Problem Set #4 Short Answer

For a ride on the ultimate roller-coaster Danica Spears, a college student from Pensacola takes a ride with one of the blue angels. See the video here:

In the last half of the video the plane makes a high velocity tight turn which is horizontal to the ground and they achieve 7.4 g's of acceleration. During the turn the plane is turned sideways (wings perpendicular to the ground) and the turn is vertical as Danica is situated. She feels herself (and you can see it on her face) pulled strongly to the floor with a gigantic force.

In order to withstand such g-forces (which in an F-18 can be greater than 9 g's) a number of coping mechanisms have been developed. See here:

You can see a more extended discussion of the problems involved in high g maneuvers here (there are two videos in sequence):

1. Considering what you know about circular motion, why are high speed and tight turns so dangerous?

2. Before losing consciousness many pilots experience loss of color vision and later tunnel vision during a high speed and tight turn. Describe two ways a pilot could reduce his g-forces if she found herself in such a situation and justify them in terms of the form of the centripetal acceleration in terms of the velocity, v, and radius, r.

First, I am jealous of Ms. Spears. Second, a bit more background is in order. Check out this video from Smarter Every Day. The most relevant bit runs from 2:15 to 3:45, but the whole thing is worth watching:

As Destin describes, in a high-g turn, blood is forced away from a pilot’s head and towards their legs. This reduces the blood pressure in the pilot’s brain, which may cause a temporary loss of consciousness. In general, this phenomenon will not hurt you, and you will recover as soon as the g-load is reduced. But, the fact that you might crash or be shot down as a result of your loss of control makes these situations extremely dangerous for fighter pilots. Thus, these situations are best avoided whenever possible, but since avoiding them is not always possible (especially in combat where a fast and tight turn is a useful maneuver), techniques and technologies have been developed to minimize the physiological impacts of high-g turns on pilots. These include tensing the muscles in your lower body to essentially “squeeze” blood out of your legs and back towards your head and wearing a device called a g-suit, which squeezes your legs externally.

But, these questions ask specifically about the physics of circular motion, so let us focus on that. From this standpoint, what happens during a high-g turn? In the cockpit, you feel a strong “gravitational force” towards your feet, which pulls blood away from your head. However, an observer on the ground sees your head accelerating towards the center of the turn and away from your blood!

That’s because an object’s direction of travel constantly changes while the object moves along a curved path. This means the object's motion is changing, since what we mean by “motion” is the rate and linear direction of an object’s change in position.

So, as long as you are turning, your motion is changing, and you are accelerating towards the center of the turn. Due to the way that high-performance aircraft perform these turns, the top of your head will always point towards the turn center, and the bottom of your feet will always point away from it. Thus, throughout the turn, your direction of acceleration points away from your feet towards your head.

But, your body is made of matter, and matter has inertia, meaning it resists acceleration/changes in its motion. Thus, your body has a natural tendency to move in a straight line at a constant speed. This goes for all the matter in your body, including your blood, and anything not directly contacting the aircraft’s seat or your rigid bones, such as your blood, will, therefore, tend to continue moving in a straight line as everything else accelerates towards the center of the turn. That’s why, from the ground observer’s perspective, your blood winds up in your feet. It moved in a straight line. The rest of you accelerated towards the center of the turn! The more you accelerate, the more your blood “gets left behind.”

Wild, huh?

I think so.

So, let us now quantify your acceleration \(a_c\) and its relationship to your speed \(v\) and the radius of your turn \(r\): \begin{equation} a_c = \frac{v^2}{r}. \end{equation}

The faster you’re moving, the more acceleration it takes to turn at a given radius. Or, the tighter the turn (and smaller the radius), the more acceleration it takes to turn at a given speed. Thus, tight, high-speed turns require tremendous amounts of acceleration, throughout which your head “runs away” from your blood. When this happens, you run the risk of losing consciousness and control, which is pretty darn dangerous.

So, what do you do if you sense the onset of a blackout? Reduce your speed or reduce your pitch, thereby increasing the radius of your turn.

Then just hope you don’t get shot down.


Problem Set #3 Short Answer

After traveling all night, you are driving down a narrow country lane. You briefly fall asleep at the wheel, but wake up as the car moves off the shoulder of the road. You wake up but rapidly lose control of the car and it zooms off the road. You are heading towards (1) a stack of bales of hay, or (2) a brick wall. You have seconds to choose, and luckily you choose to hit the hay stack and come to a stop. 

Compare in detail the cases of hitting the brick wall versus hitting the hay stack. In which scenario do you transfer more momentum?  In which scenario is your impulse larger? In which scenario is the force on the car largest? In which scenario do you experience the largest acceleration? Explain why using physical principles you have learned.

In both cases, your car start off traveling at the same initial velocity, \(v_i\). Also, in both cases, you car comes to a stop: \(v_f = 0\). Therefore, in both cases, your car experiences the same velocity change \(\Delta v\) and hence momentum change \(\Delta p\) throughout the collision: \begin{equation} m \Delta v = m \left( v_f - v_i \right) = m v_f - m v_i = p_f - p_i = \Delta p. \end{equation}

According to the impulse-momentum theorem, this momentum change is equal to the impulse \(J\) exerted on your car: \begin{equation} \Delta p = J. \end{equation}

Any impulse is proportional to the average force that causes it \(F_{avg}\) and the time interval \(\Delta t\) during which that force acts: \begin{equation} J = F_{avg} \Delta t. \end{equation}

Therefore, for a given impulse, the average force and time interval are inversely proportional to each other: \begin{equation} F_{avg} = \frac{J}{\Delta t}. \end{equation}

This means that if the time interval is longer, then the force is smaller and vice versa.

Finally, an object’s acceleration is proportional to (\(\propto\)) the force acting upon it (Newton’s 2nd law): \begin{equation} a_{avg} \propto F_{avg}. \end{equation}

This means that if the force acting on your car is greater, then your car’s acceleration is greater, and vice versa.

Combining all these statements into one expression results in \begin{equation} a_{avg} \propto F_{avg} = \frac{J}{\Delta t} = \frac{\Delta p}{\Delta t} = \frac{m \Delta v}{\Delta t}. \end{equation}

Now, imagine an object with mass \(m\) traveling at a velocity \(v\), and imagine two two scenarios in which that object collides with something and comes to a stop. In both cases, the object has the same velocity change \(\Delta v\). Therefore in both cases it has the same momentum change \(\Delta p\) resulting from the same impulse \(J\). Now image that in one case, the collision takes a long time \(\Delta t_L\), while in the collision in the other case takes a short time \(\Delta t_S\). It follows from the result above that: \begin{equation} \begin{aligned} \Delta t_S < \Delta t_L & \implies \frac{J}{\Delta t_L} < \frac{J}{\Delta t_S} \\[1ex] & \implies F_L < F_S \\[1ex] & \implies a_L < a_S. \end{aligned} \end{equation}

The amount of damage you and your car sustain will be proportional to the force exerted on you and your resulting acceleration. And, presumably, a brick wall takes less time to stop a car than does a stack of hay.

So, which one would you rather hit?


Problem Set #2 Short Answer

The Horns Rev wind farm is 15 miles off the west coast of the Danish peninsula, near Ejsberg. You can see a movie of the wind farm here: There are 171 wind turbines like this spread out over the shallow North Sea floor, some of which (the Horns Rev 2 field) can generate 6 MW of power. As we discussed in class (and also in the 4th edition) the work done by a torque is (torque x rotation angle in radians). To get the power from the wind then it is (torque x angular velocity in radians per second). Assuming no losses, the 6 MW electrical power generated must correspond to an equivalent 6 MW of power done by work from the wind on the turbines. Since there are three blades, then each blade must have 2 MW of power done on it from the wind. A typical rotation speed is 20 rotations per minute, which corresponds to 40pi or 120 radians per minute, or 2 radians per second. That means the TORQUE provided by the wind must be 2 MW/2 radians per second = 1 million Newton-meters of torque.

These turbines are absolutely gigantic.  Look here: (see "Record-Holding turbines" at the bottom). The Enercon 6 MW E-126 is 198 meters (650 ft!) tall, and the turbine is 126 meters (413 ft!) in diameter. That means each propeller blade is 63 m (206 ft!) long. If we assume that the wind force on the turbine blade acts at its center, 31.5 meters from the rotation axis, then the 1 million Newton-meters of torque from the wind, corresponds to a force from the wind on the turbine of 32,000 Newtons which is equivalent to the weight of a mass of 3200 kg, which is 6400 pounds or a little over 3 tons on each of the three propeller (turbine) blades!

In this short essay, consider that there is an extremely large force from the wind on the turbine which provides a torque which does work (and generates a lot of power). Answer the following questions in your essay:

1.  What is the net torque on the turbine? How do you know? List all the torques on the turbine which make this happen.

2.  Considering that there is a giant force from the wind ON the turbine blade, how is Newton's 3rd law satisfied? How could you tell that Newton's 3rd law is satisfied?

First, we should carefully define the components of our interacting system.

We have the turbine itself, obviously, but let's think about what the turbine actually is. It's the part of the system that looks like a fan—the blades and the hub to which they are attached. It's the part of the system that we can see rotating.

The turbine is attached to the generator, the next part of the system. In this problem, we should consider the turbine and generator to be two distinct things, though it may be more intuitive to consider them as one—don't! Instead, think of it this way: the turbine turns the generator, and the generator resists being turned. Consequently, if the turbine is turning, the generator makes it stop. We'll call this a stopping torque—a torque that causes the turbine to slow down and stop.

The third component is the air in which the turbine is immersed. When the air blows against the turbine, it will cause the turbine to turn. So, if the turbine isn't turning, blowing air makes it start. We'll call this a driving torque—a torque that causes the turbine to start and speed up.

Now, with all the components (and torques on the turbine) established, we may consider the questions being asked. Let us take them one at a time.

1. What is the net torque on the turbine? How do you know? List all the torques on the turbine which make this happen.

Well, whenever the turbine turns, the generator turns, since the generator is attached to the turbine. But remember, the generator resists being turned. Consequently, it causes the turbine to stop (via the stopping torque the generator exerts on the turbine.) However, in the video above, we see the turbines turning at constant speeds, so they are not stopping! What keeps them going? A driving torque caused by a third agent that balances out the stopping torque from the generator, perhaps?

2. Considering that there is a giant force from the wind ON the turbine blade, how is Newton's 3rd law satisfied? How could you tell that Newton's 3rd law is satisfied?

This question specifically asks about the interaction of the moving air and the turbine, but Newton's third law applies whenever any two physical objects interact with each other. So, for example, the generator exerts a stopping torque on the turbine because the turbine exerts a driving torque on the generator. Similarly, the turbine exerts a stopping torque on the moving airbecause the air exerts a driving torque on the turbine. Both of those are examples of Newton's third law in action. So, in the case of the turbine vs. the air, how can you tell that Newton's third law is being satisfied? Perhaps by measuring some change in the wind?

I hope this helps,