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Let us rotate a point $P$ an angle $\theta$ around a circle in two dimensional space.

Assume a particle is rotating around an axis in the $xy$-plane at a radius $r=\mathrm{OP}$. During $\Delta t$ the particle turns $\Delta \theta$ and travels a distance $s=r\Delta \theta$ around the curve. Now if $\Delta \theta$ becomes smaller and smaller the line segment $PQ$ turns to make a perpendicular angle with $OP$: $\mathbf{PQ} \perp \mathbf{OP}$. In the limiting case $\Delta \theta \to 0$ we have: \[ PQ=r\sin\theta=r\Delta \theta \] We derive \[ \begin{align} \Delta x &= -\mathrm{PQ} \sin\theta=-r\Delta \theta (y/r) = -y\Delta \theta \tag{2} \\ \Delta y &= \mathrm{PQ}\cos\theta=r\Delta \theta (x/r) = x\Delta \theta \tag{3} \end{align} \]

First we notice that $s=\Delta \theta r$ as before, but of course we also have $s=v\Delta t$. Combining these two equations gives: \[ \frac{\Delta \theta}{\Delta t}=\frac{v}{r} \] where $v$ is de speed in $m/s$. We define the angular velocity in $radians/s$: \[ \omega=\frac{\Delta \theta}{\Delta t} \] and we have our relationship: \[ \omega=\frac{v}{r} \]

We know already that in the case $\Delta \theta \to 0$ the velocity vectors are perpendicular to the position vector. Based on geometry we can then state that the angle between the velocity vectors at $P$ and $Q$ is equal to the angle between the position vectors at $P$ and $Q$. We can then apply the same reasoning as we did for the distance $s$ and state that $\Delta v=v\Delta \theta$. Now to go from this change in velocity to the acceleration we must divide by $\Delta t$ we get: \[ \frac{\Delta v}{\Delta t}=v \frac{\Delta \theta}{\Delta t} \] however we found already: \[ \frac{\Delta \theta}{\Delta t}=\frac{v}{r} \] so we have: \[ a=\frac{v^2}{r} \] or in angular velocity: \[ a=\omega^2 r \] note that the angular acceleration is defined by \[ \alpha=\frac{d^2\theta}{\theta^2} \]

We can also express the position vector: \[ r=x\vec{i}+y\vec{j} \] in polar coordinates $(r,\theta)$: \[ x=r\cos\theta \] \[ y=r\sin\theta \] \[ r=(r\cos\theta)\vec{i}+(r\sin\theta)\vec{j} \] In case of uniform velocity we also have: $\theta=\omega t$ and so we tranform to an equation of time: \[ r(t)=(r\cos\omega t)\vec{i}+(r\sin\omega t)\vec{j} \] based on this relation we can find $v(t), a(t)$. It also follows then algebraically: \[ v(t).r(t)=0 \] and thus they are indeed perpendicular.

That is easy, \[ \Delta W = \vec{F}\cdot \Delta \vec{x} = F_x \Delta x + F_y \Delta y \tag{1} \]

\[ \Delta W = (xF_y-y F_x) \Delta \theta \]

Well, it is the torque in the $xy$-plane. We denote is by $\tau$. Notice that it depends on the position vector and $F$. Its magnitude is equal to the area of the parallelogram spanned by these two vectors. Therefore we have following equivalent definitions of $\tau$: \begin{align*} \tau &=xF_y-y F_x \\ &=rF_{tan}\\ &=F\cdot\text{lever arm} \end{align*}

Then the work done is the sum of the work done by all the forces. We can calculate the torque for each force, $\tau_i=x_iF_{yi}-y_iF_{xi}$ add them together $\tau = \sum \tau_i$ and multiply by $\Delta \theta$: \[ \Delta W = \tau \Delta \theta \]

We apply Newton's second law, $F_x=m\tfrac{d^2x}{dt^2}$ and $F_y=m\tfrac{d^2y}{dt^2}$, and substitute this in the expression for torque: \[ \tau = xm\frac{d^2y}{dt^2}-ym\frac{d^2x}{dt^2} \]

but this is equivalent to the time derivative of something: \[ \tau=\frac{d}{dt}(xm\frac{dy}{dt}-ym\frac{dx}{dt}) \] but the time derivative of the linear momentum is equal to the force: \[ F=\frac{d}{dt}p=\frac{d}{dt}(m\frac{dx}{dt}+m\frac{dy}{dt})=\frac{d}{dt}(p_x+p_y) \] Therefore we call this something theJust as the total momentum of an object is the sum of the momenta of its parts, so is the total angular momentum the sum of the angular momenta of its parts. Then the change in this total angular momentum is the total torque: \[ \tau=\sum\tau_i=\sum \frac{dL_i}{dt}=\frac{dL}{dt} \]

The internal torques result from internal forces. But the internal forces come in action reaction pair and are directed exactly opposite along the same line. Then the two torques on the reacting objects will be equal and opposite because the lever arms for any axis are equal. The internal torques balance out pair by pair. Thus a change in the angular momentum is the result of applying an external torque. Without any external torque the angular momentum remains constant: \[ \tau=\sum\tau_i=\tau_{ext}=\frac{dL}{dt} \]

If the mass of one of its particles is $m_i$ at a distance $r_i$ from the axis of rotation, then the angular momentum for this particle is just the product of $r$ and its momentum: \[ L_i=r_ip_i=r_im_iv_i=r_i^2m_i\omega \] summing over all particles: \[ L=I\omega \] where \[ I=\sum_i m_ir_i^2 \] This is analog to the law that the momentum is mass times velocity. Velocity is replaced by angular velocity and mass by the moment of inertia $I$, which is the analog to mass. So the inertia for turning depends not only on the mass but alos on the distribution of the mass from the axis of rotation.

The torque in the $xy$ plane is $xFy-yFx$. In 3 dimensional space we have 3 coordinate axes and we can form 3 perpendicular planes with these coordinate axes: $xy,yz,xz$. We could define the torque in 3 dimensional space as a combination of the torques in these 3 planes like: \begin{align*} \tau_{xy}&=xF_y-yF_x \\ \tau_{yz}&=yF_z-zF_y \\ \tau_{zx}&=zF_x-xF_z \end{align*} The signs we used depend on the convention of the orientation of the coordinate axes. It turns out that with three axis we can have two different type of orientations: right or left. In a right handed system the direction of the $z$ axis is aligned with the direction of a right handed screw in the $xy$ plane when it is turned in the direction from the positive $x$ axis to the positive $y$ axis. If this turn is clockwise the screw moves into the plane and if the turn is counterclockwise the screw moves out of the plane. In a right-handed system the xy, yz,zx combinations have a positive sign.

Although we can express the torques in the three perpendicular planes of our coordinate axis, nothing witheld us from using three other planes. However it turn out that this new representation is just a combination of the expressions for the $xy$ $yz$ and $xz$ plane.

It seems natural to define the torque as a vector as follows: \begin{align*} \vec{\tau}&=(\tau_{yz},\tau_{zx},\tau_{xy}) &=(yF_z-zF_y,zF_x-xF_z,xF_y-yF_x) \end{align*}

To proof that this quantity is a vector it can be shown that a coordinate transformation transforms the torque in the same way as a normal vector would be tranformed. This definition for torque defines its magnitude, and direction which is perpendicular to the plane formed by $r$ and $F$. For defining its sign we assume a right handed coordinate system and define its sign with the right-screw rule.

That the torque is a vector in 3 dimensional is just a miracle of good luck that we have 3 independent planes and 3 dimensions. In 2D torque is just a number, and in 4D we would have had 6 planes but only 4 dimensions. However we must not forget that the torque is a pseudovector invented for computation. We can not measure a torque in space like the position or a force.

The relationship between $r$ and $F$ and the torque is mathematically defined as the cross product between two vectors: \[ \vec{\tau}=\vec{r} \times \vec{F} \]

But then we can also do this for the angular momentum: \[ \vec{L}=\vec{r}\times\vec{p} \] and the dynamic law: \[ \vec{\tau}=\frac{d\vec{L}}{dt} \] from which follows that without torque there is no change in the angular momentum.

**[1]**Richard P. Feynman, The Feynman Lectures on Physics Volume I,1963.

Copyright ©2012 Jacq Krol. All rights reserved. Created ; last updated .