The work done is the same in all 4 cases. The easy way of seeing this is to use the Work-Energy Theorem : work done equals change in energy. Here there is only kinetic energy, and since this depends on $\omega^2$ the minus signs make no difference, hence the change in KE is the same in each case.

However, in some cases there is a bigger change $\Delta\omega$. There is a bigger change in momentum, and the force is applied for a longer time while the object rotates through a larger angle. So isn't more work being done?

Exactly the same puzzle occurs in linear motion. It takes the same amount of work to change velocity from -2m/s to +5m/s as from +2m/s to +5m/s. An object with an initial velocity of -2m/s continues moving left retarded by the force until its velocity is brought to 0m/s. Then it moves the same distance back to the right while being accelerated, arriving back where it started but now with a velocity of +2m/s. The distance moved by the point of application of the force is greater than if the starting velocity was +2m/s. Surely that means more work is done?

No. The explanation is that work is force times **displacement**, which is a vector, not force times **distance**. The displacement is the same in both cases.

While the velocity was brought to 0m/s, the object did work on the agent which provided the force. The force did -ve work on the object. This work could have been stored, eg in a spring. The stored work was returned to the object to give it a velocity of +2m/s. It takes zero net work to change the direction of the velocity (think of work done during circular motion), only to change the magnitude of velocity.

**In mechanics no**. Torque is *not* a fundamental quantity. it's only job is to describe *where in space* a force is acting through (the line of action). Torque just describes a force at a distance. Given a force $\boldsymbol{F}$ and a torque $\boldsymbol{\tau}$ you can tell that the force acts along a line in space with direction defined by $\boldsymbol{F}$, but location defined by $\boldsymbol{\tau}$ as follows $$ \boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F} \|^2 } $$

_{In fact, you can slide the force vector anywhere along its line and it won't change the problem, so the $\boldsymbol{r}$ calculated above happens to be the point on the line closest to the origin.}

It might be easier to discuss angular momentum first, since torque is the time derivative of angular momentum, just as force is the time derivative of linear momentum.

For a single particle with linear momentum $\boldsymbol{p} = m\boldsymbol{v}$ located at some instant at a point $\boldsymbol{r}$ the angular momentum is $$ \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$

So where is the momentum line in space? The momentum line is called the axis of percussion. It is located at

$$ \boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2 } = \frac{\boldsymbol{p} \times ( \boldsymbol{r} \times \boldsymbol{p})}{\| \boldsymbol{p} \|^2} = \frac{ \boldsymbol{r} (\boldsymbol{p} \cdot \boldsymbol{p}) - \boldsymbol{p} ( \boldsymbol{p} \cdot \boldsymbol{r}) }{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \frac{ \| \boldsymbol{p} \|^2}{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \; \checkmark $$

provided that the point $\boldsymbol{r}$ is perpendicular to the momentum $\boldsymbol{p}$. Let me elaborate. Imagine the direction of the line being $\boldsymbol{\hat{e}} = \boldsymbol{p} / \| \boldsymbol{p} \|$, and consider a point $\boldsymbol{r} + t \boldsymbol{\hat{e}}$ for some arbitrary scalar $t$. The angular momentum will be $\boldsymbol{L} = ( \boldsymbol{r} + t \boldsymbol{\hat{e}}) \times \boldsymbol{p} = \boldsymbol{r} \times \boldsymbol{p} $. So where along the line (the value of $t$) doesn't matter. Finally, if $\boldsymbol{r}$ *is not perpendicular* to $\boldsymbol{p}$ you can *always* find a value of $t$ that makes the point perpendicular. Set $t = -(\boldsymbol{r} \cdot \boldsymbol{p}) / \| \boldsymbol{p} \|$ and the point *will* be perpendicular.

_{Such a point can always be found, and it is the point on the line closest to the origin.}

The conservation law for angular momentum (coupled with the conservation law for linear momentum) just states that not only the magnitude and direction of momentum is conserved **but also the line in space where moment acts through is also conserved**. So not only which direction is momentum point, by where is space it exists.

To visualize this, consider a case where you want to remove the momentum of a freely rotating body that is moving in space. You have a hammer, and you need to find out the following in order to completely stop the body. a) how much momentum to hit it with (the magnitude), b) in which direction to swing (direction) and c) where to hit it (location).

In summary, the common quantities in mechanics are interpreted as follows

$$ \begin{array}{r|l|l}
\text{concept} & \text{value} & \text{moment}\\
\hline \text{rotation axis} & \text{rot. velocity}, \boldsymbol{\omega} & \text{velocity}, \boldsymbol{v} = \boldsymbol{r}\times \boldsymbol{\omega} \\
\text{line of action} & \text{force}, \boldsymbol{F} & \text{torque}, \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \\
\text{axis of percussion} & \text{momentum}, \boldsymbol{p} & \text{ang. momentum}, \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}
\end{array} $$

The stuff under the **value** column are fundamental quantities that give us the magnitude of something (as well as the direction). The stuff under the **moment** column are secondary quantities that depend on where they are measured and give use the relative location of the fundamental quantities. Hence the terms torque = moment of force, velocity = moment of rotation and angular momentum = moment of momentum. All that means is that these quantities are $\boldsymbol{r} \times \text{(something fundamental)}$ and they describe the moment arm to this something.

The location of the line in space is always the same formula

$$ \text{(location)} = \frac{ \text{(value)} \times \text{(moment)}}{ \text{(magnitude)}^2} $$

where $\text{(magnitude)}$ is *always* the magnitude of the $\text{(value)}$ vector.

In statics, for example, we learn to balance forces and moments, which should be interpreted as balancing the force magnitude and the line of action of the force.

## Best Answer

That derivation of Feynman's is one of the best around. As you have worked out yourself,

in principleyou can think of everything in terms of force as long as you also know the position vector where the force acts. This is actually more information - oftena great dealmore - than you need to compute the dynamics and statics of a rigid body: you can slide a force vector along the straight line with its same direction and passing through its tail, and the effect of that force on arigidbody's dynamics is the same (although where the force actsisimportant for working out internal stresses on a body). The sum of forces acting at the centre of mass and the nett torque about the centre of mass isallthe information you need to computerigidbody statics and dynamics. This is in general a great deal less information (three vectors: force, torque and position of centre of mass) than a specification of all the individual forces and their positions of action. So, if you like, this is an instance ofdata compressionto make a description of dynamics more wieldy.Another way of thinking of the split between force and torque is that they correspond to the natural, intuitive split between the Euclidean isometries of

translationandrotation. Feynman's work calculation is splitting the work done by a system of forces into the resulting translational kinetic energy that results and the kinetic energy associated with rotation about the centre of mass.Ultimately you will meet the notion of Lagrangian dynamics and Noether's Theorem. Feynman from memory touches on these notions in his famous lectures in a "Symmetry In Physics" chapter. From Noether's Theorem we understand that the conservation of various quantities arises because our description of physics does not change if we impart continuous transformations on our co-ordinate systems: because most physics does not change if we shift our time co-ordinate origin, we conclude through Noether's theorem that there is a conserved quantity which we call energy. Conservation of momentum arises because our physics is invariant with respect to

translationsof our co-ordinate systems (one component of momentum conservation for each component of co-ordinate translation) and conservation of angular momentum arises because our physics is invariant with respect to co-ordinate rotations. So again we see a split of conserved quantities between those to do with translation (momentum) and those to do with rotation (angular momentum). The split thus arises very neatly and naturally from the notions of Eucledean isometries.