Work-Energy Theorem.ipynb

Kernel: VPython

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Work-Energy Theorem

PY345

Puzzle of the Day

---

A ball is released from the top of each slide. Rank the slides according to the speed of the ball at the end. Which one will reach the bottom first?

Types of Energy

• kinetic energy $T$ is energy of motion

• kinetic energy can be translational, $T_\text{translational}=\frac{1}{2}mv_{cm}^2$, or rotational, $T_\text{rotational}=\frac{1}{2}I\omega^2$

• internal energy is the energy associated with the state or configuration of a system

• potential energy $U$ is a type of internal energy associated with special kinds of forces called conservative forces

  • gravitational force: $\Delta U^g=mg\Delta y$ (near earth)

  • spring force: $\Delta U^s=\frac{1}{2}k(\Delta x)^2$ (k=spring constant, $\Delta x$=displacement from relaxed length)

• Work-Energy Theorem: the energy of a system can be changed by the application of external and/or non-conservative forces over a displacement ParseError: KaTeX parse error: Expected group after '_' at position 23: …T+\Delta U=\int_̲\limits{1}^2\ve…

• the energy added by forces is called work $W$

Conservation of energy is another way of stating Newton's Second Law. Let's start by taking the derivative of kinetic energy with respect to time.

$T=\frac{1}{2}mv^2=\frac{1}{2}m(\vec{v}\cdot\vec{v})$

$\frac{dT}{dt}=\frac{1}{2}m(\vec{v}\cdot \dot{\vec{v}}+\dot{\vec{v}}\cdot\vec{v})=m\vec{v}\cdot\dot{\vec{v}}$

We recognize that $m\dot{\vec{v}}=\vec{F}_\text{net}$.

$\frac{dT}{dt}=\vec{F}_\text{net}\cdot\vec{v}$

Multiplying both sides by $dt$ and integrating, we get

$dT=\vec{F}_\text{net}\cdot\vec{v}dt=\vec{F}_\text{net}\cdot d\vec{r}$

$\int\limits_{T_1}^{T_2}dT=\int\limits_1^2\vec{F}_\text{net}\cdot d\vec{r}$

where 1 and 2 represent locations. We can further separate $\vec{F}_\text{net}$ into internal, conservative forces and everything else.

$\Delta T=\int\limits_1^2\vec{F}_\text{int, cons}\cdot d\vec{r}+\int\limits_1^2\vec{F}_\text{ext/noncons}\cdot d\vec{r}$

$\Delta T-\int\limits_1^2\vec{F}_\text{int, cons}\cdot d\vec{r}=\int\limits_1^2\vec{F}_\text{ext/noncons}\cdot d\vec{r}$

Define $\Delta U\equiv -\int\limits_1^2\vec{F}_\text{int, cons}\cdot d\vec{r}$.

$\Delta T+\Delta U=\int\limits_1^2\vec{F}_\text{ext/noncons}\cdot d\vec{r}$

Practice

---

A ball is released from the top of each slide. Rank the slides according to the speed of the ball at the end. Which one will reach the bottom first?

Solution

---

Assumptions

• the ball is a point particle with mass $m$

• the ball starts from rest at height $y_0$

• the bottom of the ramp is $y=0$

• the earth does not move in our frame of reference

• the ball is close enough to earth that $g$ is approximately constant

Diagrams

Analysis

Normally at this point we would write down Newton's Second Law based on the free body diagram. The problem with that here is that the direction of the normal force on the ball changes for three out of the four ramps. In principle, we could work out the direction of the force if we knew the shapes of all the ramps, but the math would be rather involved. This is where conservation of energy shines.

$\Delta T+\Delta U=\int\limits_1^2\vec{F}_\text{ext/noncons}\cdot d\vec{r}$

Here are the consequences of choosing ball + earth as the system: 1) there will be two kinetic energy terms (one for each object in the system), 2) the interaction between the ball and the earth is internal (and it also happens to be conservative), and 3) the interaction with the ramp is external. The internal, conservative interaction gets counted as potential energy while the external interaction gets counted as work.

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By definition, the normal force is perpendicular to the surface of contact, and since the ball is constrained to move along the ramp then each infinitesimal displacement $d\vec{r}$ is perpendicular to the normal force. By the definition of the dot product, this means that $\vec{F}_{rb}^n\cdot d\vec{r}=0$ everywhere.

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$v=\sqrt{2gy_0}$

Check

The SI units of the answer are $\sqrt{\left [\frac{m}{s^2}\right ][m]}=\sqrt{\left [\frac{m^2}{s^2}\right ]}=\left [\frac{m}{s}\right ]$, which is what we expect for a speed.

The final speed is 0 if $y_0=0$, which makes sense. The final speed also gets bigger if $y_0$ is bigger, which also makes sense.

Interpretation

Our answer indicates that the final speed will be the same regardless of which of the ramps the ball goes down. This does not necessarily mean that they will all reach the bottom at the same time, though. Compare the black ramp and the red ramp, for example. The ball on the black ramp reaches close to its maximum speed very early on. The ball on the red ramp doesn't reach close to its maximum speed until very close to the end. Therefore the ball on the black ramp travels faster on average than the ball on the red ramp, which means it will reach the end first. The balls on the green and yellow ramps are likely to reach the bottom at roughly the same time as one another. That time is going to be smaller than that of the red ramp and larger than that of the black ramp.

from __future__ import division, print_function
import matplotlib.pyplot as plt
%matplotlib inline

#initial conditions
m=1 #mass
y=10 #starting height
g=9.8 #gravitational field strength
v=0 #initial speed
Etotal=m*g*y+0.5*m*v**2 #starting energy
dy=-0.01 #distance increment

def U(y): #define potential energy
    return m*g*y

#set up data arrays for plotting
ydata=[] #heights
Udata=[] #potential energies
Kdata=[] #kinetic energies
Edata=[] #total energy

while y>=0: #continue while above ground
    ydata.append(y) #add y to ydata array
    Udata.append(U(y)) #add U to Udata array
    Kdata.append(Etotal-U(y)) #calculate kinetic energy and add it to Kdata array
    Edata.append(Etotal) #add E to Edata array
    y=y+dy #increment height
    
#plot all the things
plt.title("types of energy as a function of height above ground") 
plt.xlabel("y (m)")
plt.ylabel("energy (J)")
plt.plot(ydata,Udata,'ro',ydata,Kdata,'b.',ydata,Edata,'kx') #U as red circles, K as blue dots, and E as black x's
plt.legend(["potential energy","kinetic energy","total energy"])

<matplotlib.legend.Legend at 0x114ef06d0>

Practice

---

Repeat the previous problem choosing the ball alone as the system.

Solution

---

Assumptions

• the ball is a point particle with mass $m$

• the ball starts from rest at height $y_0$

• the bottom of the ramp is $y=0$

• the earth does not move in our frame of reference

• the ball is close enough to earth that $g$ is approximately constant

Diagrams

Analysis

By removing the earth from the system, the gravitational force is now external. This means that it can no longer be considered as a potential energy and will instead appear on the right hand side of the conservation of energy equation. We also don't need two kinetic energy terms.

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Given that the direction of $d\vec{r}$ continually changes along the ramp, how do we evaluate $\vec{F}_{eb}^g\cdot d\vec{r}$? Here it helps to conceptually understand what the dot product does: it picks out how much of two vectors are parallel to one another. Only the vertical portion of the ball's displacement is parallel to the gravitational force. Therefore,

$\frac{1}{2}mv^2=\int\limits_{y=y_0}^{y=0}(-mg\hat{y})\cdot dy\hat{y}=-\left [mgy\right ]_{y_0}^0=mgy_0$

$v=\sqrt{2gy_0}$.

Check

The answer is the same as before, so the checks are exactly the same.

Interpretation

Thankfully, choosing a different system did not change the answer.

Practice

---

Determine the minimum height above the ground that a bungee jumper reaches in terms of the jump height, properties of the jumper, and properties of the bungee chord.

Solution

---

Assumptions

• the bungee jumper is a point mass with mass $m$

• the bungee jumper drops from rest at a height $h$

• the bungee chord is massless, has a relaxed lengths $L$, and has a spring constant $k$

• drag is negligible

• the earth doesn't move in our frame of reference

• the bungee jumper comes to rest when the chord has stretched to its maximum

Diagrams

Analysis

Initial state: Bungee jumper drops from rest from bridge

Final state: bungee chord is stretched to maximum extent

Because the bungee chord and the earth are being treated as part of the system, both of those interactions will count as potential energy.

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Even though the bridge is exerting a force on the chord, the point of application of the force does not undergo a displacement, so the right hand side is zero. Note that we have also chosen to place $y=0$ at the relaxed length of the chord. This has the advantage of simplifying the analysis quite a bit, but the disadvantage is that the value of $y$ that we solve for won't be the answer.

$mg(y-L)+\frac{1}{2}ky^2=0$

Let's unpack this a bit. Since we chose the bottom of the relaxed chord to be $y=0$, the starting height is not $h$ but rather $L$. The ending height is $y$, which should turn out to be a negative number. In the spring potential energy term, $y$ is the displacement from the relaxed length. Now we just solve for $y$.

$y^2+2\frac{mg}{k}y-2\frac{mg}{k}L=0$

$y=\frac{-\frac{2mg}{k}\pm\sqrt{4\frac{m^2g^2}{k^2}+4\left (\frac{2mgL}{k}\right )}}{2}$

$y=-\frac{mg}{k}\pm \frac{2}{2}\sqrt{\frac{m^2g^2}{k^2}+2\frac{mgL}{k}}$

$y=\frac{mg}{k}\pm\frac{mg}{k}\sqrt{1-2\frac{kL}{mg}}$

We choose the negative sign, since the bungee jumper will end up below $y=0$ (also notice that the square root will always be greater than 1, so the second term is always $>\frac{mg}{k}$). The height from the ground is:

height above ground=$h-L+y=h-L-\frac{mg}{k}-\frac{mg}{k}\sqrt{1+2\frac{kL}{mg}}=h-L-\frac{mg}{k}\left (1+\sqrt{1+2\frac{kL}{mg}}\right )$.

Check

The SI units of $\frac{mg}{k}$ are $\frac{[kg][m/s^2]}{[N/m]}=\frac{[N]}{[N/m]}=[m]$, which is what we expect for a distance. The SI units of $\sqrt{\frac{kL}{mg}}=\frac{[N/m][m]}{[kg][m/s^2]}=\frac{[N]}{[N]}=\text{no units}$, which is what we expect since it is subtracted from the number 1.

If the bungee jumper has no mass, we should expect that the chord does not stretch at all. When we substitute $m=0$, we find that the height above the ground is $h-L$, which makes sense. The same thing happens when we let $k\rightarrow \infty$.

Interpretation

In order not to hit the ground,

$h-L-\frac{mg}{k}\left (1+\sqrt{1+2\frac{kL}{mg}}\right )>0$

$h>L+\frac{mg}{k}\left (1+\sqrt{1+2\frac{kL}{mg}}\right )$.

from __future__ import division, print_function
import matplotlib.pyplot as plt
%matplotlib inline

#initial conditions with ground level as y=0
y=100 #starting height
h=y #define h for analytical calculation
v=0 #initial velocity
m=60 #mass of jumper
g=9.8 #gravitational field strength
L=10 #relaxed length of bungee chord
k=50 #spring constant
dy=-0.01 #height increment

def Ug(y): #define gravitational potential energy
    return m*g*y

def Us(y): #define spring potential energy
    if y>h-L: #spring is slack until bungee jumper reaches relaxed length
        return 0
    else:
        return 0.5*k*(h-L-y)**2

#prepare data arrays for plotting
Etotal=0.5*m*v**2+Ug(y)+Us(y) #total energy
ydata=[] #array of y-values
Ugdata=[] #array of gravitational potential energy values
Usdata=[] #array of spring potential energy values
Kdata=[] #array of kinetic energy values
Edata=[] #array of total energy values

while Etotal-Ug(y)-Us(y)>=0: #some y-values are forbidden because kinetic energy would be negative
    ydata.append(y) 
    Ugdata.append(Ug(y))
    Usdata.append(Us(y))
    Kdata.append(Etotal-Ug(y)-Us(y))
    Edata.append(Etotal)
    y=y+dy #increment the height

#plot all the things
plt.xlabel("y (m)")
plt.ylabel("energy (J)")
plt.plot(ydata,Ugdata,"ro",ydata,Usdata,"b.",ydata,Kdata,"kx",ydata,Edata,"gd")
plt.legend([r"$U^g$",r"$U^s$","K",r"$E_{total}$"])
print("The numerically calculated minimum height is", y, " m above the ground.")
print("The analytically calculated minimum height is", h-L-m*g/k-m*g/k*sqrt(1+2*k*L/(m*g))," m above the ground.")

The numerically calculated minimum height is 58.91  m above the ground.
The analytically calculated minimum height is 58.9139140021  m above the ground.

Practice

---

A box slides down a rough ramp. Determine the speed of the box at the end of the ramp, expressing your answer in terms of properties of the box and ramp. Use energy considerations instead of forces.

Solution

---

Assumptions

• the box is a point particle of mass $m$ that starts from rest at a height $y_0$ (the top of the ramp)

• the coefficient of kinetic friction between the box and the ramp is $\mu_k$

• the ramp and the earth do no move in our frame of reference

• the ramp has a length L and makes an angle $\theta$ with the horizontal

Diagrams

Analysis

We could certainly do this problem by settinng up Newton's Second Law, but let's practice it using energy instead. By choosing the earth as part of the system, the interaction between the box and the earth gets counted as potential energy instead of as work.

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Note that the coordinate system indicated in the diagram is not tilted like we would normally do when solving this using Newton's Second Law. This is because the expression for gravitational potential energy that we use is derived assuming a coordinate system where the $y$ direction points upward.

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The work done by the normal force is zero because the normal force is perpendicular to the displacement. To evaluate the friction dot product, it is easier to use the magnitude-and-angle form of the dot product (i.e. $\vec{a}\cdot\vec{b}=|vec{a}||\vec{b}|\cos\phi$) rather than the component form. The angle between the friction force and the displacement is $180^\circ$.

$\frac{1}{2}mv^2-mgy_0=\int\limits_1^2 F_{rb}^k\cos(180^\circ)dr=-F_{rb}^k\int\limits_1^2dr=-F_{rb}^kL$

To evaluate the kinetic friction force, we use $F_{rb}^k=\mu_kF_{rb}^n$, where $F_{rb}^n=mg\cos\theta$.

$\frac{1}{2}\cancel{m}v^2-\cancel{m}gy_0=-\mu_k\cancel{m}g\cos\theta L$

$v=\sqrt{2\left (gy_0-\mu_kg\cos\theta L\right )}$

Check

The SI units of $\sqrt{gy_0}$ are $\sqrt{\left [\frac{m}{s^2}\right ][m]}=\sqrt{\left [\frac{m^2}{s^2}\right ]}=\left [\frac{m}{s}\right ]$, which is a speed as expected. The SI units of $\sqrt{\mu_kg\cos\theta L}$ are $\sqrt{\left [\frac{m}{s^2}\right ][m]}=\left [\frac{m}{s}\right ]$, which is also a speed.

If the ramp angle is 90 degrees, then there is no friction and the resulting speed should be the same as for free fall.

$\lim\limits{\theta\to 90^\circ}\sqrt{2\left (gy_0-\mu_kg\cos\theta L\right )}=\sqrt{2gy_0}$

The limit $\theta\to 0$ doesn't make sense to evaluate, since the box would not be sliding and therefore there would be no kinetic friction.

Interpretation

The presence of friction clearly decreases the final speed. We also see that for a given $L$, the final speed increases with $y_0$. For a given $y_0$, the final speed decreases with $L$.

from __future__ import division, print_function
import matplotlib.pyplot as plt
import numpy as np #for arcsin
%matplotlib inline

#initial conditions
y0=5 #initial height (for analytic calculation)
y=y0 #initial height (for numerical calculation)
x=0 #initial x-coordinate
L=10 #length of ramp
mu=0.2 #coefficient of kinetic friction
m=1 #mass of box
g=9.8 #gravitational field strength
v=0 #initial speed
theta=np.arcsin(y0/L) #angle of ramp (depends on height and length)
dy=-0.01 #height increment
dL=-dy/sin(theta) #equivalent increment along length of ramp

def U(y): #define gravitational potential energy
    return m*g*y

Etotal=U(y)+0.5*m*v**2 #define starting energy

#set up arrays for plotting
ydata=[]
Udata=[]
Kdata=[]
Edata=[]

while y>=0: #continue looping while box is above ground
    Etotal=Etotal-mu*m*g*cos(theta)*dL #work done by friction changes the total energy
    Edata.append(Etotal) #append total energy to array
    Udata.append(U(y)) #append potential energy to array
    Kdata.append(Etotal-U(y)) #calculate kinetic energy and append to array
    ydata.append(y) #append height to array
    y=y+dy #increment the height
    
#plot all the things
plt.xlabel("y (m)")
plt.ylabel("energy (J)")
plt.plot(ydata, Udata,"ro", ydata, Kdata, "b.", ydata, Edata, "kx") #U as red circles, K as blue dots, Etotal as black x's
plt.legend(["U", "K", r"$E_{total}$"])

print("The final numerically calculated speed is", sqrt(2*(Etotal-U(y))/m), "m/s.") #print numerically calulated final speed 
print("The final analytically calculated speed is", sqrt(2*(g*y0-mu*g*cos(theta)*L))," m/s.") #print analytically calculated final speed
    

The final numerically calculated speed is 8.01123634528 m/s.
The final analytically calculated speed is 8.0032371058  m/s.

Practice

---

Repeat the previous problem, but replace the box with a cylinder ($I_\text{cyl}=\frac{1}{2}mR^2$).

Solution

---

Assumptions

• the cylinder has mass $m$, radius $R$, and starts from rest with its bottom at the top of the ramp

• the ramp has height $y_0$, length $L$, and makes an angle $\theta$ with the horizontal

• the cylinder does not slip as it rolls down the ramp

• the ramp and the earth do not move in our frame of reference

Diagrams

Analysis

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As before, the normal force does no work on the cylinder because the force is perpendicular to the displacement. Static friction also does no work because the point of application of the force does not move relative to the surface (i.e. the displacement is zero). Another way to think about it is that static friction does not slow the cylinder down.

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 34: …^2-\frac{1}{2}m\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{v_0^2}+\fra…

Since the center of mass of the cylinder is above the surface of the ramp, which is why I added $R$ to the initial and final heights. Whatever number you add, though, it ends up canceling out.

Another crucial thing to recognize here is that for a rolling object $\omega=\frac{v_{cm}}{R}$.

$\frac{1}{2}\cancel{m}v_{cm}^2+\frac{1}{4}\cancel{m}\cancel{R^2}\left (\frac{v_{cm}}{\cancel{R}}\right )^2-\cancel{m}gy_0$

$\frac{3}{4}v_{cm}^2=gy_0$

$v_{cm}=\sqrt{\frac{4}{3}gy_0}$

Check

The SI units of the answer are $\sqrt{\left [\frac{m}{s^2}\right ][m]}=\sqrt{\left [\frac{m^2}{s^2}\right ]}=\left [\frac{m}{s}\right ]$, which is a speed as expected.

In the limit $y_0\to 0$, the final speed is zero, which makes sense.

Interpretation

The final speed increases with $y_0$. The final speed is also smaller than that of a point particle without friction, which would be $v=\sqrt{2gy_0}$. This makes sense since some of the energy is tied up in rotation, so not as much will go toward the translational kinetic energy.

from __future__ import division, print_function
import matplotlib.pyplot as plt
import numpy as np
%matplotlib inline

y0=5
y=y0
x=0
v=0
m=1
R=0.5
g=9.8
I=0.5*m*R**2
L=10
theta=np.arcsin(y0/L)
dy=-0.01
dL=-dy/sin(theta)

def U(y):
    return m*g*y

def omega(v):
    return v/R

def Ktotal(y):
    return Etotal-U(y)

def Ktrans(y):
    return (2/3)*Ktotal(y)

def Krot(y):
    return (1/3)*Ktotal(y)

Etotal=0.5*m*v**2+0.5*I*omega(v)**2+U(y)

ydata=[]
Udata=[]
Ktransdata=[]
Krotdata=[]
Edata=[]

while y>=0:
    ydata.append(y)
    Udata.append(U(y))
    Ktransdata.append(Ktrans(y))
    Krotdata.append(Krot(y))
    Edata.append(Etotal)
    y=y+dy

plt.xlabel("y (m)")
plt.ylabel("energy (J)")
plt.plot(ydata, Udata, "ro", ydata, Ktransdata, "b.", ydata, Krotdata, "gd", ydata, Edata, "kx")
plt.legend(["U",r"$K_{trans}$", r"$K_{rot}$", r"$E_{total}$"])

print("The final numerically calculated speed is ", sqrt(2*Ktrans(y)/m), " m/s.")
print("The final analytically calculated speed is ", sqrt((4/3)*g*y0), " m/s.")
    

The final numerically calculated speed is  8.09098263501  m/s.
The final analytically calculated speed is  8.08290376865  m/s.