Center of Mass (College Board AP® Physics 1: Algebra-Based)

Study Guide

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Center of mass

  • The center of mass of a system is defined as:

The point where all of an object’s mass can be thought to be concentrated

  • When a system consists of a single object:

    • it can be modeled as a point particle that is located at the system's center of mass

    • all interactions can be considered to act at the system's center of mass

Diagram of a broom balanced on a pivot showing the center of mass, with the handle extending left and the bristles on the right.
The center of mass of a single object is the point where all of its mass is thought to act, as a result, it is the point the weight of the object acts.
  • When a system consists of multiple objects:

    • each object can be modeled as a point particle, with an associated mass and position which contributes to the system's center of mass

    • interactions between objects within a system (i.e. internal forces) do not influence the motion of a system’s center of mass

Examiner Tips and Tricks

Remember, the center of mass is a hypothetical point, so it can lie inside or outside of a body. The center of mass of a non-rigid body can change depending on its shape. For example, a person’s center of mass is lower when learning forward than when standing upright

Diagram showing a person in three poses: standing, reaching up, and bending forward, with a label indicating the center of mass shifts.

Locating center of mass

  • For systems with symmetrical mass distributions, the center of mass is located on lines of symmetry

Center of mass of a symmetrical object

Four shapes with lines indicating the center of mass: triangle, oval, trapezoid, and parallelogram, each labeled "CENTRE OF MASS."
  • The location of a system’s center of mass along a given axis can be calculated using the equation:

x with rightwards arrow on top subscript c m end subscript space equals space fraction numerator sum for i of m subscript i x with rightwards arrow on top subscript i over denominator sum for i of m subscript i end fraction

  • Where:

    • x with rightwards arrow on top subscript c m end subscript = position of the system's center of mass, in straight m

    • m subscript i = mass of each object, in kg

    • x with rightwards arrow on top subscript i = position of each object, in straight m

Calculating the center of mass of a 1D system

  • Consider a system of three objects located along the same line

  • The position of the system's center of mass can be determined using:

x with rightwards arrow on top subscript c m end subscript space equals space fraction numerator m subscript 1 x subscript 1 plus m subscript 2 x subscript 2 plus m subscript 3 x subscript 3 over denominator m subscript 1 plus m subscript 2 plus m subscript 3 end fraction

Position of the center of mass on a line

Diagram showing three masses m1, m2, and m3 on a line with distances x1, x2, x3 from the origin. A triangle marks the center of mass.
The center of mass of a system of particles along the same line will be located somewhere along that line

Calculating the center of mass of a 2D system

  • Consider a system of three objects located along the same x-y plane

  • The position of the system's center of mass can be determined using:

r with rightwards arrow on top subscript c m end subscript space equals space fraction numerator sum for i of open parentheses m subscript i r with rightwards arrow on top subscript i close parentheses over denominator sum m subscript i end fraction space equals space fraction numerator m subscript 1 r with rightwards arrow on top subscript 1 plus m subscript 2 r with rightwards arrow on top subscript 2 plus m subscript 3 r with rightwards arrow on top subscript 3 over denominator m subscript 1 plus m subscript 2 plus m subscript 3 end fraction

  • Where:

    • r with rightwards arrow on top subscript c m end subscript = position vector of the system's center of mass, in straight m

    • r with rightwards arrow on top subscript i = position vector of each particle, in straight m

Position of the center of mass on an x-y plane

Diagram of a system of particles, showing masses m1, m2, and m3 in pink shapes. The center of mass is marked with vectors and coordinates.
A system of particles along the same x-y plane can be described in terms of the position of the system's center of mass
  • The positions of the separate x and y coordinates of the system's center of mass can be determined using:

x with rightwards arrow on top subscript c m end subscript space equals space fraction numerator m subscript 1 x subscript 1 plus m subscript 2 x subscript 2 plus m subscript 3 x subscript 3 over denominator m subscript 1 plus m subscript 2 plus m subscript 3 end fraction

y with rightwards arrow on top subscript c m end subscript space equals space fraction numerator m subscript 1 y subscript 1 plus m subscript 2 y subscript 2 plus m subscript 3 y subscript 3 over denominator m subscript 1 plus m subscript 2 plus m subscript 3 end fraction

  • Using Pythagoras' theorem, the magnitude of the position vector is:

r with rightwards arrow on top subscript c m end subscript space equals space square root of open parentheses x with rightwards arrow on top subscript c m end subscript close parentheses squared space plus space open parentheses y with rightwards arrow on top subscript c m end subscript close parentheses squared end root

  • Using trigonometry, the direction of the position vector can be found from:

tan space theta space equals space y with rightwards arrow on top subscript c m end subscript over x with rightwards arrow on top subscript c m end subscript

Examiner Tips and Tricks

In AP Physics 1, you will only be expected to calculate the center of mass for systems of five or fewer particles arranged in a two-dimensional configuration or for highly symmetrical systems

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.