Work and energy

Energy

Potential energy

Force and potential energy. #ren-efp

$$ \begin{aligned}\vec{F} &= -\nabla V \\& = -\frac{\partial V}{\partial x}\hat{\imath} -\frac{\partial V}{\partial y}\hat{\jmath} - \frac{\partial V}{\partial z}\hat{k}\end{aligned} $$

Beginning with #ren-ec and generalizing it to one-dimensional displacement in the y-direction, the equation becomes:

$$ \begin{aligned} W &= \vec{F} \cdot \Delta{\vec{y}} \end{aligned} $$

Assuming the only force applied is the force $ \vec{F} $ , then the system gains potential energy, so the amount of work done is equal to $ \Delta{V} $ .

$$ \begin{aligned} \Delta{V} &= \vec{F} \cdot \Delta{\vec{y}} \end{aligned} $$

Rearranging:

$$ \begin{aligned} F &= \frac{\Delta{V}}{\Delta{y}} \end{aligned} $$

Since we are interested in the force the field exerts on the system, due to Newton's Third Law, the force is equal and opposite. Therefore, the above result becomes:

$$ \begin{aligned} F &= -\frac{\Delta{V}}{\Delta{y}} \end{aligned} $$

If the change in potential energy and displacement is infinitesimally small, this becomes a derivative:

$$ \begin{aligned} F &= -\frac{dV}{dy} \end{aligned} $$

Generalizing this result to 3 dimensions, we use the gradient to obtain the following result:

$$ \begin{aligned} \vec{F} &= -\nabla V \\ & = -\frac{\partial V}{\partial x}\hat{\imath} -\frac{\partial V}{\partial y}\hat{\jmath} - \frac{\partial V}{\partial z}\hat{k} \end{aligned} $$

Example Problem: Gravitational potential energy. #ren-xgp

Consider an object of mass $m$ in the air under the influence of gravity. Derive the potential energy $V_g$ of the object.

Using #ren-efp, we can derive the gravitational potential energy. The force due to gravity is:

$$ \begin{aligned} \vec{F} &= -mg\hat{\jmath} \end{aligned} $$

Since gravity only acts in one direction, we can simplify #ren-efp to:

$$ \begin{aligned} F &= -\frac{dV}{dy} \\ -mg &= -\frac{dV}{dy} \\ mg \, dy &= dV \\ \end{aligned} $$

Integrating both sides:

$$ \begin{aligned} \int_{V_1}^{V_2} dU &= \int_{y_1}^{y_2} mg \, dy \\ \Delta V & = mg\Delta y \end{aligned} $$

We can simplify the result by assuming $U_1$ is zero, and $y_1$ is at the reference zero-potential height. Therefore, the above becomes:

$$ \begin{aligned} V_g & = mgy = mgh \end{aligned} $$

The above result also proves that gravity is a conservative force, as it only depends on the initial and final position of the object.

Example Problem: Potential energy in a spring. #ren-xsp

Consider an object of mass $m$ attached to a spring of stiffness $k$. Derive the potential energy $V_s$ stored in the spring.

Using #ren-efp, we can derive the potential energy in a spring. From the above diagram, the spring force is:

$$ \begin{aligned} \vec{F} &= -kx\hat{\imath} \end{aligned} $$

Since the restoring force in a spring only acts in one direction, we can simplify #ren-efp to:

$$ \begin{aligned} F &= -\frac{dV}{dx} \\ -kx &= -\frac{dV}{dx} \\ kx \, dx &= dV \\ \end{aligned} $$

Integrating both sides:

$$ \begin{aligned} \int_{V_1}^{V_2} dU &= \int_{x_1}^{x_2} kx \, dx \\ \Delta V & = \frac{1}{2}k\Delta x^2 = \frac{1}{2}k(x^2_2 - x^2_1) \end{aligned} $$

We can simplify the result by assuming $U_1$ is zero, and $x_1$ is at the reference zero-potential displacement. Therefore, the above becomes:

$$ \begin{aligned} V & = \frac{1}{2}kx^2 \end{aligned} $$

The above result also proves that the restoring force in a spring is a conservative force, as it only depends on the initial and final position of the object.

Warning: Work and energy are not vectors. #wrng-vec

Work and energy are scalar quantities, as seen by the dot product that results in a scalar value. Positive and negative work/energy values represent either a gain or loss in energy in the system. For instance, in thermodynamics, work done by the system is negative, and work done on the system is positive. In mechanics, positive work represents an object speeding up (putting energy into the system), and negative work represents an object slowing down (taking energy away from the system).

Gravitational potential energy with uniform gravitational acceleration $g$. #ren-eg

$$ V_g = mgh_C $$

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

Potential energy stored in a spring with stiffness $k$. #ren-es

$$ V_s = \frac{1}{2}kx_C^2 $$

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

Work-potential energy theorem. #ren-ewp

$$ W = -\Delta V $$

Beginning with the work definition:

$$ \begin{aligned} W = \vec{F} \cdot \Delta \vec{r} \end{aligned} $$

Assuming one-dimensional motion, and using the force-potential energy relationship:

$$ \begin{aligned} W & = -\frac{\Delta V}{\Delta x}\Delta x\\ & = -\Delta V \end{aligned} $$

Kinetic energy

Kinetic energy of a point mass. #ren-ek

$$ T = \frac{1}{2}mv^2 $$

We begin with Newton's 2nd Law assuming $ \vec{F} $ is the total external force applied:

$$ \begin{aligned} \vec{F} &= m\vec{a} \\ & = m\frac{d\vec{v}}{dt} \end{aligned} $$

Using #ren-efp to relate force and energy:

$$ \begin{aligned} \frac{dU(\vec{r})}{d\vec{r}} & = m\frac{d\vec{v}}{dt} \end{aligned} $$

Note that the negative is not found here, as we're interested in the force exerted on the field, not by the field (see derivation in #ren-efp).

Rearranging then yields:

$$ \begin{aligned} dU & = m\frac{d\vec{v}}{dt} d\vec{r} \end{aligned} $$

We can then do the following:

$$ \begin{aligned} dU & = m\frac{d\vec{r}}{dt} d\vec{v} \\ & = m\vec{v} \, d\vec{v} \\ \end{aligned} $$

Integrating both sides:

$$ \begin{aligned} \int_{U_1}^{U_2} dU & = \int_{\vec{v_1}}^{\vec{v_2}}m\vec{v} \, d\vec{v} \\ \Delta U & = \frac{1}{2}m\left[\vec{v} \cdot \vec{v}\right]_{\vec{v_1}}^{\vec{v_2}} \\ \end{aligned} $$

Assuming an initial velocity of zero, and zero potential energy (i.e the object is at the reference height):

$$ \begin{aligned} U & = \frac{1}{2}m\left(\vec{v} \cdot \vec{v}\right) \end{aligned} $$

Using conservation of energy, and the fact that a vector dotted with itself equals its magnitude squared (see #rvi-eg), then:

$$ \begin{aligned} T & = \frac{1}{2}mv^2\\ \end{aligned} $$

Another perhaps less intuitive way is to begin with the work definition #ren-ef:

$$ \begin{aligned} W &= \int_{\vec{r}_1}^{\vec{r}_2} \vec{F} \cdot d\vec{r} \\ & = \int_{t_1}^{t_2} \vec{F} \cdot \dot{\vec{r}}dt \\ & = \int_{t_1}^{t_2} \vec{F} \cdot \vec{v} \, dt \end{aligned} $$

Using Newton's 2nd Law assuming $ \vec{F} $ is the total external force applied:

$$ \begin{aligned} W & = \int_{t_1}^{t_2} m\vec{a} \cdot \vec{v} \, dt \end{aligned} $$

Note that this integrand looks similar to the following:

$$ \begin{aligned} \frac{d}{dt}(\vec{u} \cdot \vec{u}) = \dot{\vec{u}} \cdot \vec{u} + \vec{u} \cdot \dot{\vec{u}} = 2\vec{u} \cdot \dot{\vec{u}} \end{aligned} $$

By the fundamental theorem of calculus, this must mean that:

$$ \begin{aligned} W = \frac{1}{2}m\left(\vec{v} \cdot \vec{v}\right) \end{aligned} $$

And we've arrived at the last step of the previous derivation.

Kinetic energy of an arbitrary body. #ren-ekb

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

Kinetic energy of a rigid body about the center of mass $C$. #ren-ekr

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

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 $M$. #ren-em

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

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 terms are eliminated.

Kinetic energy of a rigid body about an arbitrary body point $Q$. #ren-ea

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

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 axis through point $Q$.

Work-kinetic energy theorem. #ren-wk

$$ W = \Delta T $$

Skipping the derivation shown in #ren-ep, taking the last step with the integral:

$$ \begin{aligned} \int dW & = \int_{\vec{v_1}}^{\vec{v_2}}m\vec{v} \, d\vec{v} \\ W & = \frac{1}{2}m\left[\vec{v} \cdot \vec{v}\right]_{\vec{v_1}}^{\vec{v_2}} \\ & = \Delta T \end{aligned} $$

Work-energy theorem. #ren-eco

$$ W = Delta T + Delta V $$

where $W$ is the work done by non-conservative forces. If non-conservative forces are not present, then it becomes conservation of energy.

Summary

Total energy

$$ \begin{aligned} E &= T + V \\ Total \ Energy &= Kinetic + Potential \end{aligned} $$

Kinetic

Point Mass. #sum-pm

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

Rigid Body. #sum-rb

$$ \begin{aligned} T &= \frac{1}{2} m \ {v_c}^2 + \frac{1}{2} I_c \ \omega^2 \quad \text{(about center of mass C)} \\ T &= \frac{1}{2} \ I_O \ \omega^2 \quad \text{(if point O is pinned)} \\ T &= \frac{1}{2} \ I_M \ \omega^2 \quad \text{(if M is the instantaneous center)} \end{aligned} $$

Potential

Gravitational. #sum-pg

$$ V = m \ g \ h_c \quad \text{ (h is the height of center of mass)} $$

Elastic. #sum-pe

$$ V = \frac{1}{2}\ k \ x^2 \quad \text{ (k is the spring stifness and x is elongation)} $$

Work

Work-Energy Principal #wan-we

$$ W = \Delta{E} = E_f - E_0 $$

Work done by a force

Work done by a force $\vec{F}$. #ren-wf

$$ \begin{aligned}W &= \int_{\vec{r}_1}^{\vec{r}_2} \vec{F} \cdot d\vec{r} \\& = \int_{t_1}^{t_2} \vec{F} \cdot \dot{\vec{r}}dt\end{aligned} $$

Work done by a constant force $\vec{F}$. #ren-wc

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

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

Work done by a constraint force $\vec{F}$. #ren-wt

$$ W = 0 $$

The work $W$ is by a force of constraint force $ \vec{F} $ .

By definition a constraint force $ \vec{F} $ does not permit movement in the direction that it applies. This means that the constraint force and the displacement are always orthogonal and the dot product between orthogonal vectors is zero.

Work done by a moment $M$. #ren-wm

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

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

Example Problem: Constant Force. #wfe-cf

Work done by $ \vec{F} $ is $ W = 6 J $. Assume no gravity, frictionless, constraint force $ \vec{R} $ of the slot of the mass.

What is the increase in energy of the mass?

$ > 6 J $
$ 6 J $
$ < 6 J $
Need to first find $ \vec{R} $

B. $ 6 J $

$$ \Delta E = W = \vec{R} \ \cdot \Delta \ \vec{r} \quad , \quad \vec{R} \perp \vec{r} $$

Conservative vs non-conservative forces

A conservative force is one such that the work done by the force only depends on the initial and final positions of the object (i.e independent of the path taken by the object). Examples of conservative forces are gravity and spring force, where only the initial and final displacements are needed to calculate the work done by those forces. Examples of a non-conservative force are any type of drag forces (e.g friction, air resistance).

The diagram below illustrates different paths and the work done by a conservative forces vs. a non-conservative force.

If the force moving an object from point $A$ to point $B$ was conservative, the work done by that force in all paths is equal. If the force was non-conservative, the work done by that force is not equal, and it depends on the path length.

Work done by friction

$$ W = \Delta{E} = E_f - E_i $$

Work done by friction can be positive, zero, or negative.

Example Problem: Friction during rolling wihtout slip. #wdf-rws

Which direction does friction force $ \vec{F} $ act?

Right
Zero
Left

What is the work done by friction?

$ > 0 $
Zero
$ < 0 $

A.
B.

$$ W = \int \vec{F} \cdot \vec{r} \,dt = 0 \quad \text{as} \quad \vec{r} = 0 $$
- Friction prevents slip
- Converts rotation motion into translational

Work done by a moment

$$ W = \int_{\theta_0}^{\theta_f} M_z \,d\theta = \int_{t_0}^{t_f} M_z \ \dot{\theta} \,d\theta $$

Example Problem: Rod with friction moment. #wdm-rm

Rod shown above starts from rest, falls to horizontal, friction moment of 2 Nm. What is the work done by friction?

$ \geq 2 J $
$ 0 < W < 2 J $
$ = 0 $
$ W \leq -2 J $

D.

$ W = \int_{\frac{\pi}{4}}^0 (2 Nm) \, d\theta \\ = \mu \ (\Delta \theta) \\ = (2 Nm)(0- \frac{\pi}{4}) = -\frac{\pi}{2} J $

Friction

Stick	Transition	Slip
No relative motion $ v_{px} = 0 $ and $ a_{px} = 0 $ $ F \leq \mu N $	No relative motion $ v_{px} = 0 $ and $ a_{px} = 0 $ Critical friction force $ F = \mu N $	Relative motion $ v_{px} \neq 0 $ or $ a_{px} \neq 0 $ $ F = \mu N $ $ \hat{F} $ opposed motion

Solution procedure

Known Case

Determine case: stick, transition, slip (and direction)
FBD, equations of motion (or statics), solve

Unknown Case

Try stick

Assume $ v_{px} = 0 $ and $ a_{px} = 0 $
Solve for N, F
Check $ |F| \leq \mu |N| $

Try slip

Assume $ F = \mu N $ (in one direction)
Solve for motion of contact point and N, F
Check $ v_{px} \neq 0 $ or $ a_{px} \neq 0 $ and $ \hat{F} $ opposed motion ($ \hat{v} $ or $ \hat{a} $)

Otherwise

Assume $ F = \mu N $ (in opposite direction) and resume from step B.2