Contemporary Challenges: Fall-2024
Homework 1 (SOLUTION): Due 5 Friday 10/4

  1. Travel by car or bike S1 5082S

    (Remember to read the Homework-Write-Up Guide)

    In this question you will compare the energy used by (i) an electric bicycle traveling 15 miles at 15 mph to (ii) the energy used by an electric car traveling the same distance at 60 miles per hour.

    1. Find the ratio of the energies used by the two options. I'm looking for a numerical value of the ratio. Use the simplest coarse-grained model for transport (kinetic energy of the wind tail) and clearly state your assumptions (for example, assume flat roads).
    2. Use a more refined model by including rolling resistance. The rolling resistance for cars and bicyles is equivalent to climbing an uphill grade of approximately 1%. (The exact value of the equivalent uphill grade depends on the tire pressure, and the viscoelasticity of the tire material).

    As we saw in class, the energy in the wind tail of a car is given by

    \begin{align*} \text{Energy in car's wind trail} &= \frac{1}{2}\rho C_{\text{D,car}} A_{\text{car}} v_{\text{car}}^2 d. \\ \end{align*}

    For a bike, I can then write down

    \begin{align*} \text{Energy in bike's wind trail} &= \frac{1}{2}\rho C_{\text{D,bike}} A_{\text{bike}} v_{\text{bike}}^2 d. \\ \end{align*}

    The ratio of these energies is \begin{align*} \frac{\text{Energy in car's wind trail}}{\text{Energy in bike's wind trail}} = \frac{\cancel{\tfrac{1}{2}}C_{\text{D,car}}A_{\text{car}} v_{\text{car}}^2\cancel{d_{car}}}{\cancel{\tfrac{1}{2}}C_{\text{D,bike}} A_{\text{bike}} v_{\text{bike}}^2\cancel{d_{bike}}} \\ \end{align*}

    since I'm considering the car and the bike going the same distance. The ratio of the speeds is 4 (given in the question prompt). As discussed in class, \(C_{\text{D,car}}A_{\text{car}} \approx 1 \text{ m}^2\). An adult riding a commuter bike presents a frontal area of about 0.5 m2 and a drag coefficient of \(\approx\) 1. Substituting these values in, I get

    \begin{align*} \frac{\text{Energy in car's wind trail}}{\text{Energy in bike's wind trail}} \approx \frac{1\text{ m}^2}{0.5\text{ m}^2}(4)^2=32 \\ \end{align*}

    This tells me that the car uses about 32 times more energy than the bike.

    Optional: If I want to include rolling resistance in my model, I need to add the work done by rolling (being careful to do true algebra): \begin{align*} \frac{E_{\text{car}}}{E_{\text{bike}}} &= \frac{\text{Energy in car's wind trail} + \text{Work done on car by rolling resistance}}{\text{Energy in bike's wind trail} + \text{Work done on bike by rolling resistance}} \\ &= \frac{\tfrac{1}{2}C_{\text{D,car}}A_{{\text{car}}} \rho_{\text{air}} v_{\text{car}}^2\cancel {d_{\text{car}}} + c_{\text{rr,car}}m_{\text{car}}g \cancel{d_{\text{car}}}}{\tfrac{1}{2}C_{\text{D,bike}} A_{\text{bike}} \rho_{\text{air}} v_{\text{bike}}^2\cancel{d_{\text{bike}}}+ c_{\text{rr,bike}}m_{bike}g \cancel{d_{\text{bike}}}} \\ \end{align*}

    A typical car mass might be 2000kg and a typical bike mass might be 10 kg.

    From the SEWTHA text (p.258), the coefficients of rolling resistance are \(c_{\text{rr,car}} = 0.010\) and \(c_{\text{rr,bike}} = 0.050\).

    The density of air is \(\rho_{air} = 1.3 \times 10^{-3}\text{ kg/m}^3\)

    Making converstions to SI units: \begin{align*} v_{\text{car}} = 60 \text{ mph} \left(0.44 \frac{\text{m/s}}{\text{mph}}\right) = 27 \text{ m/s}\\ v_{\text{bike}} = 15 \text{ mph} \left(0.44 \frac{\text{m/s}}{\text{mph}}\right) = 6 \text{ m/s}\\ \end{align*}

    Plugging in \begin{align*} \frac{E_{\text{car}}}{E_{\text{bike}}} &= \frac{\tfrac{1}{2}(1 \text{m}^2) (1.3 \text{ kg/m}^3)(27 \text{ m/s})^2 + 0.010(2000 \text{ kg})(10 \text{ N/kg}) }{\tfrac{1}{2} (0.5 \text{m}^2) (1.3 \text{ kg/m}^3) (6 \text{ m/s})^2+ 0.050(10 \text{ kg})(10 \text{N/kg})} \\ &= 30 \end{align*} So, in this simple rolling resistance model, the car uses 30 times more energy than a bicycle. In this calculation, the energy is dominated by energy in the wind trail.

  2. Speed of a solar car S1 5082S

    This self-driving solar car is travelling on a flat road on a windless day. The sun is directly overhead.

    (a) Draw an energy flow diagram to describe the system. An arrow at the top of the flow diagram will represent incoming solar energy (landing on the solar panel). One circle will represent the solar panel, and one circle will represent the electric motor. Label each arrow with quantitative values of the energy flow (joules per second).

    (b) Estimate how fast this self-driving solar car can travel on a flat road on a windless day when the sun is directly overhead. Give you answer in meters per second.

    Use the following parameters for the system:

    • The car is 1.8 m wide, 1 m tall and 3 m long. The top surface of the car is entirely covered with solar panels.
    • The sun is directly overhead and the intensity of the sunlight is 1000 J/(s.m2).
    • The electric motors are powered directly by the solar panels (no battery power).
    • The solar panels convert sunlight energy into electrical energy with 20% efficiency (the other 80% of sunlight energy is heating the solar panel).
    • The electric motors convert electrical energy into mechanical work with efficiency 90% efficiency.
    • The drag coefficient is 0.2.
    • The energy dissipation associated with the tires rolling on the road can be neglicted.

    We want to extimate how fast a new solar car design can travel.

    The 810 J/s goes into the Kinetic Energy of wind behind the car. \begin{align*} \frac{KE}{\Delta t} &= 810 \text{ J/s} \\ &= \frac{\frac{1}{2} \rho_{\text{air}} C_{\text{drag}} A v^{2}d}{\Delta t}\\ &= \frac{1}{2} \rho_{\text{air}} C_{\text{drag}} A v^{2}\frac{d}{\Delta t}\\ &= \frac{1}{2} \rho_{\text{air}} C_{\text{drag}} A v^{2}v \\ &= \frac{1}{2} \rho_{\text{air}} C_{\text{drag}} A v^{3} \end{align*} In this case, \(\rho_{\text{air}} = 1.3 \text{ kg/m}^3\), \(C_{\text{drag}} = 0.2\), and \(A = 1.5 \text{ m}^2\). We can use equation 1 to solve for \(v\) and substitute in the numerical values. \begin{align*} v = \sqrt[3]{\frac{2(810 \text{ J/s})}{(1.3 \text{ kg/m}^3) (1.5 \text{ m/s}) (0.2)}} = 16 \text{ m/s} \end{align*}

    Sense-making

    From class, we know KE\(_\text{of air}\)/\(\Delta\)t = 20,000 J/s when driving a car at 70 mph. The solar car is going 36 mph (about half the speed) and for the solar car KE\(_\text{of air}\)/\(\Delta\)t is about 20 times less. 70 mph \(\rightarrow\) 36 mph is consistent with reducing \(v^3\) by a factor of about 8. KE\(_\text{of air}\)/\(\Delta\)t has also been reduced by aerodynamic design.

  3. Piano tuners in Chicago S1 5082S

    (Long answer format, please read the Homework-Write-Up Guide)

    In a fabled story about Enrico Fermi (famous physicist), Fermi was asked how many people work as piano tuners in Chicago. Fermi did some mental arithmetic and quickly answered the question with surprising accuracy. Your task is to recreate Fermi's calculation.

    Fermi's method of problems has spread far beyond the physics community. Today, tech companies and business consulting companies expect their employees to do Fermi problems: https://www.youtube.com/watch?v=KAo6Vn5bDF0.

    Background: Pianos were popular when Fermi was living in Chicago in the 1940s. The population of Chicago was about 2 million people. Approximately 1 in 10 households had a piano. Pianos got out of tune at regular intervals (about 2 or 3 years), so the piano owner would call a technician (the piano tuner) to tighten/loosen the 88 strings inside the piano. Each tuning job took at least an hour.

    Fermi used his general knowledge to estimate proportionality constants: For example, the number of pianos in Chicago was proportional to the number of households (the proporitionality constant was 0.1.).

    To recreate Fermi's calculation make your own quantitaive estimates of proportionality constants (practice using your reasoning skills; avoid using Google). Each proportionality constant will be approximate; that is the essence of this estimation technique. To organize your calculation in a logical, easy-to-follow fashion, set up each line of math with one proportionality constant. For example,

    \begin{align} ``(2 \times 10^6 \text{ people}) \div (3 \text{ people per household}) = 0.7 \times 10^6 \text{ households}'' \end{align}

    Keep track of units as you go along: households, pianos, hours, etc. Use round numbers at each step of the calculation because a 5% calculational “error” will be smaller than the 10-30% uncertainty in the proportionality constants. How many piano tuners do you think were working in Chicago in the 1940s?

    In the work below, I bold faced the five proportionality constants. I start with the population and estimate the number of households \begin{align} \text{(population of Chicago)} \div \textbf{(persons per household)} = \text{households} \end{align} Using the number of households, I calculate the number of pianos \begin{align} \text{households} \div \textbf{(households per piano)} = \text{pianos} \end{align} Using the number of pianos, I calculate the number of tunings per year, \begin{align} \text{pianos} \times \textbf{(tunings per piano per year)} = \text{tunings per year} \end{align} I need to know how many tunings a technician can perform each year: \begin{align} \textbf{(hours per technician per year)} \div \textbf{(hours per tuning)} = \text{tunings per technician per year} \end{align} Finally, I can calculate the number of technicians, \begin{align} \text{tunings per year} \div \text{tunings per technician per year} = \text{technicians} \end{align} Let's put in some numbers. Well assume a population of 2 million people. A typical household in the US is probably one kid and two parents (3 people), so there are \(\sim\)0.7 million households in Chicago.

    The fraction of households that have a piano is less than 100% and probably more than 1%. I'll guess 10%. That means about 70,000 pianos in Chicago.

    How often do people tune pianos? They shouldn't wait 10 years, but they can't afford every month. I'll guess once per year. Therefore, we expect 70,000 tuning jobs per year. (Note by David Roundy: I think we've had our piano tuned twice in 15 years, so this is perhaps a bit optimistic.)

    It probably takes more than 1 hour to tune a piano but less than 10 hours (there are 88 keys and it probably takes about a minute for each string). I'll guess it takes 2 hours.

    A typical worker works 40 hours/week \(\approx\)2000 hours/year. That means one technician can tune 1000 pianos per year. In conclusion, there must be about 70 piano tuners in Chicago.

    Note: There are absolutely other ways you could have solved this problem had we not asked you to estimate hours. You could instead have estimated the salary of a piano tuner and the cost of a piano tuning. That's probably harder, if you've never paid to have your piano tuned.

  4. Three ideas for the term project S1 5082S

    Read the description of the term project on the class website at “Introduction to term project”. Identify three (3) subjects that you find interesting/intriguing (for example, solar energy, exoplanets, ...). Within each subject, pose a question that might have an interesting quantitative answer: “Since it requires energy to make a solar panel, how long does it take to recoup that energy?”, “How far away could we see an Earth-like planet orbiting a Sun-like star?” ... You should turn in 3 different subjects and 3 different quantitative questions (quantitative means “quantities that can be calculated and/or measured”)

    Let your mind wander as broadly as possible. Subjects and questions are not restricted to the topics taught in PH315. During this exploratory stage, be bold and daring; you are not committing yourself to solve all 3 questions. To spark your imagination, there is a list of ideas on the class website. The instructor will read your ideas and give you feedback. Whenever possible, the feedback will point you towards a coarse-grained model that is helpful for answering your question. Use the feedback to help decide which question you will develop further (or whether you need to go back to the drawing board).