Systems and Center of Mass: Study Guide (Physics C: Mechanics 2.1)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses, and both have a Topic 2.1 with the same title. This guide is the calculus-based one. As well as sums over particles, it uses integrals to find the mass and center of mass of objects whose density changes from place to place. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Choosing a system

In Unit 1 you treated every moving thing as a point. From now on you must first decide what you are analysing. A system is the object or group of objects inside a boundary that you choose. Everything outside is the environment.

The choice is yours, and a good choice makes a problem easier. A car can be one system. So can the car plus its passengers, or just one wheel. Always say it in words, for example “system: cart and the brick on it”.

Four ideas follow from that choice.

1. A system’s properties come from the interactions inside it. A foam mattress is springy because of how its cells push on one another. A stack of magnets holds its shape because of the forces between the magnets. Change those interactions and the system behaves differently.
2. Systems can be open. Mass or energy can cross the boundary. A rocket loses mass as exhaust leaves it. A pan on a hob gains energy from the flame. If nothing crosses, the system is closed for that quantity.
3. Parts can behave differently from the whole. When a diver tucks during a dive, her arms and legs move in quite different ways, but the system “whole diver” follows one smooth path through the air.
4. Internal structure can matter, and it can change. A full, sealed water tank on a lorry behaves like a solid block. A half-full tank does not: when the lorry brakes, the water surges forward. Changing something outside the system (the lorry’s braking) has changed the arrangement inside it.

When is a system one object?

If the parts and their interactions do not affect the question you are asking, you can model the system as a single object. You do not need to know how a falling sack of rice is packed to find when it lands. You do need the inner details to decide whether a tall load on a trolley tips over when the trolley stops suddenly.

When you do treat a system as one object, you place it at one special point: its center of mass.

Center of mass of particles

For particles of mass mᵢ at positions xᵢ along one axis, the center of mass is the mass-weighted average position:

x_cm = Σmᵢxᵢ / Σmᵢ

Repeat it for each axis: y_cm = Σmᵢyᵢ / Σmᵢ, and z_cm the same way. In vector form, r_cm = Σmᵢrᵢ / M, where M = Σmᵢ is the total mass.

Three habits keep you safe:

• Positions are signed. A mass at x = −0.30 m pulls x_cm towards negative x.
• Divide by the total mass, not by the number of particles.
• The answer does not depend on where you put the origin. Move the origin and every xᵢ and x_cm shift by the same amount, so the physical point stays put. Pick the origin that makes the arithmetic easiest.

For two particles, the center of mass lies on the line joining them, closer to the heavier one. The distances from each particle are in the inverse ratio of the masses.

Symmetry and uniform pieces

For an object with a symmetrical mass distribution, the center of mass lies on every line (or plane) of symmetry. So a uniform rod has its center of mass at its midpoint, and a uniform rectangle, disc or sphere at its geometric center.

This lets you split a complicated object into uniform pieces. Replace each piece by a particle, with the piece’s mass, at the piece’s own center of mass. Then use the particle formula. A missing piece (a hole) can be handled as a piece with negative mass.

Center of mass of a continuous object

Now let the density change along the object. Cut the object into tiny elements, each of mass dm at position x. The sum becomes an integral:

x_cm = ∫x dm / ∫dm, where ∫dm = M, the total mass.

For a rod or other thin object along the x-axis, describe the mass with the linear mass density:

λ = dm/dx (unit kg/m)

So each element has dm = λ(x) dx, and

M = ∫λ(x) dx and x_cm = (1/M) ∫x λ(x) dx, both over the length of the rod.

For a uniform rod, λ is constant and x_cm comes out at the midpoint, as symmetry says it must.

The same idea works in two and three dimensions. With an area density σ (kg/m²) the mass is M = ∫σ dA. With a volume density ρ (kg/m³) it is M = ∫ρ dV. In this course the useful trick is to choose elements that make the integral one-dimensional: for a flat plate whose density changes only with x, use thin strips parallel to y, each of area (height × dx).

The system as one object at its center of mass

Once you have the center of mass, you can replace the whole system by a single particle of mass M placed there. This is the model behind every free-body diagram in the rest of the unit (Topic 2.2): the dot you draw stands for the center of mass.

The model has limits. It tells you about the motion of the system as a whole. It does not tell you how the parts move relative to each other. A thrown spanner spins as it flies, and any point on its handle traces a looping path, but, with air resistance ignored, its center of mass follows a smooth parabola, like the projectiles of Topic 1.5. Why that happens is shown in Unit 4 with momentum; here, just notice that the parts and the whole behave differently.

Worked example 1: an L-shaped bracket

Question. A bracket is made from two thin uniform steel bars welded at a right angle. The horizontal bar is 0.40 m long with mass 1.2 kg. The vertical bar is 0.30 m long with mass 0.90 kg. Take the origin at the corner, +x along the horizontal bar and +y up the vertical bar. (a) Find the center of mass of the bracket. (b) A 0.30 kg clamp is then fixed at the free end of the horizontal bar. Find the new center of mass.

1. Use symmetry for each bar. Each bar is uniform, so its center of mass is at its midpoint: horizontal bar at (0.20 m, 0); vertical bar at (0, 0.15 m).
2. Total mass: M = 1.2 + 0.90 = 2.1 kg.
3. x-coordinate: x_cm = (1.2 × 0.20 + 0.90 × 0) / 2.1 = 0.24 / 2.1 = 0.114 m.
4. y-coordinate: y_cm = (1.2 × 0 + 0.90 × 0.15) / 2.1 = 0.135 / 2.1 = 0.064 m.
5. (b) Add the clamp as a particle at (0.40 m, 0). New total mass 2.4 kg. x_cm = (0.24 + 0.30 × 0.40) / 2.4 = 0.36 / 2.4 = 0.150 m; y_cm = 0.135 / 2.4 = 0.056 m.

Check. The center of mass lies on the line joining the two bar centers, dividing it in the ratio 0.90 : 1.2, nearer the heavier bar. It is not on either bar: the bracket’s center of mass is in empty space. Adding the clamp at the far end of the horizontal bar moved x_cm towards it (0.114 m → 0.150 m) and pulled y_cm down a little, because the new mass sits at y = 0.

Worked example 2: a non-uniform rod by integration

Question. A wooden pointer is 0.80 m long and thicker at one end. Take the origin at the thin end and +x along the pointer towards the thick end. Its linear mass density is λ(x) = 0.50 + 1.5x, with λ in kg/m and x in m. Find (a) its mass and (b) the position of its center of mass.

1. Units first. The 0.50 is in kg/m. The 1.5 multiplies a length, so it is in kg/m²; then 1.5x is in kg/m too.
2. (a) Mass: M = ∫₀^0.80 (0.50 + 1.5x) dx = [0.50x + 0.75x²]₀^0.80 = 0.40 + 0.48 = 0.88 kg.
3. (b) Moment integral: ∫₀^0.80 x(0.50 + 1.5x) dx = [0.25x² + 0.50x³]₀^0.80 = 0.16 + 0.256 = 0.416 kg·m.
4. Divide by the mass: x_cm = 0.416 / 0.88 = 0.47 m from the thin end (0.473 m to 3 s.f.).

Check with a limiting case. If the 1.5x term were zero, the pointer would be uniform. The general result for λ = 0.50 + bx over 0.80 m reduces to 0.40 m when b = 0, the midpoint, as symmetry requires. With b > 0 the extra mass is towards the thick end, so x_cm moves past 0.40 m, which is what we found.

A point to notice. The mass to the left of x_cm is ∫₀^0.473 λ dx ≈ 0.40 kg, and to the right ≈ 0.48 kg. They are not equal. The point with half the mass on each side is at about 0.50 m. The center of mass weights each element by its distance, so the heavier elements far out count for more.

Common misconceptions

• “The center of mass is always at the geometric center.” Only for a uniform, symmetric object. Worked example 2 shows a rod whose center of mass is 0.07 m past its midpoint.
• “The center of mass must be inside the material.” The bracket in Worked example 1, a ring and a horseshoe all have it in empty space.
• “Half the mass is on each side of the center of mass.” Not in general. For the pointer, the split is about 0.40 kg and 0.48 kg.
• Dividing by the number of particles. The formula divides by the total mass, Σmᵢ.
• Dropping signs. A particle at negative x must enter the sum as a negative term.
• Confusing λ with mass. λ is mass per unit length. You must integrate λ dx to get a mass; λ × (length) works only when λ is constant.
• Integrating x λ and stopping there. ∫x λ dx has unit kg·m. Divide by M to get a position.
• “A system can always be treated as one point.” Only when the inner details do not matter for the question. A sloshing tank or a tipping load needs its internal structure.

Where this leads

The center of mass is the dot you draw in every free-body diagram, and choosing a system decides which forces appear on it. That is the next step: Topic 2.2, Forces and Free-Body Diagrams. The same “cut into dm and integrate” method returns in Unit 5 for rotational inertia, ∫r² dm, and in Unit 4 you will see why the center of mass of an isolated system keeps moving steadily however its parts move. Earlier: Topic 1.5, Motion in Two or Three Dimensions. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.