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Center of Mass & Linear Momentum

Source lecture(s): SC133 Lec 10

Intuition

Toss a hammer spinning through the air: every point traces a wild curve — except one. One special point rides a perfect projectile parabola as if the hammer were a single particle. That point is the center of mass (CM): the mass-weighted average position, the handle by which Newton's laws grab an extended object.

Definition

\[\vec r_\text{cm} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i} \qquad\text{or, for continuous bodies,}\qquad \vec r_\text{cm} = \frac{1}{M}\int \vec r\, dm\]

For symmetric uniform bodies the CM sits at the geometric center. It need not lie inside the material at all — a doughnut's CM is in the hole, and a high-jumper's CM can pass under the bar while the body arcs over it (the Fosbury flop).

Why the CM is special

Differentiate twice and use Newton's third law: all internal forces cancel in pairs, leaving

\[\boxed{\,M\vec a_\text{cm} = \sum \vec F_\text{ext}\,}\]

The CM of any system — hammer, exploding firework, diving cat — obeys the single-particle Newton's second law under the external forces alone. Internal rearrangement, however violent, cannot move the CM.

Linear momentum of a system

Total momentum is the CM in disguise:

\[\vec P = \sum_i m_i\vec v_i = M\vec v_\text{cm}, \qquad \frac{d\vec P}{dt} = \sum\vec F_\text{ext}\]

If the net external force is zero, \(\vec P\) is constant — the conservation law that powers all of collision physics. More on the momentum concept itself: linear momentum.

Worked example: walking on a boat

A 60 kg person walks 3 m toward the shore-end of a 120 kg boat (frictionless water). How far does the boat move?

No external horizontal force ⇒ the CM stays put. Let the boat shift \(d\) backward; the person moves \(3 - d\) forward in the ground frame:

\[60(3 - d) = 120\,d \Rightarrow d = 1\,\text{m}\]

The boat slides a metre backward — and no amount of walking, jumping, or shoving can carry the system's CM ashore.

Worked example: exploding projectile

A shell following a parabola explodes at its apex into two equal fragments; one drops straight down. Where does the other land?

The CM continues on the original parabola and would land at range \(R\). With one fragment at \(R/2\) (below the apex), symmetry of the CM average puts the other at \(\tfrac{3R}{2}\).

Common mistakes

  • Thinking internal forces can shift the CM. Rockets work by throwing mass backward — the exhaust's momentum is real; the CM of (rocket + exhaust) never accelerates without external force.
  • Placing the CM inside the body by reflex — check the geometry (L-shapes, rings, crescents).
  • Forgetting the CM velocity is momentum ÷ total mass — a useful instant sanity check in collision problems.

Knowledge graph position

Prerequisites: Newton's laws. Leads to: Collisions, Rotation.

Quiz

Q1 (computational). Masses 2 kg at \(x = 0\) and 6 kg at \(x = 4\) m. Where is the CM?

Answer

\(x_\text{cm} = (2\cdot0 + 6\cdot4)/8 = 3\,\text{m}\) — three times closer to the heavier mass, as the inverse-ratio rule demands.

Q2 (conceptual). An astronaut floating at rest in space throws a wrench. Describe the CM of (astronaut + wrench) afterward.

Answer

Still at rest, forever. The throw is internal; astronaut and wrench carry equal and opposite momenta, and the CM stays fixed — which is exactly why throwing the wrench propels the astronaut.

Q3 (multiple choice). A firework explodes mid-flight. Immediately after, the CM of all fragments: (a) stops (b) continues on the pre-explosion trajectory (c) scatters unpredictably

Answer

(b). The explosion is internal; only gravity (external) acts on the CM, which continues its parabola until fragments start landing.