Physics - Rotation of point mass (particle)

On this page we derive the rotation values from a point mass (particle). The point mass is not necessarily rotating about its own axis (although it could, subatomic particles have spin). What we are interested in here is the contribution of the particle to the rotational properties of a bigger mass about some fixed point. If I am not making myself very clear try reading numerical methods.

The following pages will then go on to derive quantities for finite solid bodies by integrating across the volume. Most of these quantities are vectors of dimension 3 which has a component in the x,y and z directions. To denote a vector quantity we show an arrow above the quantity, for more information about vectors see here.

Angular Velocity ()

See under kinematics here.

Angular Momentum ()

The instantaneous angular momentum of a particle is the product of its mass and its angular velocity.

= x = x m =m

h the instantaneous angular momentum about PC bivector kg m2/s
x the cross product operator (see here for definition)    
r the instantaneous position of the particle relative to point. pv-pc vector m
p the instantaneous linear momentum of the particle = mv vector kg m/s
m the mass of the particle. scalar kg

As with rotation velocity, the angular momentum of a point is not an absolute value, but it depends on which point that the rotation is measured about.

As discussed here, angular momentum of any closed system is conserved. However the particle under consideration here, may have external forces acting on it, so it is not a closed system. When we go on to consider a rigid body rotating about its centre of mass, then the angular momentum of the whole system (about any point) is conserved, this is the sum of the angular momentums of the particles that make up the body. So as it rotates, the instantaneous angular momentum of each particle that makes up the body, will be continuously varying, but the total angular momentum is constant.

Angular Acceleration ()

See under kinematics here.

Torque ()

The torque, about point , caused by a force is the cross product of distance from and the force .

= x

t the instantaneous torque about PC bivector N m
r instantaneous position of the particle relative to point. pv-pc vector m
f instantaneous force on the particle. vector N
x vector cross product operator(see here for definition)    

Note that this is both a linear force and a torque about . If we want a 'pure' torque without any net linear force then, we can apply two equal and opposite forces to two particles which are offset. If we want a 'pure' linear force without any net torque, then this will only happen if the torque is measured about the centre of mass of a system of particles, and the force is applied in line with the centre of mass. However on this page we are considering a single particle only, we will go on to consider solid bodies later.

Inertia [I]

We now want to derive the relationship between and .

For, linear motion we have =m from Newtons second law.

It turns out that the equivalent of Newtons second law for rotation is:

= [I]

t the instantaneous torque about PC bivector kg m/s2
[I] moment of inertia tensor kg m2
a the instantaneous angular acceleration about PC bivector m/s2

But how do we derive this for a particle? and why is the rotational equivalent of the scalar m (mass) a matrix [I]? what are the components of [I] and how are they related to m?

We will try to answer these questions as follows, start with the linear version of Newtons second law:


cross multiply both sides by :

x = x (m)

substituting = x and = x


= x ( x m)

It is not easy to combine these two consecutive cross multiplications by because vector are not associative under cross product { x * (y * z) is not (x * y) * z }. however there is a trick which can be done here:

Method using Matrices

The vector cross product can by replaced by matrix multiplication with a skew symmetric matrix as defined below.

So define

[~r] is equivalent to x

so if:



[~r] =
0 -r3 r2
r3 0 -r1
-r2 r1 0

So applying this to = x ( x m) gives:


Since matrix multiplication is associative we can calculate [~r][~r] first to give:

[~r][~r] =
0 -r3 r2
r3 0 -r1
-r2 r1 0
0 -r3 r2
r3 0 -r1
-r2 r1 0
-r3*r3 - r2*r2 r2*r1 r3*r1
r1*r2 -r3*r3 -r1*r1 r3*r2
r1*r3 r2*r3 -r2*r2 - r1*r1

So substituting this gives the relationship between and

-r3*r3 - r2*r2 r2*r1 r3*r1
r1*r2 -r3*r3 -r1*r1 r3*r2
r1*r3 r2*r3 -r2*r2 - r1*r1

This is for a single particle, in the next pages we will show how to integrate these across a solid volume:

alternative derivation of inertia matrix using multivectors

Multivectors are defined here. If we start with the equation derived above:

= x ( x m)

but replace the vector cross produce 'x' with the outer product then the equation becomes:

= ( m)

So if we expand out the vectors we get:

tx 1 +ty 2 +tz 3 = (rx 1 +ry 2 +rz 3) ((rx 1 +ry 2 +rz 3) m(ax 1 +ay 2 +az 3))

tx 1 +ty 2 +tz 3 = m(rx 1 +ry 2 +rz 3) ((ry*ax - rx*ay) 12 +(rx*az - rz*ax) 31 +(rz*ay - ry*az) 23)

so separating out the 1, 2 and 3 basis gives:

tx = - (ry*ax - rx*ay) ry + (rx*az - rz*ax) rz
ty = (ry*ax - rx*ay) rx - (rz*ay - ry*az) rz
tz = -(rx*az - rz*ax) rx + (rz*ay - ry*az) ry

grouping the terms together gives:

tx = (-ry^2 -rz^2)ax + (rx*ry) ay + (rx*rz) az
ty = (ry*rx)ax + (-rx^2 -rz^2) ay + (ry*rz) az
tz = (rx*rz)ax + (ry*rz) ay + (-rx^2 -ry^2) az

Which is the same result as the matrix method, but it does not give us any more information, what I would really like to do is combine this rotational expression with the equivalent linear equation F=ma. Can anyone help me find a way to do that?

Angular momentum principles

  1. The angular momentum of a solid object depends on the point that we are measuring to rotation around.
  2. Even a non-rotating object, which is travelling in a straight line in the inertial frame has a non zero angular momentum if its trajectory does not pass through the point we are measuring to rotation around.
  3. Angular momentum is only conserved when we are measuring the momentum about the same unchanging point for all objects.
  4. Angular momentum is only conserved for closed systems, we have to be careful to take into account all external influences on the system.

Further Inertia topics

Angular Energy (E)

Energy is a scalar quantity

E = 1/2 m ()

w the instantaneous angular velocity about . (units radians / sec) bivector s-1
the dot product operator (see here for definition)    
m the mass of the particle scalar kg
v the instantaneous linear velocity vector m/s

E = 1/2 m (( x )• ( x ))

if I = m then

E = 1/2 I ()

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Correspondence about this page

Book Shop - Further reading.

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