Newton's Second Law in Polar Coordinates.ipynb

Kernel: VPython

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Newton's Second Law in Polar Coordinates

PY345

Derivation of Newton's Second Law in Polar Coordinates

In cartesian coordinates

$\dot{\vec{r}}=\frac{d}{dt} (x\hat{x}+y\hat{y}+z\hat{z})$

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$=\dot{x}\hat{x}+\dot{y}\hat{y}+\dot{z}\hat{z}$.

In polar coordinates,

$\vec{r}=r\hat{r}$,

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As the position of an object changes in polar coordinates, the direction of $\hat{r}$ changes to always point toward the object. Therefore $\dot{\hat{r}}\ne 0$. To determine what it is, imagine a point moving along a unit circle from $\vec{r}_i=\hat{r}_i$ to $\vec{r}_f=\hat{r}_f$.

The change in position $\Delta \hat{r}=\hat{r}_f-\hat{r}_i$ has a length of $\Delta \phi$ and points in the $\hat{\phi}$ direction (counterclockwise, though technically it only points exactly in the $\hat{\phi}$ direction when $\Delta \phi$ approaches zero).

Therefore,

$\Delta \hat{r}=\Delta \phi \hat{\phi}$

$\dot{\hat{r}}=\lim_{\Delta t \to 0}\frac{\Delta \hat{r}}{\Delta t}=\lim_{\Delta t\to 0}\frac{\Delta \phi\hat{\phi}}{\Delta t}=\dot{\phi}\hat{\phi}$.

Now we can evaluate $\dot{\vec{r}}$ and $\ddot{\vec{r}}$ in polar coordinates.

$\dot{\vec{r}}=\dot{r}\hat{r}+r\dot{\hat{r}}=\dot{r}\hat{r}+r\dot{\phi}\hat{\phi}$

$\ddot{\vec{r}}=\ddot{r}\hat{r}+\dot{r}\dot{\hat{r}}+\dot{r}\dot{\phi}\hat{\phi}+r\ddot{\phi}\hat{\phi}+r\dot{\phi}\dot{\hat{\phi}}$

We are not done yet, though, because now we have to figure out what $\dot{\hat{\phi}}$ is. To do that, let's imagine a point moving along a unit circle again.

It may not be obvious how $\Delta \hat{\phi}$ was determined in this figure, so let's move $\hat{\phi}_f$ so that its tail touches that of $\hat{\phi}_i$.

You can see that the two $\Delta \hat{\phi}$ vectors are parallel, which means that they are the same vector (recall that vectors are determined by length and orientation, not by where they start and end). You may also notice that those two triangles are similar triangles, which means that the angle between $\hat{\phi}_i$ and $\hat{\phi}_f$ is $\Delta \phi$. Lastly, when drawn from the midpoint of the arc, $\Delta\hat{\phi}$ points toward the center of the circle. Therefore,

$\Delta \hat{\phi}=-\Delta\phi\hat{r}$

$\dot{\hat{\phi}}=\lim_{\Delta t\to 0}\frac{\Delta \hat{\phi}}{\Delta t}=\lim_{\Delta t\to 0}\frac{-\Delta \phi\hat{r}}{\Delta t}=-\dot{\phi}\hat{r}$.

Putting it all together,

$\ddot{\vec{r}}=\ddot{r}\hat{r}+\dot{r}\dot{\phi}\hat{\phi}+\dot{r}\dot{\phi}\hat{\phi}+r\ddot{\phi}\hat{\phi}-r\dot{\phi}^2\hat{r}$

$=\left (\ddot{r}-r\dot{\phi}^2\right )\hat{r}+\left (r\ddot{\phi}+2\dot{r}\dot{\phi}\right )\hat{\phi}$.

From this we can write Newton's Second Law in polar coordinates (cylindrical coordinates for three dimensions).

$$\begin{equation}F_r=m\left (\ddot{r}-r\dot{\phi}^2\right )\end{equation}$$$$F_\phi=m\left (r\ddot{\phi}+2\dot{r}\dot{\phi}\right )$$$$F_z=m\ddot{z}$$

Practice

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Determine expressions for the acceleration of the puck in polar coordinates.

Solution

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Assumptions

• the puck and the block are point masses

• there is no friction between the puck and the table

• there is no friction between the string and the table

Diagrams

Analysis

According to Newton's Second Law in polar coordinates

$F_r=m\left (\ddot{r}-r\dot{\phi}^2\right )$

and

$F_\phi=m\left (r\ddot{\phi}+2\dot{r}\dot{\phi}\right )$.

There are no forces in the $\phi$-direction. There are, however, forces in the $r-$ and $z$-directions. First, let's look at the puck.

radial direction: $-F_{sp}^t=m\left (\ddot{r}-r\dot{\phi}^2\right )$

$\phi$-direction: $0=m\left (r\ddot{\phi}+2\dot{r}\dot{\phi}\right )$

z-direction: ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 20: …p}^n-F_{ep}^g=m\̲c̲a̲n̲c̲e̲l̲t̲o̲{\ddot{z}}{0}

For the hanging block,

$F_{sb}^t-F_{eb}^g=M\ddot{z}$.

Note that $\ddot{z}_b=\ddot{r}$ because if the hanging block accelerates upward then the puck accelerates outward. This allows us to combine equations since we also know that the tensions are the same.

$-m\left (\ddot{r}-r\dot{\phi}^2\right )-Mg=M\ddot{r}$

$-m\ddot{r}+mr\dot{\phi}^2-Mg=M\ddot{r}$

$\ddot{r}\left (M+m\right )=mr\dot{\phi}^2-Mg$

$\ddot{r}=\frac{mr\dot{\phi}^2-Mg}{M+m}$

From the $\phi$-direction, we can solve for $\ddot{\phi}$.

$\ddot{\phi}=-\frac{2\dot{r}\dot{\phi}}{r}$

Check

The SI units of $\ddot{r}$ are $\frac{[kg][m][s]^{-2}}{[kg]}=\frac{[m]}{[s]^2}$, which is what we expect. The SI units of $\ddot{\phi}$ are $\frac{[m/s][s]^{-1}}{[m]}=[s]^{-2}$, which is also what we expect.

One special case is when $M\gg m$, in which we would expect that the block would be pretty much in free fall. If that is the case, then any terms with $m$ are negligible.

$\lim\limits_{M\gg m} \ddot{r}=\frac{-Mg}{M}=-g$

Another special case is when $\dot{\phi}=0$, in which this should reduce to the Half Atwood machine discussed previously.

$\lim\limits_{\dot{\phi}\to 0} \ddot{r}=\frac{-Mg}{M+m}$

Interpretation

We can see that the puck is more likely to accelerate outward if it is spinning around very fast. We also see that it is more likely to accelerate inward if the hanging block is very heavy.

The equation for $\ddot{\phi}$ shows us that it depends on $\dot{r}$. That is, the rate at which the puck swings around depends on how the puck is moving inward or outward.

#this program is under construction
from __future__ import division, print_function
from vpython import *

#initial values
phi=0 #initial angle
r=1 #radius of circle
dr=0 #initial radial velocity
dphi=2 #angular speed in rad/s
g=9.8 #gravitational field strength
t=0 #initial time
y=0.8 #initial height of block
m=1 #mass of puck
M=2 #mass of block
dt=0.001 #time step

#draw objects (note that the vertical direction is y, whereas we used z in the analysis)
scene=canvas(center=vec(0,1.2,0))
puck=cylinder(pos=vec(r*cos(phi),1,r*sin(phi)),axis=vec(0,0.1,0),radius=0.2,color=color.black, make_trail=True)
string=cylinder(pos=vec(0,1.05,0),axis=vec(r*cos(phi),0,r*sin(phi)),radius=0.01)
table=box(pos=vec(0,0.9,0),length=4*r,width=4*r,height=0.1)
#block=box(pos=vec(0,y,0),length=0.3,height=0.3,width=0.3,v=vec(0,dr,0))

#create graph
xygraph=graph(title="x vs. t", xtitle="t (s)", ytitle="x (m)")
xdata=gcurve(color=color.blue)

while r>=0.1 and r<=2: #stop animation when puck reaches center or gets too far out
    rate(1/dt) #make sure looping rate does not exceed 100 loops per second
    ddr=(m*r*dphi**2-M*g)/(M+m) #second derivative of radial coordinate
    ddphi=-2*dr*dphi/r #angular acceleration derived in analysis
    dr=dr+ddr*dt #update radial velocity
    dphi=dphi+ddphi*dt #update angular velocity
    r=r+dr*dt #update radial coordinate
    phi=phi+dphi*dt #update angle
    puck.pos=vec(r*cos(phi),1,r*sin(phi)) #move the puck
    string.axis=vec(r*cos(phi),0,r*sin(phi)) #move the string
    #block.v=block.v+vec(0,ddr,0)*dt
    #block.pos=block.pos+block.v*dt
    xdata.plot(t,puck.pos.x) #plot the x-coordinate
    t=t+dt #increment the time

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Practice

---

A skateboard of mass $m$ is on a semicircular half-pipe with radius $R$. Determine an expression for $\phi(t)$.

Solution

---

Assumptions

• the skateboard rolls without friction

• the skateboard starts from rest at an angle $\phi_0$

• air resistance is negligible

Diagrams

Analysis

Newton's Laws in Polar Coordinates are

$$F_r=m\left (\ddot{r}-r\dot{\phi}^2\right )$$$$F_\phi=m\left (r\ddot{\phi}+2\dot{r}\dot{\phi}\right )$$

Applying these to the free body diagram we get,

$F_{es}^g\cos\phi-F_{hs}^n=m(\ddot{r}-r\dot{\phi}^2)$

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for the r-direction and

$-F_{es}^g\sin\phi=m(r\ddot{\phi}+2\dot{r}\dot{\phi})$

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 32: …\phi=\cancel{m}\̲c̲a̲n̲c̲e̲l̲t̲o̲{R}{r}\ddot{\ph…

$\ddot{\phi}=-\frac{g}{R}\sin\phi$

in the $\phi$-direction. This does not have any obvious solutions. One option is to determine the resulting motion numerically, which we will do soon. Another option is to make another assumption about the motion, e.g. that $\phi$ is always small. If $\phi\ll 1\text{rad}$, then

$\sin\phi\approx \phi$

$\ddot{\phi}\approx -\frac{g}{R}\phi$.

For what function do you get the same thing back with a negative sign after differentiating it twice? Sine and Cosine.

$\phi(t)=A\sin\left (\sqrt{\frac{g}{R}}t\right )+B\cos\left (\sqrt{\frac{g}{R}}t\right )$

Let's check that this does in fact satisfy $\ddot{\phi}= -\frac{g}{R}\phi$. Let $\omega=\sqrt{\frac{g}{R}}$ for brevity.

$\dot{\phi}(t)=A\omega\cos(\omega t)-B\omega\sin(\omega t)$

$\ddot{\phi}(t)=-A\omega^2\sin(\omega t)-B\omega^2\cos(\omega t)=-\omega^2(A\sin(\omega t)+B\cos(\omega t))=-\frac{g}{R}\phi(t)$

It works, so now let's use the initial conditions ($\phi(0)=\phi_0$ and $\dot(\phi)(0)=0$) to determine the constants A and B.

$\phi(0)=A\sin(0)+B\cos(0)=\phi_0$

$B=\phi_0$

$\dot{\phi}(0)=A\omega\cos(0)-\phi_0\omega\sin(0)=0$

$A\omega=0\Rightarrow A=0$

Therefore,

$\phi(t)=\phi_0\cos(\sqrt{\frac{g}{R}}t)$.

Check

The units of $\sqrt{\frac{g}{R}}$ are $\sqrt{\frac{\text{distance}/\text{time}^2}{\text{distance}}}=\frac{1}{\text{time}}$, so the argument of the cosine function has no units. Whatever the units of $\phi_0$ will also be the units of $\phi(t)$.

Let's check some limits to see if the answer makes sense. If this were to happen in the absence of gravity (i.e. $g=0$), then $\phi(t)=\phi_0$, which makes sense. You also get that same result if $R\rightarrow \infty$, which corresponds to a completely flat halfpipe.

Interpretation

Our answer indicates to us that the skateboard oscillates back and forth, which makes sense. It also tells us that the frequency of the oscillation depends on the strength of gravity and the size of the halfpipe.

from __future__ import division, print_function
from vpython import *

#initial conditions
R=3.0
phi=pi/4.0
dphi=0
g=9.8
t=0
tmax=10
dt=0.01

#draw halfpipe and skateboard
scene=canvas(center=vec(0,1,0))
halfpipe=extrusion(path=[vec(0,R-0.1,-1),vec(0,R-0.1,1)], shape=shapes.arc(radius=3,angle1=-pi, angle2=0))
skateboard=box(pos=vec(R*sin(phi),R*(1-cos(phi)),0),axis=vec(cos(phi),sin(phi),0),length=0.5, width=0.2, height=0.1,color=color.yellow)

#create graphs
xGraph=graph(title="x vs. t (blue=exact, red=approximate)", xtitle="t (s)", ytitle="x (m)", xmin=0, xmax=tmax)
xdata=gcurve(color=color.blue)
approximatedata=gcurve(color=color.red)

while t<=tmax:
    rate(1/dt) #make sure the looping speed doesn't exceed 100 iterations per second
    ddphi=-(g/R)*sin(phi) #calculate new value of angular acceleration
    dphi=dphi+ddphi*dt #calculate new value of angular speed
    phi=phi+dphi*dt #calculate new angle
    skateboard.pos=vec(R*sin(phi),R*(1-cos(phi)),0) #redraw skateboard
    skateboard.axis=vec(cos(phi),sin(phi),0) #reorient skateboard
    xdata.plot(t,skateboard.pos.x) #plot the x-coordinate
    approximatedata.plot(t,R*sin(pi/4*cos(sqrt(g/R)*t))) #plot the approximate x-coordinate
    t=t+dt #increment the time

<IPython.core.display.Javascript object>

Practice

---

A puck connected to a string swings in a circle on a rough, horizontal table. Determine an expression for the angle of the string as a function of time.

Solution

---

Assumptions

• the puck is a point mass with mass $m$

• the string has no mass

• the puck starts at $\phi(0)=\phi_0$ with an initial angular speed $\dot{\phi}_0$

• the coefficient of kinetic friction is $\mu_k$

• the puck is moving fast enough to be sliding

Diagrams

Analysis

According to the free body diagrams, the equations for Newton's Second Law are

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in the r-direction,

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 62: …r\ddot{\phi}+2m\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{\dot{r}}\do…

$-\mu_kF_{fp}^n=mr\ddot{\phi}$

in the $\phi$-direction, and

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 20: …p}^n-F_{ep}^g=m\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{\ddot{z}}\R…

in the z-direction. Combining the z-equation with the $\phi$-equation, we get

$-\mu_k\cancel{m}g=\cancel{m}r\ddot{\phi}$

$\ddot{\phi}=-\frac{\mu_kg}{r}$.

We integrate to find $\dot{\phi}(t)$.

$\ddot{\phi}=\frac{d\dot{\phi}}{dt}=-\frac{\mu_kg}{r}\Rightarrow \int\limits_{\dot{\phi}_0}^{\dot{\phi}(t)}=-\int\limits_0^t\frac{\mu_kg}{r}dt'$

$\dot{\phi}(t)=-\frac{\mu_kg}{r}t+\dot{\phi}_0$

Repeat to find $\phi(t)$.

$\dot{\phi}(t)=\frac{d\phi}{dt}=-\frac{\mu_kg}{r}t+\dot{\phi}_0\Rightarrow \int\limits_{\phi_0}^{\phi(t)}d\phi=\int\limits_0^t\left (-\frac{\mu_kg}{r}t'+\dot{\phi}_0\right )dt'$

$\phi(t)=-\frac{1}{2}\frac{\mu_kg}{r}t^2+\dot{\phi}_0t+\phi_0$

Notice that we didn't even have to use the equation for the r-direction. All that would do is tell us the tension in the string (which we could do now that we know $\phi(t)$).

Check

The units of $-\frac{\mu_kg}{r}t^2$ are $\frac{\text{distance}/\text{time}^2}{\text{distance}}\times \text{time}^2$, which cancel out completely. The same is true for the second terms. This may seem wrong since we are expecting an angle, but radians is a unit that doesn't really count as a unit in the normal sense, so this is actually fine.

In the limits $\mu_k\rightarrow 0$ and/or $g\rightarrow 0$, the first terms goes away. This makes the puck continue to go around without slowing down, which makes sense.

Interpretation

The fact that the first term is negative and the others are positive means that at some point the puck will stop and start to move in the opposite direction. This unphysical scenario occurs when

$\dot{\phi}(t)=0=-\frac{\mu_kg}{r}t+\dot{\phi}_0\Rightarrow t=\frac{\dot{\phi}_0r}{\mu_kg}$.

After this point, our model breaks down because the kinetic friction force would disappear.

from __future__ import division, print_function
from vpython import *

#initial values
phi=0 #initial angle
r=1 #radius of circle
dphi=10 #angular speed in rad/s
mu=0.1 #coefficient of kinetic friction
g=9.8 #gravitational field strength
t=0 #initial time
dt=0.01 #time step

#draw objects (note that the vertical direction is y, whereas we used z in the analysis)
scene=canvas(center=vec(0,1,0))
puck=cylinder(pos=vec(r*cos(phi),0,r*sin(phi)),axis=vec(0,0.1,0),radius=0.2,color=color.black)
string=cylinder(pos=vec(0,0.05,0),axis=vec(r*cos(phi),0,r*sin(phi)),radius=0.01)
floor=box(pos=vec(0,-0.1,0),length=3*r,width=3*r,height=0.1)

#create graph
xygraph=graph(title="x vs. t", xtitle="t (s)", ytitle="x (m)")
xdata=gcurve(color=color.blue)

while dphi>=0: #continue loop until puck stops
    rate(1/dt) #make sure looping rate does not exceed 100 loops per second
    ddphi=-mu*g/r #angular acceleration derived in analysis
    dphi=dphi+ddphi*dt #update angular velocity
    phi=phi+dphi*dt #update angle
    puck.pos=vec(r*cos(phi),0,r*sin(phi)) #move the puck
    string.axis=vec(r*cos(phi),0,r*sin(phi)) #move the string
    xdata.plot(t,puck.pos.x) #plot the x-coordinate
    t=t+dt #increment the time

<IPython.core.display.Javascript object>

Practice

---

How fast does the table need to spin in order for Einstein to remain pinned to the wall?

Solution

---

Assumptions

• Einstein is a point mass with mass $m$

• the table is turning fast enough that Einstein is not moving vertically

• the coefficient of static friction between Einstein and the wall is $\mu_s$

• Einstein stays at a constant distance from the center of the disk

• the table turns at a constant rate

Diagrams

One unusual aspect of this situation is that the static friction force points upward (what would happen if the wall were slippery?) whereas the normal force points inward.

Analysis

According to the free body diagram, the equations for Newton's Second Law are

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in the r-direction,

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 64: …r\ddot{\phi}=-2\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{\dot{r}}\do…

in the $\phi$-direction, and

ParseError: KaTeX parse error: Undefined control sequence: \cancelto at position 20: …E}^s-F_{eE}^g=m\̲c̲a̲n̲c̲e̲l̲t̲o̲{0}{\ddot{z}}\R…

in the z-direction.

The $\phi$-direction equation just confirms that the table will turn at a constant speed (since $\ddot{\phi}=0$). Since $F^s\le \mu_sF^n$ for static friction, the equation for the r-direction becomes

$\frac{F_{wE}^s}{\mu_s}\le mr\dot{\phi}^2\Rightarrow \frac{\cancel{m}g}{\mu_s}\le \cancel{m}r\dot{\phi}^2$

$\dot{\phi}\ge\sqrt{\frac{g}{\mu_sr}}$

Check

The units of the right hand side are $\sqrt{\frac{\text{distance}/\text{time}^2}{\text{distance}}}=\frac{1}{\text{time}}$, which is the proper unit for an angular speed.

If $g$ is really big, then $\dot{\phi}$ must be large in order to keep Einstein up against the wall, which makes sense. Also, if $g=0$, the table doesn't need to spin at all to keep him up.

Interpretation

The main limitation of this analysis is that it doesn't apply for turning speeds that are small enough that Einstein would fall.