## Prerequisites

If you are not familiar with this subject you may like to look at the following pages first:

## Kinetic Energy of Solid Body

For purely linear motion the kinetic energy of a solid body is:

T = 0.5 m v^{2}

For purely rotational motion the kinetic energy, about its centre of mass, of a solid body is:

T = 0.5 w^{t} [I] w

So if we have combined linear and rotational motion can we just add the kinetic energy due to these parts separately? I think the answer is no because the energy of a particle is proportional to the square of its velocity. We need to add on another factor:

T = 0.5 m v^{2} + 0.5 w^{t} [I] w + m(v•(w x r))

This can be derived as follows:

T for particle on solid body = 0.5 sum(mi vi^{2} )

but the velocity of a particle on a solid body is (v + w x r) so the kinetic energy is:

T = 0.5 sum(mi (v + w x r)•(v + w x r))

this can be expanded out to give:

T = 0.5 sum(mi (v^{2} + (w x r)•(w x r) +2v•(w x r))

So, there is a part that is due purely to linear motion, a part that is due purely to rotational motion, and a part that is due to the product of linear motion and rotational motion.

For further information see this page.

## Symmetrical body

It might be simplest to start with a symmetrical body, that is for every element of mass on one side of the centre of mass there is an identical element of mass on the other side.

## Rotational Energy

The rotational energy is the sum linear energy

rotational energy = linear energy of A + linear energy of B

= 0.5 (m/2) v^{2} + 0.5 (m/2) (-v)^{2}

= 0.5 (m/2) (w r)^{2} + 0.5 (m/2) (-w r)^{2}

= 0.5 m w^{2} r^{2}

= 0.5 I w^{2}

where I = m * r^{2}

In the more general case I is the moment of inertia which in the two dimensional case is the second moment of mass

So

energy of rotation = K = 0.5 I w^{2}

## Rotational Momentum

As explained here, the rotational momentum is a different quantity from linier momentum.

L = r x p

where:

- p = m * v
- x = cross product

So, L =m * (r x v)

In two dimensions:

I = m * r^{2}

and v = w r

Which gives, L = I w

So in the example above the inertia for each object is I=m/2 * r^{2}
So the total inertia is I=m * r^{2}

So the total angular momentum for the example above is: L=m * r^{2}
* w

In this example both w and L are vectors pointing toward the viewer since we are using a right hand coordinate system as explained here.

## Next

Combined linier and rotational energy and momentum of solid body