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Contents

Section 9.1 - Conceptual framework

Section 9.2 - Uniform circular motion

Section 9.3 - Nonuniform circular motion

Section 9.4 - Summary

Section 9.1 - Conceptual framework

Section 9.2 - Uniform circular motion

Section 9.3 - Nonuniform circular motion

Section 9.4 - Summary

I now live fifteen minutes from Disneyland, so my friends and family in my native Northern California think it's a little strange that I've never visited the Magic Kingdom again since a childhood trip to the south. The truth is that for me as a preschooler, Disneyland was not the Happiest Place on Earth. My mother took me on a ride in which little cars shaped like rocket ships circled rapidly around a central pillar. I knew I was going to die. There was a force trying to throw me outward, and the safety features of the ride would surely have been inadequate if I hadn't screamed the whole time to make sure Mom would hold on to me. Afterward, she seemed surprisingly indifferent to the extreme danger we had experienced.

My younger self's understanding of circular motion was partly right and partly wrong. I was wrong in believing that there was a force pulling me outward, away from the center of the circle. The easiest way to understand this is to bring back the parable of the bowling ball in the pickup truck from chapter 4. As the truck makes a left turn, the driver looks in the rearview mirror and thinks that some mysterious force is pulling the ball outward, but the truck is accelerating, so the driver's frame of reference is not an inertial frame. Newton's laws are violated in a noninertial frame, so the ball appears to accelerate without any actual force acting on it. Because we are used to inertial frames, in which accelerations are caused by forces, the ball's acceleration creates a vivid illusion that there must be an outward force.

In an inertial frame everything makes more sense. The ball has no force on it, and goes straight as required by Newton's first law. The truck has a force on it from the asphalt, and responds to it by accelerating (changing the direction of its velocity vector) as Newton's second law says it should.

Another interesting example is an insect organ called the halteres, a pair of small knobbed limbs behind the wings, which vibrate up and down and help the insect to maintain its orientation in flight. The halteres evolved from a second pair of wings possessed by earlier insects. Suppose, for example, that the halteres are on their upward stroke, and at that moment an air current causes the fly to pitch its nose down. The halteres follow Newton's first law, continuing to rise vertically, but in the fly's rotating frame of reference, it seems as though they have been subjected to a backward force. The fly has special sensory organs that perceive this twist, and help it to correct itself by raising its nose.

I was correct, however, on a different point about the Disneyland ride. To make me curve around with the car, I really did need some force such as a force from my mother, friction from the seat, or a normal force from the side of the car. (In fact, all three forces were probably adding together.) One of the reasons why Galileo failed to refine the principle of inertia into a quantitative statement like Newton's first law is that he was not sure whether motion without a force would naturally be circular or linear. In fact, the most impressive examples he knew of the persistence of motion were mostly circular: the spinning of a top or the rotation of the earth, for example. Newton realized that in examples such as these, there really were forces at work. Atoms on the surface of the top are prevented from flying off straight by the ordinary force that keeps atoms stuck together in solid matter. The earth is nearly all liquid, but gravitational forces pull all its parts inward.

Circular motion always involves a change in the direction of
the velocity vector, but it is also possible for the
magnitude of the velocity to change at the same time.
Circular motion is referred to as *uniform* if \(|\mathbf{v}|\) is
constant, and *nonuniform* if it is changing.

Your speedometer tells you the magnitude of your car's velocity vector, so when you go around a curve while keeping your speedometer needle steady, you are executing uniform circular motion. If your speedometer reading is changing as you turn, your circular motion is nonuniform. Uniform circular motion is simpler to analyze mathematically, so we will attack it first and then pass to the nonuniform case.

Which of these are examples of uniform circular motion and which are nonuniform?

(1) the clothes in a clothes dryer (assuming they remain against the inside of the drum, even at the top)

(2) a rock on the end of a string being whirled in a vertical circle

(answer in the back of the PDF version of the book)Figure c showed the string pulling in straight along a radius of the circle, but many people believe that when they are doing this they must be “leading” the rock a little to keep it moving along. That is, they believe that the force required to produce uniform circular motion is not directly inward but at a slight angle to the radius of the circle. This intuition is incorrect, which you can easily verify for yourself now if you have some string handy. It is only while you are getting the object going that your force needs to be at an angle to the radius. During this initial period of speeding up, the motion is not uniform. Once you settle down into uniform circular motion, you only apply an inward force.

If you have not done the experiment for yourself, here is a theoretical argument to convince you of this fact. We have discussed in chapter 6 the principle that forces have no perpendicular effects. To keep the rock from speeding up or slowing down, we only need to make sure that our force is perpendicular to its direction of motion. We are then guaranteed that its forward motion will remain unaffected: our force can have no perpendicular effect, and there is no other force acting on the rock which could slow it down. The rock requires no forward force to maintain its forward motion, any more than a projectile needs a horizontal force to “help it over the top” of its arc.

Why, then, does a car driving in circles in a parking lot
stop executing uniform circular motion if you take your foot
off the gas? The source of confusion here is that Newton's
laws predict an object's motion based on the *total*
force acting on it. A car driving in circles has three forces on it

(1) an inward force from the asphalt, controlled with the steering wheel;

(2) a forward force from the asphalt, controlled with the gas pedal; and

(3) backward forces from air resistance and rolling resistance.

You need to make sure there is a forward force on the car so that the backward forces will be exactly canceled out, creating a vector sum that points directly inward.

Since experiments show that the force vector points directly inward, Newton's second law implies that the acceleration vector points inward as well. This fact can also be proven on purely kinematical grounds, and we will do so in the next section.

Suppose that we want to subject the first law to
a stringent experimental test.^{1} The law predicts that
if we use a clock to measure the rate of rotation of an object spinning frictionlessly, it won't
“naturally” slow down as Aristotle would have expected. But what is a clock but something with hands
that rotate at a fixed rate? In other words, we are comparing one clock with another. This is called
a clock-comparison experiment. Suppose that the laws of
physics weren't purely Newtonian, and there really was a very slight Aristotelian tendency for
motion to slow down in the absence of friction. If we compare two clocks, they should both slow
down, but if they aren't the same type of clock,
then it seems unlikely that they would slow down at exactly the same rate, and over time they should
drift further and further apart.

High-precision clock-comparison experiments have been done using a variety of clocks. In atomic clocks,
the thing spinning is an atom. Astronomers can observe the rotation of collapsed stars called pulars,
which, unlike the earth, can rotate with almost no disturbance due to geological activity or friction
induced by the tides. In these experiments, the pulsars are observed to match the rates of the atomic
clocks with a drift of less than about \(10^{-6}\) seconds over a
period of 10 years.^{2}
Atomic clocks using atoms of different elements drift relative to one another by no more than
about \(10^{-16}\) per year.^{3}

It is not presently possible to do experiments with a similar level of precision using human-scale rotating objects. However, a set of gyroscopes aboard the Gravity Probe B satellite were allowed to spin weightlessly in a vacuum, without any physical contact that would have caused kinetic friction. Their rotation was extremely accurately monitored for the purposes of another experiment (a test of Einstein's theory of general relativity, which was the purpose of the mission), and they were found to be spinning down so gradually that they would have taken about 10,000 years to slow down by a factor of two. This rate was consistent with estimates of the amount of friction to be expected from the small amount of residual gas present in the vacuum chambers.

A subtle point in the interpretation of these experiments is that if there was a slight tendency
for motion to slow down, we would have to decide what it was supposed to slow down relative to.
A straight-line motion that is slowing down in some frame of reference can always
be described as *speeding up* in some other appropriately chosen frame (problem 12,
p. 89). If the laws of physics did have this slight Aristotelianism
mixed in, we could wait for the anomalous acceleration or deceleration to stop. The object we were
observing would then define a special or “preferred” frame of reference. Standard theories of
physics do not have such a preferred frame, and clock-comparison experiments can be viewed as
tests of the existence of such a frame.
Another test for the existence of a
preferred frame is described on p. 265.

◊

In the game of crack the whip, a line of people stand holding hands, and then they start sweeping out a circle. One person is at the center, and rotates without changing location. At the opposite end is the person who is running the fastest, in a wide circle. In this game, someone always ends up losing their grip and flying off. Suppose the person on the end loses her grip. What path does she follow as she goes flying off? (Assume she is going so fast that she is really just trying to put one foot in front of the other fast enough to keep from falling; she is not able to get any significant horizontal force between her feet and the ground.)

◊

Suppose the person on the outside is still holding on, but feels that she may loose her grip at any moment. What force or forces are acting on her, and in what directions are they? (We are not interested in the vertical forces, which are the earth's gravitational force pulling down, and the ground's normal force pushing up.) Make a table in the format shown in section 5.3.

◊

Suppose the person on the outside is still holding on, but feels that she may loose her grip at any moment. What is wrong with the following analysis of the situation? “The person whose hand she's holding exerts an inward force on her, and because of Newton's third law, there's an equal and opposite force acting outward. That outward force is the one she feels throwing her outward, and the outward force is what might make her go flying off, if it's strong enough.”

◊

If the only force felt by the person on the outside is an inward force, why doesn't she go straight in?

◊

In the amusement park ride shown in the figure, the cylinder spins faster and faster until the customer can pick her feet up off the floor without falling. In the old Coney Island version of the ride, the floor actually dropped out like a trap door, showing the ocean below. (There is also a version in which the whole thing tilts up diagonally, but we're discussing the version that stays flat.) If there is no outward force acting on her, why does she stick to the wall? Analyze all the forces on her.

◊

What is an example of circular motion where the inward force is a normal force? What is an example of circular motion where the inward force is friction? What is an example of circular motion where the inward force is the sum of more than one force?

◊

Does the acceleration vector always change continuously in circular motion? The velocity vector?

In this section I derive a simple and very useful equation for the magnitude of the acceleration of an object undergoing constant acceleration. The law of sines is involved, so I've recapped it in figure i.

The derivation is brief, but the method requires some explanation and justification. The idea is to calculate a \(\Delta\mathbf{v}\) vector describing the change in the velocity vector as the object passes through an angle \(\theta \). We then calculate the acceleration, \(\mathbf{a}=\Delta\mathbf{v}/\Delta t\). The astute reader will recall, however, that this equation is only valid for motion with constant acceleration. Although the magnitude of the acceleration is constant for uniform circular motion, the acceleration vector changes its direction, so it is not a constant vector, and the equation \(\mathbf{a}=\Delta\mathbf{v}/\Delta t\) does not apply. The justification for using it is that we will then examine its behavior when we make the time interval very short, which means making the angle \(\theta \) very small. For smaller and smaller time intervals, the \(\Delta\mathbf{v}/\Delta t\) expression becomes a better and better approximation, so that the final result of the derivation is exact.

In figure j/1, the object sweeps out an angle \(\theta \). Its direction of motion also twists around by an angle \(\theta\), from the vertical dashed line to the tilted one. Figure j/2 shows the initial and final velocity vectors, which have equal magnitude, but directions differing by \(\theta \). In j/3, I've reassembled the vectors in the proper positions for vector subtraction. They form an isosceles triangle with interior angles \(\theta\), \(\eta\), and \(\eta\). (Eta, \(\eta\), is my favorite Greek letter.) The law of sines gives

\[\begin{equation*}
\frac{|\Delta\mathbf{v}|}{\sin\theta} = \frac{|\mathbf{v}|}{\sin\eta} .
\end{equation*}\]

This tells us the magnitude of \(\Delta \mathbf{v}\), which is one of the two ingredients we need for calculating the magnitude of \(\mathbf{a}=\Delta\mathbf{v}/\Delta t\). The other ingredient is \(\Delta t\). The time required for the object to move through the angle \(\theta \) is

\[\begin{equation*}
\Delta t = \frac{\text{length of arc}}{|\mathbf{v}|} .
\end{equation*}\]

Now if we measure our angles in radians we can use the definition of radian measure, which is \((\text{angle})=(\text{length of arc})/(\text{radius})\), giving \(\Delta t=\theta r/|\mathbf{v}|\). Combining this with the first expression involving \(|\Delta v|\) gives

\[\begin{align*}
|\mathbf{a}| &= |\Delta \mathbf{v}|/\Delta t \\
&= \frac{|\mathbf{v}|^2}{r} \: \cdot \: \frac{\sin\theta}{\theta} \: \cdot \: \frac{1}{\sin\eta} .
\end{align*}\]

When \(\theta \) becomes very small, the small-angle approximation \(\sin \theta\approx \theta\) applies, and also \(\eta \) becomes close to 90°, so \(\sin \eta \approx 1\), and we have an equation for \(|\mathbf{a}|\):

\[\begin{equation*}
|\mathbf{a}| = \frac{|\mathbf{v}|^2}{r} . \shoveright{\text{[uniform circular motion]}}
\end{equation*}\]

\(\triangleright\) A bicyclist is making a turn along an arc of a circle with radius 20 m, at a speed of 5 m/s. If the combined mass of the cyclist plus the bike is 60 kg, how great a static friction force must the road be able to exert on the tires?

\(\triangleright\) Taking the magnitudes of both sides of Newton's second law gives

\[\begin{align*}
|\mathbf{F}| &= |m\mathbf{a}| \\
&= m|\mathbf{a}| .
\end{align*}\]

Substituting \(|\mathbf{a}|=|\mathbf{v}|^2/r\) gives

\[\begin{align*}
|\mathbf{F}| &= m|\mathbf{v}|^2/r \\
&\approx 80\ \text{N}
\end{align*}\]

(rounded off to one sig fig).

\(\triangleright\) You're driving on a mountain road with a steep drop on your right. When making a left turn, is it safer to hug the center line or to stay closer to the outside of the road?

\(\triangleright\) You want whichever choice involves the least acceleration, because that will require the least force and entail the least risk of exceeding the maximum force of static friction. Assuming the curve is an arc of a circle and your speed is constant, your car is performing uniform circular motion, with \(|\mathbf{a}|=|\mathbf{v}|^2/r\). The dependence on the square of the speed shows that driving slowly is the main safety measure you can take, but for any given speed you also want to have the largest possible value of \(r\). Even though your instinct is to keep away from that scary precipice, you are actually less likely to skid if you keep toward the outside, because then you are describing a larger circle.

\(\triangleright\) The period can be related to the speed as follows:

\[\begin{align*}
|\mathbf{v}| &= \frac{\text{circumference}}{T} \\
&= 2\pi r/T .
\end{align*}\]

Substituting into the equation \(|\mathbf{a}|=|\mathbf{v}|^2/r\) gives

\[\begin{equation*}
|\mathbf{a}| = \frac{4\pi^2r}{T^2} .
\end{equation*}\]

\(\triangleright\) We can solve this by finding the period and plugging in to the result of the previous example. If it makes 48 revolutions in one minute, then the period is 1/48 of a minute, or 1.25 s. To get an acceleration in mks units, we must convert the radius to 0.35 m. Plugging in, the result is 8.8 \(\text{m}/\text{s}^2\).

\(\triangleright\) In a discussion question in the previous section, we made the assumption that the clothes remain against the inside of the drum as they go over the top. In light of the previous example, is this a correct assumption?

\(\triangleright\) No. We know that there must be some minimum speed at which the motor can run that will result in the clothes just barely staying against the inside of the drum as they go over the top. If the clothes dryer ran at just this minimum speed, then there would be no normal force on the clothes at the top: they would be on the verge of losing contact. The only force acting on them at the top would be the force of gravity, which would give them an acceleration of \(g=9.8\ \text{m}/\text{s}^2\). The actual dryer must be running slower than this minimum speed, because it produces an acceleration of only \(8.8\ \text{m}/\text{s}^2\). My theory is that this is done intentionally, to make the clothes mix and tumble.

◊ Solved problem: The tilt-a-whirl — problem 6

◊ Solved problem: An off-ramp — problem 7

◊

A certain amount of force is needed to provide the acceleration of circular motion. What if were are exerting a force perpendicular to the direction of motion in an attempt to make an object trace a circle of radius \(r\), but the force isn't as big as \(m|\mathbf{v}|^2/r\)?

◊

Suppose a rotating space station, as in figure l on page 243, is built. It gives its occupants the illusion of ordinary gravity. What happens when a person in the station lets go of a ball? What happens when she throws a ball straight “up” in the air (i.e., towards the center)?

What about nonuniform circular motion? Although so far we have been discussing components of vectors along fixed \(x\) and \(y\) axes, it now becomes convenient to discuss components of the acceleration vector along the radial line (in-out) and the tangential line (along the direction of motion). For nonuniform circular motion, the radial component of the acceleration obeys the same equation as for uniform circular motion,

\[\begin{equation*}
a_r = v^2/r ,
\end{equation*}\]

where \(v=|\mathbf{v}|\), but the acceleration vector also has a tangential component,

\[\begin{equation*}
a_t = \text{slope of the graph of $v$ versus $t$} .
\end{equation*}\]

The latter quantity has a simple interpretation. If you are going around a curve in your car, and the speedometer needle is moving, the tangential component of the acceleration vector is simply what you would have thought the acceleration was if you saw the speedometer and didn't know you were going around a curve.

\(\triangleright\) When you're making a turn in your car and you're afraid you may skid, isn't it a good idea to slow down?

\(\triangleright\) If the turn is an arc of a circle, and you've already completed part of the turn at constant speed without skidding, then the road and tires are apparently capable of enough static friction to supply an acceleration of \(|\mathbf{v}|^2/r\). There is no reason why you would skid out now if you haven't already. If you get nervous and brake, however, then you need to have a tangential acceleration component in addition to the radial one you were already able to produce successfully. This would require an acceleration vector with a greater magnitude, which in turn would require a larger force. Static friction might not be able to supply that much force, and you might skid out. The safer thing to do is to approach the turn at a comfortably low speed.

◊ Solved problem: A bike race — problem 5

*uniform circular motion* — circular motion in which the
magnitude of the velocity vector remains constant

*nonuniform circular motion* — circular motion in which the
magnitude of the velocity vector changes

*radial* — parallel to the radius of a circle; the in-out direction

*tangential* — tangent to the circle, perpendicular to
the radial direction

\(a_r\) — radial acceleration; the component of the acceleration vector along the in-out direction

\(a_t\) — tangential acceleration; the component of the acceleration vector tangent to the circle

{}

If an object is to have circular motion, a force must be exerted on it toward the center of the circle. There is no outward force on the object; the illusion of an outward force comes from our experiences in which our point of view was rotating, so that we were viewing things in a noninertial frame.

An object undergoing uniform circular motion has an inward acceleration vector of magnitude

\[\begin{equation*}
|\mathbf{a}| = v^2/r ,
\end{equation*}\]

where \(v=|\mathbf{v}|\). In nonuniform circular motion, the radial and tangential components of the acceleration vector are

\[\begin{align*}
a_r &= v^2/r \\
a_t &= \text{slope of the graph of $v$ versus $t$} .
\end{align*}\]

\begin{homeworkforcelabel}{mixer}{1}{}{1}When you're done using an electric mixer, you can get
most of the batter off of the beaters by lifting them out of
the batter with the motor running at a high enough speed.
Let's imagine, to make things easier to visualize, that we
instead have a piece of tape stuck to one of the beaters.

(a) Explain why static friction has no effect on whether or
not the tape flies off.

(b) Analyze the forces in which the tape participates, using
a table the format shown in section 5.3.

(c) Suppose you find that the tape
doesn't fly off when the motor is on a low speed, but
at a greater speed, the tape won't stay on. Why would the
greater speed change things? [Hint: If you don't invoke any
law of physics, you haven't explained it.]
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{circularaccelunits}{1}{}{2}Show that the expression \(|\mathbf{v}|^2/r\) has the units of acceleration. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{loop}{1}{}{3}A plane is flown in a loop-the-loop of radius 1.00 km. The plane starts out flying upside-down, straight and level, then begins curving up along the circular loop, and is right-side up when it reaches the top. (The plane may slow down somewhat on the way up.) How fast must the plane be going at the top if the pilot is to experience no force from the seat or the seatbelt while at the top of the loop? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{circularcalculus}{1}{1}{4}In this problem, you'll derive the equation
\(|\mathbf{a}|=|\mathbf{v}|^2/r\) using calculus. Instead of comparing
velocities at two points in the particle's motion and then
taking a limit where the points are close together, you'll
just take derivatives. The particle's position vector is
\(\mathbf{r}=(r \cos\theta)\hat{\mathbf{x}} + (r\sin\theta)\hat{\mathbf{y}}\), where \(\hat{\mathbf{x}}\) and \(\hat{\mathbf{y}}\)
are the unit vectors along the \(x\) and \(y\) axes. By the
definition of radians, the distance traveled since \(t=0\) is
\(r\theta \), so if the particle is traveling at constant
speed \(v=|\mathbf{v}|\), we have \(v=r\theta \)/t.

(a) Eliminate \(\theta\) to get the particle's position vector as a function of
time.

(b) Find the particle's acceleration vector.

(c) Show
that the magnitude of the acceleration vector equals \(v^2/r\).
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{cyclists}{1}{}{5}(solution in the pdf version of the book) Three cyclists in a race are rounding a semicircular curve. At the moment depicted, cyclist A is using her brakes to apply a force of 375 N to her bike. Cyclist B is coasting. Cyclist C is pedaling, resulting in a force of 375 N on her bike Each cyclist, with her bike, has a mass of 75 kg. At the instant shown, the instantaneous speed of all three cyclists is 10 m/s. On the diagram, draw each cyclist's acceleration vector with its tail on top of her present position, indicating the directions and lengths reasonably accurately. Indicate approximately the consistent scale you are using for all three acceleration vectors. Extreme precision is not necessary as long as the directions are approximately right, and lengths of vectors that should be equal appear roughly equal, etc. Assume all three cyclists are traveling along the road all the time, not wandering across their lane or wiping out and going off the road. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{tilt-a-whirl}{2}{}{6}The amusement park ride shown in the figure consists
of a cylindrical room that rotates about its vertical axis.
When the rotation is fast enough, a person against the wall
can pick his or her feet up off the floor and remain
“stuck” to the wall without falling.

(a) Suppose the rotation results in the person having a
speed \(v\). The radius of the cylinder is \(r\), the person's
mass is \(m\), the downward acceleration of gravity is \(g\),
and the coefficient of static friction between the person
and the wall is \(\mu_s\). Find an equation for the speed,
\(v\), required, in terms of the other variables. (You will
find that one of the variables cancels out.)

(b) Now suppose two people are riding the ride. Huy is
wearing denim, and Gina is wearing polyester, so Huy's
coefficient of static friction is three times greater. The
ride starts from rest, and as it begins rotating faster and
faster, Gina must wait longer before being able to lift her
feet without sliding to the floor. Based on your equation
from part a, how many times greater must the speed be before
Gina can lift her feet without sliding down?(solution in the pdf version of the book)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{off-ramp}{1}{}{7}(solution in the pdf version of the book) An engineer is designing a curved off-ramp for a freeway. Since the off-ramp is curved, she wants to bank it to make it less likely that motorists going too fast will wipe out. If the radius of the curve is \(r\), how great should the banking angle, \(\theta \), be so that for a car going at a speed \(v\), no static friction force whatsoever is required to allow the car to make the curve? State your answer in terms of \(v\), \(r\), and \(g\), and show that the mass of the car is irrelevant. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{train}{1}{}{8}Lionel brand toy trains come with sections of track in standard lengths and shapes. For circular arcs, the most commonly used sections have diameters of 662 and 1067 mm at the inside of the outer rail. The maximum speed at which a train can take the broader curve without flying off the tracks is 0.95 m/s. At what speed must the train be operated to avoid derailing on the tighter curve?(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{conicalpendulum}{1}{}{9}The figure shows a ball on the end of a string of length
\(L\) attached to a vertical rod which is spun about its
vertical axis by a motor. The period (time for one rotation) is \(P\).

(a) Analyze the forces in which the ball participates.

(b) Find how the angle \(\theta \) depends on \(P,g\), and \(L\).
[Hints: (1) Write down Newton's second law for the vertical
and horizontal components of force and acceleration. This
gives two equations, which can be solved for the two
unknowns, \(\theta \) and the tension in the string. (2) If
you introduce variables like \(v\) and \(r\), relate them to the
variables your solution is supposed to contain, and eliminate them.](answer check available at lightandmatter.com)

(c) What happens mathematically to your solution if the
motor is run very slowly (very large values of \(P\))?
Physically, what do you think would actually happen in this case?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{rodent}{1}{}{10}Psychology professor R.O. Dent requests funding for an
experiment on compulsive thrill-seeking behavior in
guinea pigs, in which the subject is to be attached to the end
of a spring and whirled around in a horizontal circle. The
spring has relaxed length \(b\), and obeys Hooke's law
with spring constant \(k\). It is stiff enough to keep from
bending significantly under the guinea pig's weight.

(a) Calculate the length of the spring when it is
undergoing steady circular motion in which one rotation
takes a time \(T\). Express your result in terms of \(k\), \(b\), \(T\), and the
guinea pig's mass \(m\).(answer check available at lightandmatter.com)

(b) The ethics committee somehow fails to veto the
experiment, but the safety committee expresses concern. Why?
Does your equation do anything unusual, or even spectacular,
for any particular value of \(T\)? What do you think is the
physical significance of this mathematical behavior?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{governor}{2}{}{11}The figure shows an old-fashioned device called a
flyball governor, used for keeping an engine running at the
correct speed. The whole thing rotates about the vertical
shaft, and the mass \(M\) is free to slide up and down. This
mass would have a connection (not shown) to a valve that
controlled the engine. If, for instance, the engine ran too
fast, the mass would rise, causing the engine to slow back down.

(a) Show that in the special case of \(a=0\), the angle
\(\theta \) is given by

\[\begin{equation*}
\theta = \cos^{-1}\left(\frac{g(m+M)P^2}{4\pi^2mL}\right) ,
\end{equation*}\]

where \(P\) is the period of rotation (time required for
one complete rotation).

(b) There is no closed-form solution for \(\theta \) in the
general case where \(a\) is not zero. However, explain how the
undesirable low-speed behavior of the \(a=0\) device would be
improved by making \(a\) nonzero.

[Based on an example by J.P. den Hartog.]
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{circling-blocks}{1}{}{12}The figure shows two blocks of masses \(m_1\) and \(m_2\) sliding in circles on a frictionless table. Find the tension in the strings if the period of rotation (time required for one rotation) is \(P\).(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{paradox}{1}{}{13}The acceleration of an object in uniform circular motion can be given either by \(|\mathbf{a}|=|\mathbf{v}|^2/r\) or, equivalently, by \(|\mathbf{a}|=4\pi^2r/T^2\), where \(T\) is the time required for one cycle (example 6 on page 241). Person A says based on the first equation that the acceleration in circular motion is greater when the circle is smaller. Person B, arguing from the second equation, says that the acceleration is smaller when the circle is smaller. Rewrite the two statements so that they are less misleading, eliminating the supposed paradox. [Based on a problem by Arnold Arons.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{sirius-centrifugal}{1}{}{14}The bright star Sirius has a mass of \(4.02\times10^{30}\ \text{kg}\) and lies at a distance of \(8.1\times10^{16}\ \text{m}\) from our solar system. Suppose you're standing on a merry-go-round carousel rotating with a period of 10 seconds, and Sirius is on the horizon. You adopt a rotating, noninertial frame of reference, in which the carousel is at rest, and the universe is spinning around it. If you drop a corndog, you see it accelerate horizontally away from the axis, and you interpret this as the result of some horizontal force. This force does not actually exist; it only seems to exist because you're insisting on using a noninertial frame. Similarly, calculate the force that seems to act on Sirius in this frame of reference. Comment on the physical plausibility of this force, and on what object could be exerting it.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{wall-of-death}{1}{}{15}In a well known stunt from circuses and carnivals, a motorcyclist rides around inside
a big bowl, gradually speeding up and rising higher. Eventually the cyclist can get
up to where the walls of the bowl are vertical. Let's estimate the conditions under
which a running human could do the same thing.

(a) If the runner can run
at speed \(v\), and her shoes have a coefficient of static friction \(\mu_s\), what is
the maximum radius of the circle?(answer check available at lightandmatter.com)

(b) Show that the units of your answer make sense.

(c) Check that its dependence on the variables makes sense.

(d) Evaluate your result numerically for
\(v=10\ \text{m}/\text{s}\) (the speed of an olympic sprinter) and \(\mu_s=5\). (This is
roughly the highest coefficient of static friction ever achieved for surfaces that
are not sticky. The surface has an array of microscopic fibers like a hair brush,
and is inspired by the hairs on the feet of a gecko. These assumptions are not
necessarily realistic, since the person would have to run at an angle, which would
be physically awkward.)(answer check available at lightandmatter.com)

\end{homeworkforcelabel}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

[2] Matsakis *et al.*, Astronomy and Astrophysics 326 (1997) 924.
Freely available online at adsabs.harvard.edu.