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**Definition/Summary**The moment of a vector is the cross-product of the position vector with that vector (e.g., a force or momentum vector).

So it is is a vector (strictly, a pseudovector), and obeys the rules of vector addition.

Moment of momentum is more usually called angular momentum.

In most examination questions, all the vectors are in the same plane (the plane of the examination paper), and so all the moments are parallel to each other (and perpendicular to the paper), and so they can all be treated as numbers instead of vectors.

**Equations**Moment of force = Torque: [itex]\mathbf{\tau}= \mathbf{r} \times \mathbf{F}[/itex]

Moment of momentum = Angular Momentum: [itex]\mathbf{\mathcal{L}}=\mathbf{r}\times \mathbf{p} [/itex]

Cross-product of Newton's second law with a fixed position vector [itex]\mathbf{r}[/itex] gives:

[tex]\mathbf{r} \times \sum \mathbf{F} = \mathbf{r} \times \frac{d \mathbf{p}}{dt}[/tex]

But [itex]d/dt(\mathbf{r} \times \mathbf{p}) = \mathbf{r} \times (d\mathbf{p}/dt)[/itex] since [itex](d \mathbf{r}/dt )\times \mathbf{p} =0[/itex], giving:

[tex]\mathbf{\tau}_{total}\equiv\sum\mathbf{r} \times \mathbf{F}=\frac{d}{dt}(\mathbf{r} \times \mathbf{p})= \frac{d \mathbf{\mathcal{L}}}{dt}[/tex]

i.e., total moment of force = rate of change of total angular momentum

A rigid body at each instant of time rotates about an axis.

Its total Moment of momentum (Angular Momentum) about any point on that axis is its Moment of Inertia about that axis times its angular velocity:

[tex]\mathbf{\mathcal{L}} = I \mathbf{\omega}[/tex]

By the Parallel Axis Theorem, this is equal to:

[tex]\mathbf{\mathcal{L}} = (I_C\,+\,Md^2) \mathbf{\omega}[/tex]

where M is the total mass, d is the distance from the centre of mass to that axis, and [itex]I_C[/itex] is the Moment of Inertia about the parallel axis through the centre of mass.

**Extended explanation**Every force

**F**has a moment about any

*point*P.

To find the moment, draw R, the point of application of the force, and L, the line of the force, and draw the perpendicular line PQ from P to L (so both Q and R lie on L).

For the Moment of a velocity, R is the position of the centre of mass, and L is the line of the velocity.

Then the moment of

**F**about P is the vector written "

**r**x

**F**" (pronounced "r cross F"), where

**r**is the position vector PR. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is PQ times F.

PQ is sometimes called the lever arm.

Note that if P is

*on*the line L, then P = Q, so PQ = 0, so the moment of the force is 0.

In nearly all exam problems, everything is in the same plane (the plane of the examination paper!), so

*all*the moment vectors are vertically out of the page.

In other words, they're all parallel to each other, so we can forget that they're vectors, and treat them simply as numbers, PQ times F.

We can take moments about

*any*point, so we always choose whatever point makes the calculations easiest.

Usually, it's the point of application of an unknown force, so that the moment of

*that*force is 0, making the equation shorter!

**Moment about an axis:**

Moment about a point is a vector, and so it has a component in any direction:

[tex](\mathbf{r}\times\mathbf{F})\cdot\hat{\mathbf{k}}\ [/tex]

Moment about an axis is simply the component

*in the direction of that axis*of the moment about any point on that axis.

It is independent of which point on the axis is chosen:

[tex]((\mathbf{r}\,+\,a\hat{\mathbf{k}})\times\mathbf{F})\cdot\hat{\mathbf{k}}\ =\ (\mathbf{r}\times\mathbf{F})\cdot\hat{\mathbf{k}}\ +\ a(\hat{\mathbf{k}}\times\mathbf{F})\cdot \hat{\mathbf{k}}\ =\ (\mathbf{r}\times\mathbf{F})\cdot\hat{\mathbf{k}}[/tex]

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