# Final 2015 Practice Exam 2 Solutions

## Question 1

To get around the world quickly Santa relies on high speed circular orbits around the Earth, in which the centripetal force is provided by gravity. (The initial acceleration to get him to the required velocity for this orbit is provided by Rudolf).

a)(15 points) If he wants to complete a full orbit in 2 hrs at what height above the surface of the earth must he fly?

$\frac{mv^{2}}{r}=\frac{Gmm}{r^{2}}$

$\frac{2\pi r}{v}=2\times60\times60=7200\mathrm{s}$

$v=\frac{2\pi r}{7200}$

$\frac{(2\pi)^2r^{2}}{7200^{2}r}=\frac{GM}{r^{2}}$

$r^{3}=\frac{7200^{2}GM}{(2\pi)^{2}}$

$r^{3}=5.24\times10^{20}\mathrm{m^{3}}$

$r=8061\mathrm{km}$

$h=r-r_{e}=1681\mathrm{km}$

b) (5 points) At what speed is Santa moving while he is in this orbit?

$v=\frac{2\pi r}{7200}=\frac{2\pi 8.061\times10^{6}}{7200}=7034\mathrm{m/s}$

c) (5 points) Santa has 1000 kg of presents in his sleigh. What is the difference in the gravitational potential energy for those presents when they are in orbit compared to when they are on the ground?

$-\frac{GMm}{8.061\times10^6}+\frac{GMm}{6.380\times10^6}=1.3\times10^{10}\mathrm{J}$

## Question 2

An elf in Santa's workshop raises a $4\,\mathrm{kg}$ present from the workshop conveyer belt to Santa's loading bay by pulling on a rope that passes over a pulley which has mass $2\,\mathrm{kg}$ and radius $20\,\mathrm{cm}$. The bearing of the pulley is frictionless. For a solid disk the moment of inertia is $\frac{1}{2}MR^{2}$. The 4 kg present is initially at rest and the elf would like to lift it at a constant speed of $0.5\,\mathrm{ms^{-1}}$.

a) (5 points ) In order to have the weight be moving with this velocity $1\,\mathrm{s}$ after the elf starts pulling on the rope, with what force $F$ should the elf pull on the rope?(Assume uniform acceleration of the object.)

$a=0.5\mathrm{ms^{-2}}$

$Fr-Tr=\frac{1}{2}m_{pulley}r^{2}\alpha=\frac{1}{2}m_{pulley}r^{2}\frac{a}{r}$

$F-T=\frac{1}{2}m_{pulley}a$

$m_{gift}a=T-m_{gift}g\to T=m_{gift}(a+g)$

$F=m_{gift}(a+g)+\frac{1}{2}m_{pulley}a$

$F=4(10.3)+\frac{1}{2}\times2\times0.5=41.7\mathrm{N}$

b) (5 points) What is the angular acceleration of the pulley while the elf is doing this?

$\alpha=\frac{a}{r}=\frac{0.5}{0.2}=25\mathrm{s^{-2}}$

c) (5 points) What is the magnitude of the normal force exerted by the ground on the elf, who weighs 20kg, during parts (a) and (b)?

$F_{N}=m_{elf}g-F=20\times9.8-41.7=154.3\mathrm{N}$

d) (5 points) Once the object has reached a constant speed of $0.5\,\mathrm{ms^{-1}}$ what force should the elf now apply to maintain that speed?

$F=m_{gift}g=4\times9.8=39.2\mathrm{N}$

e) (5 points) What is the angular velocity (in rpm) of the pulley while the elf is raising the object with a constant speed of $0.5\,\mathrm{ms^{-1}}$?

$\omega=\frac{v}{r}=\frac{0.5}{0.2}=2.5\mathrm{s^{-1}}=23.8\,\mathrm{rpm}$

## Question 3

After numerous unfortunate incidents involving polar bears and tourists who have seen too many Coke$^{\mathrm{TM}}$ ads, Santa has decided to install a new sign at the North Pole. To do this he uses a uniform metal rod of length 0.8 m and a cable which he attaches 20 cm from the end of the metal rod. The new sign has mass 3 kg and hangs from the end of the rod, while the rod itself has mass 1 kg. The cable is attached the top of the pole and the rod is held by a hinge 30 cm down from the top of the pole.

a) (5 points) What angle $\theta$ does the cable make with the rod?

$\theta=\tan^{-1}\frac{30}{60}=26.57^{o}$

b) (10 points) What is the tension in the cable?

Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$

Horizontal forces, right is positive $-T\cos\theta+Fh_{x}=0$

Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$

$T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$

c) (5 points) What is the vertical component of the force exerted by the hinge on the rod? Does this force component point up or down?

$Fh_{y}=4.981-102.24\sin26.57^{o}=-6.53\mathrm{N}$

So force points down.

d) (5 points) What is the horizontal component of the force exerted by the hinge on the rod? Does this force component point to the left or to the right?

$Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$

Force points to the right

## Question 4

Santa wants to give a child a ball that will float in water with exactly half of the ball above the surface and the other half below when the temperature is 20$\mathrm{^{o}C}$. The ball will be 20 cm in diameter and made from rubber which is 1 cm thick and has density 1800 $\mathrm{kg\, m^{-3}}$. The density of air at 20$\mathrm{^{o}C}$ under normal conditions, such as those that exist above the surface of the water, is 1.2 $\mathrm{kg\, m^{-3}}$. The density of water is 1000 $\mathrm{kg\, m^{-3}}$. The inside of the ball is to be filled with pressurized air, which will have a higher mass density than that outside the ball.

A. (10 points) What should the mass density of air inside the ball be so that ball floats as desired?

Weight of displaced fluid

$(\frac{1}{2}1000+\frac{1}{2}1.2)\frac{4}{3}\pi0.1^{3}$

Weight of ball including dense air

$1800\frac{4}{3}\pi(0.1^{3}-0.09^{3})+\rho_{air}\frac{4}{3}\pi0.09^{3}$

At equilibrium these two are equal, so

$\rho_{air}=\frac{(\frac{1}{2}1000+\frac{1}{2}1.2)\frac{4}{3}\pi0.1^{3}-1800\frac{4}{3}\pi(0.1^{3}-0.09^{3})}{\frac{4}{3}\pi0.09^{3}}$

$\rho_{air}=17.55\mathrm{kg\,m^{-3}}$

b) (5 points) What is the mass of the air inside the ball?

$m_{air}=17.55\times\frac{4}{3}\pi0.09^{3}=.054\mathrm{kg}$

c) (5 points) The molar mass of air is 29 g/mol. How many moles of air are there inside the ball?

$n=\frac{54}{29}=1.85 \mathrm{moles}$

d) (5 points) If we treat the air as an ideal gas, to what pressure should the ball be inflated to at a temperature of 20$\mathrm{^{o}C}$? Give your final answer in atm.

$PV=nRT$

$P=\frac{1.85\times8.314\times293}{\frac{4\pi}{3}0.09^{3}}=1471\mathrm{kPa}=14.6\mathrm{atm}$

## Question 5

A spring of spring constant $k=10 \,\mathrm{N/m}$ is hanging from the ceiling. A 500 g wooden Santa is attached to the spring changing it's equilibrium position by $x_{0}$. In all of the following questions you may neglect the mass of the spring.

a) (5 points) Find $x_{0}$.

$kx_{0}=mg$

$x_{0}=\frac{0.5\times9.8}{10}=0.49\mathrm{m}$

b) (5 points) The Santa is pushed back up so the spring has the same length as it had before the Santa was attached. How much potential energy has been added to the system?

$\Delta PE=\int_{x_{0}}^0 F_{Net}\,dx=\int_{x_{0}}^0(kx-mg)\,dx=-\frac{1}{2}kx_{0}^{2}+mgx_{0}=-\frac{1}{2}\times10\times0.49^{2}+0.5\times9.8\times0.49=1.2\mathrm{J}$

If we took the spring in the other direction then

$\Delta PE=\int_{x_{0}}^{2x_{0}} F_{Net}\,dx=\int_{x_{0}}^{2x_{0}}(kx-mg)\,dx=\frac{3}{2}kx_{0}^{2}-mgx_{0}=\frac{3}{2}\times10\times0.49^{2}-0.5\times9.8\times0.49=1.2\mathrm{J}$

So you can see that there is the same amount of energy stored in either direction, which is important for simple harmonic motion to occur.

c) (5 points) The Santa is then released and executes simple harmonic motion.What is the frequency of the simple harmonic motion?

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{10}{0.5}}=0.71\mathrm{Hz}$

d) (5 points) What is the magnitude of the maximum velocity during simple harmonic motion? Give the position of the point or points at which this occurs in terms of displacement from $x_{0}$.

$v_{max}=\omega A= 2\pi\times0.71\times0.49=2.19\mathrm{m/s}$ occurs at displacement $x=0\mathrm{m}$ from $x_{0}$

e) (5 points) What is the magnitude of the maximum acceleration during simple harmonic motion? Give the position of the point or points at which this occurs in terms of displacement from $x_{0}$.

$a_{max}=\omega^{2} A=(2\pi\times0.71)^{2}\times0.49=9.8\mathrm{m/s^{2}}$ occurs at displacements $x=-0.49\mathrm{m},x=0.49\mathrm{m}$ from $x_{0}$

## Question 6

Santa slides down a chimney which has a volume of 10\,m$^{3}$. The chimney is sealed at the bottom by a floor, and sealed at the top by Santa's belly. We can treat the air in the chimney as an ideal diatomic gas. (Note: Santa's suit has very good thermal insulation!)

a) (5 points) When Santa slides down the chimney from top to bottom he reduces the volume of the air by a factor of 5 (ie. $V_{final}=\frac{V_{initial}}{5}=2\,\mathrm{m^{3}}$). If the initial pressure is standard atmospheric pressure what is the final pressure. (Hint: No heat is added to the air in this process and we can approximate it is quasistatic.)

$P_{i}V_{i}^{\gamma}=P_{f}V_{f}^{\gamma}$ where in this case $\gamma=\frac{7}{5}$

$P_{f}=P_{i}(\frac{V_{i}}{V_{f}})^{\gamma}=101.3\mathrm{kPa}\times{5}^{7/5}=964.2\mathrm{kPa}$

b) (5 points) If the initial temperature of the air in the chimney was $2\mathrm{^{o}C}$ what is the final temperature of the air in the chimney?

As the gas is ideal

$P_{i}V_{i}=nRT_{i}$

$P_{f}V_{f}=nRT_{f}$

$\frac{T_{f}}{T_{i}}=\frac{P_{f}V_{f}}{P_{i}V_{i}}$

$T_{f}=\frac{964.2}{101.3}\times\frac{1}{5}\times275=523.5\mathrm{K}$

c) (5 points) To get Santa out of the chimney a fire is lit under it which adds heat to the air, so that it expands isothermally. How much work is done by the gas as it expands isothermally back to it's original volume.

For an isothermal process

$W=nRT\ln{\frac{V_{f}}{V_{i}}}$

$PV=nRT$

$W=964.2\times10^{3}\times 2\ln(5)=3103\mathrm{kJ}$

d) (5 points) What is the pressure in the chimney when Santa has returned to his original position?

For an isothermal process

$P_{i}V_{i}=P_{f}V_{f}$

$P_{f}=\frac{V_{f}}{V_{i}}P_{i}=\frac{964.2}{5}=192.8\mathrm{kPa}$

e) (5 points) Sketch a pressure-volume diagram which shows both processes, making sure to label each process and the direction in which they occur.

Red is first process, green is second 