Central Forces: Spring-2025
HW 05 (SOLUTION): Due W3 D2

1. Lines in Polar Coordinates S1 5240S

   (Algebra involving trigonometric functions) Purpose: Practice with polar equations.

   The general equation for a straight line in polar coordinates is given by: \begin{equation} r(\phi)=\frac{r_0}{\cos(\phi-\delta)} \end{equation} where \(r_0\) and \(\delta\) are constant parameters. Find the polar equation for the straight lines below. You do NOT need to evaluate any complicated trig or inverse trig functions. You may want to try plotting the general polar equation to figure out the roles of the parameters.

   1. \(y=3\)

      Our answer in each case should be a specific polar equation \(r(\phi)\) that does not have \(x\) or \(y\) in it.

      In general, the geometric interpretation of this equation is that \(r_0\) represents the closest distance of the line from the origin (you can see this because largest the denominator can be is 1, making \(r_0\) the smallest value of r(\(\phi\))), and \(\delta\) is the angle that the line makes with the y-axis measured counterclockwise (you can see this by plotting several lines with different values of \(\delta\)).
      For this particular line, \(y=3\), the point of the line closest to the origin is (0,3), so \(r_0 = 3\). The line is a horizontal straight line, so \(\delta = \frac{\pi}{2}\). \begin{equation} r(\phi)=\frac{3}{\cos(\phi-\frac{\pi}{2})} \end{equation}

   2. \(x=3\)

      This is a vertical straight line (the angle relative to the y-axis is zero), and the closet point to the origin is \((3,0)\). \begin{equation} r(\phi)=\frac{3}{\cos\phi} \end{equation}

   3. \(y=-3x+2\)

      This is a line with negative slope that crosses the y-axis at \((0,2)\) and crosses the x-axis at \(\left(0, \frac{2}{3}\right)\). The angle between the line and the y-axis is:

      \begin{align} \tan(\delta) &= \frac{\left(\frac{2}{3}\right)}{2} \\ \delta &= \arctan\left(\frac{1}{3}\right) \ \ \end{align} The point of closest approach is where there is an intersection between this line and the line that is perpendicular to it and also passes through the origin. Lines that are perpendicular to \(y=-3x+2\) have slope \(m=+\frac{1}{3}\). The line with slope \(m=+\frac{1}{3}\) and passes through the origin is: \begin{equation} y=\frac{1}{3}x \end{equation} The point where these two lines intersect is: \begin{align} \frac{1}{3}x &= -3x+2 \\ \frac{10}{3}x &= 2 \\ x &= \frac{3}{5}, \quad y = \frac{1}{5} \\ \end{align} The value of \(r_0\) is then: \begin{align} r_0 &= \sqrt{x^2+y^2}\\[12pt] &= \sqrt{\left(\frac{3}{5}\right)^2+\left(\frac{1}{5}\right)^2}\\ &= \frac{\sqrt{10}}{5} \end{align} \begin{equation} r(\phi)=\frac{\sqrt{10}}{5\cos\left(\phi-\arctan\ \left(\frac{1}{3}\right) \right)} \end{equation}

2. Find Force Law: Logarithmic Spiral Orbit S1 5240S

   (Use the equation for orbit shape.) Gain experience with unusual force laws.

   In science fiction movies, characters often talk about a spaceship “spiralling in” right before it hits the planet. But all orbits in a \(1/r^2\) force are conic sections, not spirals. This spiralling in happens because the spaceship hits atmosphere and the drag from the atmosphere changes the shape of the orbit. But, in an alternate universe, we might have other force laws.

   In class, we discussed how to calculate the shape of the orbit for an inverse square potential. More generally, the equation for the orbit of a mass \(\mu\) under the influence of a central force \(f(r)\) is given by: \begin{align} \frac{d^2 u}{d\phi^2} + u &=-\frac{\mu}{\ell^2}\frac{1}{u^2}f\left(\frac{1}{u}\right)\\ \Rightarrow f\left(\frac{1}{u}\right)&=-\frac{\ell^2}{\mu}u^2 \left(\frac{d^2 u}{d\phi^2} + u\right) \end{align} where \(u=r^{-1}\).

   Find the force law for a mass \(\mu\), under the influence of a central-force field, that moves in a logarithmic spiral orbit given by \(r = ke^{\alpha \phi}\), where \(k\) and \(\alpha\) are constants.

   If we want the orbit \(r = ke^{\alpha \phi}\), then \(u = \frac{1}{k}e^{-\alpha \phi}\), \(\frac{d^2 u}{d\phi^2} = \frac{\alpha^2}{k}e^{-\alpha \phi}\) and \begin{align} f\left(\frac{1}{u}\right)&= -\frac{\ell^2}{\mu}\frac{1}{k^2}e^{-2\alpha \phi} \left(\frac{\alpha^2}{k}+\frac{1}{k}\right)e^{-\alpha \phi}\\ &=-\frac{\ell^2}{\mu}\frac{1}{k^3}(\alpha^2+1)e^{-3\alpha \phi}\\ \Rightarrow f(r)&=-\frac{\ell^2}{\mu}(\alpha^2+1)\frac{1}{r^3} \end{align} This is an attractive force (notice the overall minus sign.)

3. Hockey S1 5240S

   (Synthesis Problem: Brings together several different concepts from this unit.) Use effective potential diagrams for other than \(1/r^2\) forces.

   Consider the frictionless motion of a hockey puck of mass \(m\) on a perfectly circular bowl-shaped ice rink with radius \(a\). The central region of the bowl (\(r < 0.8a\)) is perfectly flat and the sides of the ice bowl smoothly rise to a height \(h\) at \(r = a\).

   1. Sketch the potential energy for this system (just the potential energy, not the effective potential). Set the zero of potential energy at the top of the sides of the bowl.

      

      See Figure 1. Consider the frictionless motion of a hockey puck of mass \(m\) on a perfectly circular bowl-shaped ice rink with radius \(a\). The central region of the bowl (\(r < 0.8a\)) is perfectly flat and the sides of the ice bowl smoothly rise to a height \(h\) at \(r = a\).

   2. Situation 1: the puck is initially moving radially outward from the exact center of the rink. What minimum velocity does the puck need to escape the rink?

      The puck must have enough energy to reach the top of the rink:
      \begin{align} E &= \frac{1}{2} mv_0^2-mgh=0\\ \Rightarrow v_0 &= \sqrt{2gh} \end{align}

   3. Situation 2: a stationary puck, at a distance \(\frac{a}{2}\) from the center of the rink, is hit in such a way that it's initial velocity \(\vec v_0\) is perpendicular to its position vector as measured from the center of the rink. What is the total energy of the puck immediately after it is struck?

      The total energy is still \(E=\frac{1}{2} mv_0^2-mgh\) since only the direction of the initial velocity has changed.

   4. In situation 2, what is the angular momentum of the puck immediately after it is struck?

      \(\vec L=\vec r\times \vec p=\frac{a}{2}m v_0 \hat{z}\) in the direction out of the plane of the orbit.

   5. Draw a sketch of the effective potential for situation 2.

      

      See Figure 2.

   6. In situation 2, for what minimum value of \(\vec v_0\) does the puck just escape the rink?

      The total energy of the system is: \begin{equation} E=\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}+U(r) \end{equation} and it must be high enough for the puck to escape the potential well, i.e. for there to be unbound orbits. Use conservation of angular momentum (\(l=\frac{a}{2}mv_0\)) and conservation of energy, where the initial state is just after the puck is hit (part c) and the final state is just after the puck escapes the bowl (\(r=a\) and \(\dot r=0\)). \begin{align} E_f &= E_i\\ 0+\frac{l^2}{2ma^2}+0 &= \frac{1}{2}mv_0^2-mgh\\ \frac{(\frac{a}{2}mv_0)^2}{2ma^2} &= \frac{1}{2}mv_0^2-mgh\\ \frac{1}{8}mv_0^2&=\frac{1}{2}mv_0^2-mgh\\ v_0&=\sqrt{\frac{8}{3} gh} \end{align}

4. Confidence Rating S1 5240S After solving each problem on the assignment, indicate your answers to the following questions for each problem. Answer for the problem as a whole, even if the problem has multiple parts.
   1. Question Confidence How confident are you that you are interpreting the problem the way the instructor intends?
      
      For the rest of the questions, assume you have interpreted the problem correctly
   2. Problem Confidence How confident are you that you could independentl come up with a correct solution process to a similar problem on a future problem set?
      
   3. Answer Confidence How confident are you that your final answer to this question is correct (not solution process)?
      
   4. Makes Sense To what degree do you understand how your answer fits (or does not fit) the physical or mathematical situation of the problem?