Chapter 5: Dynamics of Uniform Circular Motion Tuesday, September 17, 2013 10:00 PM Rotational Kinematics We'll discuss the basics of rotational kinematics in this chapter; the kinematics equations for constant angular acceleration are discussed in Chapter 8. The basic quantities of rotational kinematics are angular position, angular displacement, angular velocity, and angular acceleration; period and frequency. Let's first discuss the period and frequency for circular motion: period T: amount of time needed to complete one revolution (unit: s, min, h, etc.) frequency f: number of revolutions per unit time (unit: s -1 (i.e., "revolutions per second"), rpm, etc.) basic formula relating T and f: Remember that the basic formula relating period and frequency is only valid when T and f have compatible units. For example, if you would like T to be measured in minutes (which really means minutes per revolution), then f must be measured in revolutions per minute (rpm). Example: Your car's tachometer reads 2500 rpm. How long does it take for the engine to make one cycle? Solution: Ch5 Page 1
Thus, the engine completes one cycle every 24 thousands of a second. Example: The earth rotates once on its axis every day. Determine the earth's frequency in rpm. Solution: Degrees and radians Ch5 Page 2
Angular position is measured in degrees or radians; here is a review of angle measure in degrees and radians; remember that the radian is a "unitless" unit: Remember that for a complete circle, the circumference is and therefore the angle of a complete circle is But we know that for a complete circle, = 360. Thus, Using these two conversion factors allows one to convert from degrees to radians, or vice versa. For example, Ch5 Page 3
Angular displacement and angular velocity Connecting linear and angular kinematic quantities Ch5 Page 4
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Centripetal acceleration A confusing example: driving around a circular curve in a car. We previously discussed the fact that we viscerally feel that we are thrown outwards when we go around a curve in a car; however, the acceleration is actually inward. It's understandable why this is confusing; our bodies are telling us that something is pushing us outwards, yet the following arguments indicate that the acceleration is actually inwards. It's difficult for the brain to understand something when the body feels the opposite. (You can understand why it took a few millennia for the greatest minds in humanity to figure out what is going on here.) The vector diagrams below argue that the acceleration in uniform (that is, constant speed) circular motion is really towards the centre of the circle. (As we argued in a previous chapter, the acceleration is not exactly towards the centre if the speed is increasing or decreasing in circular motion.) Because the acceleration is exactly towards the centre in uniform circular motion, it is called centripetal acceleration ("centripetal" just means "towards the centre"). Ch5 Page 7
The following figure is relevant (and the conclusions are valid) for circular motion with CONSTANT speed only: Ch5 Page 8
formula for centripetal acceleration Ch5 Page 9
Let's derive a neat formula for the magnitude of the centripetal acceleration. In Part (a) of the figure, let s represent the length of the circular arc (dashed arc), which is nearly the same as the length of the vector labelled d. Then, From the first figure in Part (c), We can't be sure of this formula for centripetal acceleration because of the approximations used, and because some of the quantities were average values. Nevertheless, the formula is correct, although it will take calculus to prove that it is exact (calculus lovers are welcome to read the proof that is further below in the notes). Getting an intuitive feel for the formula for centripetal acceleration: Ch5 Page 10
Why does the formula for centripetal acceleration that we just wrote down have the square of the speed in the numerator, and not just the speed? One way to see this is using what is called dimensional analysis. In essence, you check the units on both sides of the equation and the units match if the numerator of the right side has the square of the speed, but not just the speed. Such unit checks cannot prove that the formula is correct (the formal proof is below, in blue), but they can help you quickly spot incorrect formulas. Thus, if the units of each term of an equation match, the formula is not necessarily correct (some unitless factors might be missing), but if the units of each term are not the same then you can be sure the formula is incorrect. OK, that's good, but it would be better to have an intuitive feel for why the square of the speed appears in the numerator. Think of it in this way: For a car trip around a circular track at a constant speed, the "job" of the acceleration vector is to change the direction of the velocity vector. If you double the speed, then the change in velocity over a certain time interval is twice as great, because both the initial and final velocity vectors are twice as long as before. However, the same distance is traversed in half the time. Because the acceleration is the change in velocity divided by the time interval, by doubling the speed going around the circle you pick up a factor of 2 because the change in velocity has doubled, and also a factor of 2 because the time interval is half as much. Thus, doubling the speed around the circle requires a centripetal acceleration that is 4 times as great. You can readily understand why the radius of the circle appears in the denominator by thinking about going around a circular track that is very large. Because the curve of such a large circle is so gentle, the rate at which the velocity vector turns is very small, and so the centripetal acceleration is also very small. The larger the circle, the smaller the centripetal acceleration for a given speed. The discussion in the last few paragraphs is based on the assumption that the motion around a circle is at a constant speed; if the speed is not constant, then the situation is more complicated. We'll get into some of the complications in a later chapter, and we'll save other complications for second-year physics. Example: A flywheel with radius 30 cm rotates at a constant rate of 200 Ch5 Page 11
rpm. a. Determine the period of rotation. b. Determine the speed of a point on the rim of the flywheel. c. Determine the acceleration of a point on the rim of the flywheel. We can also simply say that the period is T = 0.3 s, because the "per revolution" is understood as part of the definition of the period. (b) The distance d covered by a point on the rim of the wheel in one revolution is Because the speed is constant, the speed of a point on the rim of the wheel is Ch5 Page 12
For calculus lovers only, here is a precise derivation of the formula for centripetal acceleration. For an object moving in a circle of radius r at a constant speed, the position function can be written as Now differentiate twice to obtain an expression for the acceleration, using the chain rule. Remember that theta is a function of time and the radius of the circle is constant. You'll also note that we've used a fact from vector calculus, that to differentiate a vector function you just differentiate each component separately. You might like to think about why this is true. (Further discussion to come in second-year calculus.) Note that the acceleration is in the direction opposite to the position vector. The position vector points from the centre of the circle to the location of the moving object; thus, the acceleration vector points towards the centre of the circle, which is why it's called a "centripetal" (centre-seeking) acceleration. The magnitude of the centripetal acceleration is Note that Ch5 Page 13
Thus Example: maximum speed for a car turning around a curve on a level road with friction A car of mass 1200 kg moves around a curve on level ground that has a a radius of 20 m. Determine the maximum speed for which the car can safely move around the curve. The coefficient of friction is 0.5. Ch5 Page 14
Thus, the maximum acceleration that friction between the tires and the road can produce is 4.9 m/s 2. Notice that the maximum safe speed through the curve is independent of the mass of the vehicle; thus, a speed limit sign can be used that is appropriate for all vehicles, whether they are light motorcycles or heavy transport trucks. Ch5 Page 15
Also note that static friction between the tires and the road is the origin of the force pushing the vehicle towards the centre of the circle (why is it not kinetic friction?). Because the static friction force points towards the centre of the circle, most textbooks call it a centripetal force. Note, however, that this is not a new kind of force, but is simply a reminder that the force acts towards the centre of the circle. In this example, it's a static friction force, nothing new. Example: banking angle for a highway curve Determine the ideal banking angle for a highway curve that has a (horizontal) radius of 20 m. Suppose that the typical driving speed around the curve is about equal to the speed determined in the previous example. Solution: There is no friction on the road; presumably it's very icy. Ch5 Page 16
Notice that the ideal banking angle is independent of the mass of the vehicle; this is nice. It means that one can design a banked highway that will be appropriate for all vehicles, no matter their mass, so it will be just as safe for light motorcycles and heavy transport trucks (at the specified speed). Example: apparent weight for motion in a vertical circle Consider a car going around a vertical "loop-the-loop" of radius 3 m. Determine the minimum speed the car needs to make it through the loop. Ch5 Page 17
Alternatively: Consider a bucket of water spinning in a vertical circle and determine the minimum speed (or angular speed) so that the water does not fall out of the bucket. Solution: Draw a free-body diagram for the car when it is at the top of the loop: This means that if the car is to complete the circle, the force must be provided by the normal force from the loop and gravity. As the speed increases, the normal force has to increase to provide the necessary force. On the other hand, if the speed of the car decreases, then the normal force will also decrease, until at a critical speed, the weight of the car will be sufficient to provide the centripetal force. If the speed were to decrease below this critical minimum value, the car will leave the loop and crash down. Thus, the minimum speed for the car to make it through the loop corresponds to n = 0. Setting n = 0 and solving for the speed of the car, we obtain: Ch5 Page 18
This may not seem like a very high speed, but remember that the loop is not very big. I once saw a "cirque" stunt where motorcycles were flying around the inside of a spherical metal structure, and the radius might have been this big and the speeds seemed about this fast or a bit faster. For a much bigger loop, a larger speed is required. Now solve the problem of the water in the bucket yourself, using a string of reasonable length. How fast do you have to swing a bucket around so that the water doesn't fall out? centrifuges Read about centrifuges in the text book; they provide a nice practical example of circular motion. (Also, you'll think about physics the next time you use a lettuce spinner, which is a sort of centrifuge.) Newton's law of gravity Ch5 Page 19
Example: gravitational force between Earth and Moon The following diagram is meant to help us understand the "inverse-square" nature of the gravitational force law. The same inverse-square nature is present in the decrease in the intensity of light or sound from a point source, which is the inspiration for the diagram. Imagine "spraying" a substance, such as paint from a paint-sprayer. If the spray is uniform, the area covered by the spray increases by a factor of 4 if the distance from the sprayer doubles; similarly, the area covered by the spray increases by a factor of 9 if the distance from the sprayer triples. In general, the area covered increases by a factor of r 2 when the distance increases by a factor of r. But, if the area covered doubles, then the "intensity" of the spray decreases by a factor of 2, because the amount of spray doesn't change, it's just spread over a larger area. Ch5 Page 20
The gravitational force behaves similarly; as you get farther away from the source by a factor of r, the gravitational force decreases by a factor of r 2, because it is spread over an area that has increased by a factor of r 2. Newton's law of gravity is an inverse-square force law, and has the same structure as Coulomb's law for the force between two charged particles at rest, as we'll learn in second semester. The diagram above is intended to illustrate that the force decreases by a factor of 4 when the distance between the objects increases by a factor of 2. Surface gravity of a planet Example: gravitational force between Earth and a small object of mass m at the Earth's surface Ch5 Page 21
This provides insight into our assumption earlier in the course that the acceleration due to gravity g is constant; we can see by the previous equation that this is not exactly true. Close to the Earth's surface it is approximately true, but as you move away from the Earth's surface the value of the acceleration due to gravity decreases. The equation above also gives us a way to "weigh" the Earth. The acceleration due to gravity can be measured in a laboratory (in fact you did so in the pendulum experiment in this course), and so can the gravitational constant G (look up the famous Cavendish experiment for details). The radius of the Earth can be determined using an ingenious geometrical method first devised by Eratosthenes (you can also look this up); then the previous equation can be solved for the mass of the Earth. The same formula can be used to determine the acceleration due to gravity on other planets, moons, asteroids, etc. Just replace the mass and radius of Earth by the mass and radius of the other planet. Also note that some books call the acceleration due to gravity at the surface (i.e., "g") by the term "surface gravity." Example: Determine the Moon's surface gravity. Solution: Look up the following data in the textbook: Then the surface gravity of the Moon is Ch5 Page 22
Thus, the surface gravity of the Moon is about one-sixth the surface gravity of the Earth. What are the consequences of this? What would it be like to walk about on the Moon? Example: Determine the acceleration due to gravity (a) 1 km above the Earth's surface and (b) 100 km above the Earth's surface. Compare each value to the value of the acceleration due to gravity at the Earth's surface. Solution: Look up the following data in the textbook: Using these data, the acceleration due to gravity at the Earth's surface is Ch5 Page 23
(a) The acceleration due to gravity 1 km above the Earth's surface is (b) The acceleration due to gravity 100 km above the Earth's surface is At a height of 1 km above the Earth's surface, the value of the acceleration due to gravity has decreased by 0.031%, and so it is quite reasonable for ordinary calculations to assume that the acceleration due to gravity is constant over this range of distances, as we did in Chapters 2 and 3. At a height of 100 km above the Earth's surface, the value of the acceleration due to gravity has decreased by 3.1%, and so it may not be a reasonable assumption to treat it as constant over this range of distances. These calculations also emphasize the point that gravity does not suddenly disappear when you are out in space; spacecraft in orbit around the Earth are in free fall, and so the apparent weight of objects and people in orbit is zero. However, their real weight is not zero, as the Earth's gravity is still quite strong at altitudes at which satellites orbit. Ch5 Page 24
Satellite motion orbital motion of a satellite around Earth direction of gravitational forces at various points of the orbit gravitational acceleration is approximately constant near surface, but the direction is clearly not constant over larger scales, nor is the magnitude constant over larger scales "Weightlessness" in space satellites in orbit are in free fall hence occupants are "weightless;" that is their apparent weight is zero, even though their real weight is not check the textbook for details (It's really apparent weightlessness; the objects still have weight, because there is still a gravitational force acting on them, but because they are in free fall they feel weightless.) Kepler's third law of planetary motion Using Newton's law of gravity and Newton's second law of motion, we can derive Kepler's third law of planetary motion. Ch5 Page 25
If the orbit of the planet is elliptical instead of circular, a more complex analysis shows that Kepler's third law is still valid provided that we use the "semi-major axis" of the orbit in place of r. The semi-major axis of the elliptical orbit is the distance from the centre of the ellipse to the most distant point on the orbit. Example: Use Kepler's third law of planetary motion to determine the distance between the Earth and Sun, given that the mass of the Sun is about 2 10 30 kg. Solution: Make sure to convert the period of the Earth into seconds: Ch5 Page 26
Dark matter: One of today's unsolved puzzles about the universe As we discussed in class, if you are deep below the surface of the Earth, let's say in a very deep mine shaft, your weight is less than at the surface of the Earth. Only the mass "interior" to you (i.e., at radii smaller than yours) is effective in exerting a force on you; the force exerted on you by the mass of the Earth that is at larger radii cancels. This means that if you were anywhere inside a hollow spherical shell of mass, provided the shell has constant density, the gravitational force on you is zero. (If you wish to learn more about this, look up "Gauss's law" for gravity (there is a version of Gauss's law for electrostatic forces as well); to understand the mathematical argument, you'll need to have some integral calculus under your belt.) The same principle can be applied to the motions of stars in our galaxy. If you analyze the motion of stars at various positions in our galaxy, you can deduce the amount of mass in the galaxy that lies closer to the galactic centre than Ch5 Page 27
the given star (using Newton's law of gravity and Newton's laws of motion). Repeating this kind of analysis for many stars gives us a good idea for the distribution of mass in the galaxy. And this leads to a puzzle: The amount of mass that we detect by usual means (regular light telescopes, radio telescopes, etc.) is nowhere near enough to account for the mass that we know must be there by analyzing motions in the galaxy. That is, the "visual matter" does not account for all the matter that must be present; there must be some "dark matter." What on Earth can this dark matter be? Nobody knows. It is highly unlikely that it could be simply ordinary matter that can't be detected (such as "cold" dust particles or gas, abandoned TV sets, etc.), so scientists have turned to more speculative possibilities. Maybe dark matter is some exotic new form of matter. If so, such forms of matter have not been detected in laboratories, which leaves us no closer to resolving the puzzle. This is an example of the type of unresolved puzzle that is found at the frontier of every branch of science. There are always unsolved puzzles, which means there is always room for new ideas, and creative researchers have plenty of opportunities for making interesting new discoveries. Maybe one of you will devote the time and work necessary to reach one of the research frontiers; it will take a lot of time and work to reach the frontier, but for the right kind of person (i.e., one who is persistent and willing to put up with a certain amount of failure and frustration) the journey will be a lot of fun and very satisfying. Geostationary satellite orbits It's convenient to have communications satellites that orbit Earth above its equator with a period equal to Earth's rotational period; in this way, they "hover" over the same geographical point on Earth. Using Kepler's third law we can calculate the radius of the orbit of such "geostationary" satellites. Ch5 Page 28
This is the distance from the centre of the Earth, so the distance of such a satellite from the surface of the Earth is 6400 km less, which is 35,850 km above the Earth's surface. The International Space Station orbits Earth at an altitude of about 400 km, which is considered "low Earth orbit;" geosynchronous satellites are in "high Earth orbit." Additional exercises: Example: A turntable rotates counterclockwise at 78 rpm. A speck of dust on the turntable is at = 0.45 rad at t = 0 s. Determine the angle of the speck at t = 8.0 s. (The result should be between 0 and 2.) Solution: Ch5 Page 29
Example: A satellite orbiting the Moon very near the surface has a period of 110 min. Use this information, together with the radius of the Moon (which is 1.74 10 6 m), and the mass of the Moon (which is 7.36 10 22 kg), to calculate the free-fall acceleration on the Moon's surface. Solution: Ch5 Page 30
This means that the surface gravity (which is another word for the free-fall acceleration at the surface) on the Moon is about 1/6 as much as the surface gravity on the Earth. How would this change life for you if you lived on the Moon for a while? Example: A 500 g ball swings in a vertical circle at the end of a 1.5-mlong string. When the ball is at the bottom of the circle, the tension in the string is 15 N. Determine the speed of the ball at this point. Solution: Ch5 Page 31
An interesting question was asked in class: How does the Earth's gravity vary as you go inside the Earth? You might imagine going down a very deep mine shaft and wonder about the magnitude of the acceleration due to gravity at the bottom of the mine shaft. Measurements have been made, and they are in accord with the following reasoning, which requires calculus to understand in full mathematical detail. The first point to understand is that the gravitational force inside a hollow spherical shell with constant density is zero. Calculus lovers can go here for a proof: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html or here: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/gravfield.htm Non-calculus people can carefully examine the following diagram, taken from the previous link. Consider an imaginary double-cone drawn from the point P and intersecting the shell in the two areas indicated. By symmetry, the force on a small mass at P due to the mass at each red area is directed along the red arrows. The mass at A 2 is closer to P than the mass at A 1, but the amount of mass at A 2 is less than the amount of mass at A 1. Both influences cancel exactly, and so the force on a small mass at P from each of the masses at the red areas is equal and opposite, and so the net force on the small mass at P due to the masses at the red areas is zero. (In a little more detail, the area is proportional to the square of the distance from P, but the gravitational force at P due to the mass at the area is inversely proportional to the square of the distance to P.) The same argument applies no matter which double-cone we draw, and so the net force at P due to the entire mass of the spherical shell is zero. Conclusion: The gravitational force anywhere inside a spherical shell of constant density is zero. Ch5 Page 32
If you are at the bottom of a deep mineshaft, then, the gravitational force on you due to the mass of the Earth that lies at radii greater than your radius is zero. This is because the part of the Earth farther from the centre than you can be considered to be a collection of thin spherical shells, and you lie inside all of these shells. We've concluded that at any position inside a spherical shell of constant density the gravitational force is zero, so the force on you due to the mass of the Earth that lies at greater radii than you is zero. The net gravitational force on you can be calculated by considering only the mass that is interior to you (shaded in red in the following figure). Let's suppose that you are at a distance r from the centre of the Earth. How much of the Earth's total mass lies closer to the centre than you (shaded in red)? Assuming that the Earth's density is constant, the mass M that is closer to the Earth's centre than you is Ch5 Page 33
If your mass is m, your weight at the bottom of the mine shaft is mg, which is equal to the gravitational force exerted on you by the mass M of the part of the Earth interior to your position: Thus, the acceleration due to gravity at the bottom of the mine shaft is Inserting the expression for M derived above, we obtain a convenient expression for the acceleration due to gravity at the bottom of the mine shaft: Ch5 Page 34
This is interesting: The acceleration due to gravity in a mine shaft inside the Earth varies linearly with distance from the centre of the Earth. Using the expression for the acceleration due to gravity at the surface of the Earth, we can write the acceleration due to gravity in the mine shaft as Thus, if you're half way down to the centre of the Earth, the acceleration due to gravity would be half of the surface value. Similarly, if you've only gone down 1 percent of the Earth's radius, the acceleration due to gravity will be 1 percent less than the surface value. Of course, for any realistic mine shaft, the actual change in g is very small indeed. For example, if the mine shaft is 2 km deep, then the value of g at the bottom of the mine shaft will be 6398/6400 times the surface value, which amounts to a little more than 3/100 of a percent difference. We learned earlier in the chapter that the acceleration due to gravity decreases as 1/r 2 as you go away from the surface of the Earth, so a graph of the variation of g both inside and outside the Earth looks something like this: Ch5 Page 35
I haven't drawn the graph very well, but you can see the linear portion inside the Earth, and then the portion outside the Earth is supposed to decrease asymptotically towards zero, never reaching zero. Ch5 Page 36