight down). What is the velocity of the other piece?

Horizontal momentum is conserved at the moment of the explosion (gravity is vertical, the explosion is very short):
10.0 · 30.0 = 4.00 · 0 + 6.00 · v   →   v = 300 / 6.00 = 50.0 m/s (in the same direction)
K before: ½ · 10.0 · 30.0² = 4.50 × 10³ J
K after:  ½ · 6.00 · 50.0² = 7.50 × 10³ J   (the chemical energy of the explosion was added)

In an explosion the kinetic energy increases. This is the opposite of an inelastic collision: internal energy is converted into kinetic energy. The momentum is conserved again.

Worked example 6 (boat). A 60.0 kg person walks 5.00 m relative to the boat from one end to the other of a 140 kg boat resting on still water.

Motion of the boat relative to the shore: d = m L / (m + M) = 60.0 · 5.00 / 200 = 1.50 m (backward)
Motion of the person relative to the shore: 5.00 - 1.50 = 3.50 m (forward)
Check: 60.0 · 3.50 = 210 = 140 · 1.50   ✓ (CM in place)

Common mistakes