# Energy and work

Kinetic and potential energy.

\begin{aligned} E &= T + V \\ \text{total energy} &= \text{kinetic energy} + \text{potential energy} \end{aligned}

## Kinetic energy

Kinetic energy of a point mass.

$T = \frac{1}{2} m v^2$

Kinetic energy of an arbitrary body.

$T = \iiint_{\mathcal{B}} \frac{1}{2} \rho v^2 \, dV$

Kinetic energy of a rigid body about the center of mass $C$.

$T = \frac{1}{2} m v_C^2 + \frac{1}{2} I_{C,\hat\omega} \omega^2$

Using #ren-ea below, in the case that the point $Q$ is the center of mass $C$ we see that $\vec{r}_{QC} = \vec{r}_{CC} = 0$ and so the middle term in #ren-ea is eliminated, leaving the result.

Kinetic energy of a rigid body about the instantaneous center $M$.

$T = \frac{1}{2} I_{M,\hat\omega} \omega^2$

Using #ren-ea below, in the case that the point $Q$ is the instantaneous center $M$ we have that $\vec{v}_Q = \vec{v}_M = 0$ and so the first two germs are eliminated.

Kinetic energy of a rigid body about an arbitrary body point $Q$.

$T = \frac{1}{2} m v_Q^2 + m \vec{v}_Q \cdot \left( \vec\omega \times \vec{r}_{QC} \right) + \frac{1}{2} I_{Q,\hat\omega} \omega^2$

We start with the general expression #rem-eb: \begin{aligned} T &= \iiint_{\mathcal{B}} \frac{1}{2} \rho v_P^2 \, dV, \end{aligned} where we integrate over the body with a location $P$. We choose a point $Q$ fixed to the body and use #rkg-er to express the velocity of $P$ in terms of $\vec{v}_Q$ and $\vec\omega$, giving \begin{aligned} T &= \iiint_{\mathcal{B}} \frac{1}{2} \rho \|\vec{v}_Q + \vec\omega \times \vec{r}_{QP}\|^2 \, dV \\ &= \iiint_{\mathcal{B}} \frac{1}{2} \rho \left(\vec{v}_Q + \vec\omega \times \vec{r}_{QP}\right) \cdot \left(\vec{v}_Q + \vec\omega \times \vec{r}_{QP}\right) \, dV \\ &= \iiint_{\mathcal{B}} \frac{1}{2} \rho \left( \vec{v}_Q \cdot \vec{v}_Q + 2 \vec{v}_Q \cdot (\vec\omega \times \vec{r}_{QP}) + (\vec\omega \times \vec{r}_{QP}) \cdot (\vec\omega \times \vec{r}_{QP}) \right) \, dV. \end{aligned} We next use the fact that $\vec{v}_Q$ and $\vec\omega$ do not depend on the integration point within the body to pull them outside of the integrals, which results in \begin{aligned} T &= \frac{1}{2} \|\vec{v}_Q\|^2 \iiint_{\mathcal{B}} \rho \, dV + \vec{v}_Q \cdot \bigg( \vec\omega \times \iiint_{\mathcal{B}} \rho \vec{r}_{QP} \, dV \bigg) + \iiint_{\mathcal{B}} \frac{1}{2} \rho (\vec\omega \times \vec{r}_{QP}) \cdot (\vec\omega \times \vec{r}_{QP}) \, dV \\ &= \frac{1}{2} m v_Q^2 + \vec{v}_Q \cdot \left( \vec\omega \times m \vec{r}_{QC} \right) + \iiint_{\mathcal{B}} \frac{1}{2} \rho \| \vec\omega \times \vec{r}_{QP} \|^2 \, dV \\ &= \frac{1}{2} m v_Q^2 + m \vec{v}_Q \cdot \left( \vec\omega \times \vec{r}_{QC} \right) + \iiint_{\mathcal{B}} \frac{1}{2} \rho \omega^2 r_{QP}^2 \sin^2\theta \, dV \\ &= \frac{1}{2} m v_Q^2 + m \vec{v}_Q \cdot \left( \vec\omega \times \vec{r}_{QC} \right) + \frac{1}{2} \omega^2 \iiint_{\mathcal{B}} \rho (r_{QP} \sin\theta)^2 \, dV, \end{aligned} where $\theta$ is the angle between $\vec\omega$ and $\vec{r}_{QP}$. But $r_{QP}\sin\theta$ is simply the orthogonal distance to point $P$ from the line through $Q$ in direction $\vec\omega$, so from #rem-ei we see that the final integral above is the moment of inertia $I_{Q,\hat\omega}$ about the $\hat\omega$ axis through point $Q$.

## Potential energy

Gravitational potential energy with uniform gravitational acceleration $g$.

$V = m g h_C$

The distance $h_C$ is the height of the center of mass of the body above the reference height.

## Work and power

Work done by a force $\vec{F}$.

$W = \int_{\vec{r}_1}^{\vec{r}_2} \vec{F} \cdot d\vec{r}$

Work done by a constant force $\vec{F}$.

$W = \vec{F} \cdot \Delta \vec{r}$

The change in position is $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$.

Compute the work definition for a constant force by pulling $\vec{F}$ out of the integral, so the resulting trivial integral is just the difference between the endpoints.

Forces of constraint do zero work.

$W = 0$

The work $W$ is by a force of constraint $F$.

By definition a constraint force $\vec{F}$ does not permit movement in the direction that it applies, so the incremental work $\vec{F} \cdot d\vec{r}$ is always zero and the total work is also zero.

Power transferred by a force $\vec{F}$.

$P = \vec{F} \cdot \vec{v}$

The velocity $\vec{v}$ is the velocity of the point where $\vec{F}$ is applied.

Work in terms of power.

$W = \int_{t_1}^{t_2} P \, dt = \int_{t_1}^{t_2} \vec{F} \cdot \vec{v} \, dt$

The velocity $\vec{v}$ is the velocity of the point where $\vec{F}$ is applied.

We start from the expression #ren-ef for work as the integral of incremental work $\vec{F} \cdot d\vec{r}$. We change the variable of integration from $\vec{r}$ to $t$, and we substitute $d\vec{r} = \vec{v} dt$ which is simply a rearrangement of the definition #rkv-ev of velocity as $\vec{v} = \frac{d\vec{r}}{dt}$.

Work done by a moment $M$.

\begin{aligned} W &= \int_{\theta_1}^{\theta_2} M \, d\theta \\ &= \int_{t_1}^{t_2} M \dot\theta \, dt \\ &= M \, \Delta \theta \text{ (for constant M)} \end{aligned}

The rotation angle $\theta$ is measured around the same axis $\vec{a}$ about which the moment $M$ is applied.

Work-energy principle.

$W = \Delta E = E_2 - E_1$

The increase in energy $\Delta E$ is the change from the initial energy $E_1$ to final energy $E_2$, and $W$ is the total external work done on the system.

Assume $\vec{F}$ is the total external force acting on a point mass $m$. Then using work in terms of power and Newton's equations gives \begin{aligned} W &= \int_{t_1}^{t_2} \vec{F} \cdot \vec{v} \, dt \\ &= \int_{t_1}^{t_2} m \vec{a} \cdot \vec{v} \, dt \\ &= \left[ \frac{1}{2} m \vec{v} \cdot \vec{v} \right]_{t_1}^{t_2} \\ &= E_2 - E_1. \end{aligned} The kinetic energy is the total energy of the system in this case because all external forces (including those from potentials) are included in $\vec{F}$.