Angular Momentum. Brown University Physics 0030 Physics Department Lab 4

Transcription

1 Angular Momentum Introduction In this experiment, we use a specially designed air table on which we cause the collisions of a ball and a disk, and so observe the consequence of angular momentum conservation. Notice as you read the description of the collision that it is inherently inelastic, so that the kinetic energy will not be conserved through the collision. The kinetic energy does serve, however, to define the initial state, in particular the velocity, and hence the initial angular momentum. Since no external torques act on the system, the angular momentum is conserved, in analogy with linear momentum conservation when no external forces act on the system. To elaborate on the analogy with linear momentum in one dimension, we can define for rotational motion the quantities L, the angular momentum, and K, the rotational kinetic energy. For the rotation of a rigid body about a fixed axis, the vector nature of angular momentum is expressed by simply choosing clockwise rotations as positive and counter-clockwise rotations as negative (or vice versa, if one prefers). Notice the correspondence: Translation Rotation Velocity v ω Momentum p = mv L = Iω Kinetic energy T = mv! /2 K = Iω! /2 Here I is the moment of inertia, m the mass, v the linear velocity and ω the angular velocity. The angular velocity is measured in radians/sec. The rolling condition, when a body (e.g. a disk or ball) of radius r rolls or rotates without slipping, is that v = rω. Basis of the experiment A steel ball rolls from rest down a ramp to an air table, acquiring both rotational and translational kinetic energy. Because of the curvature of the ramp at the bottom, the ball leaves it horizontally, with a velocity that can be measured. (It can also be calculated, but only approximately, from the change in potential energy.) 1

2 Figure 1: Equipment setup. The ball then makes an inelastic collision with a steel disk free to rotate on the air table (it collides with a low-mass aluminum catcher attached to the disk). The ball is locked to the catcher at some known distance from the axis of rotation of the disk. Disk and ball now rotate together, with an angular velocity that can be measured with reasonable precision. The experiment requires careful alignment (it e.g., the ball must enter the catcher perpendicular to the disk radius) to maintain the simplicity of the data analysis. The experiment will be carried out for four values of initial angular momentum L!. The apparatus is shown in Figure 1. You will find a description of the theory in Appendix I. Here you will also find relevant equations that will be referenced throughout the procedure and analysis sections later in the manual. Procedure 1. Begin by measuring the steel disk radius and the mass of the steel ball. 2. Work out the moments of inertia. Formulas for moments of inertia for bodies of various standard shapes given in the reference (Young, Chapter 9). Do not take apart the apparatus the masses of the catcher and disk have been carefully measured and are recorded on the air table. 3. Level the air table carefully. Notice that, with the triangular arrangement of the screws, the two parallel to one side will control leveling in a direction parallel to that side. Then the single remaining screw will adjust the leveling in a direction perpendicular to that side. 2

3 NOTE THE FOLLOWING IN USING DISKS ON THE AIR TABLE: DO NOT ROTATE THE DISKS AGAINST EACH OTHER OR ROTATE THEM ON THE TABLE WITHOUT APPLYING THE COMPRESSED AIR THAT SEPARATES THE COMPONENTS. SERIOUS WEAR AND ABRASION CAN OCCUR UNLESS THE SUPPORTING AIR LAYERS ARE IN MAINTAINED. Because of a height limitation in the apparatus, it will be necessary to use two disks. The bottom (steel) disk will be held tightly by tape and serves only as a spacer. The top (steel) disk is free to rotate. 4. Turn on the air supply and test that the parts turn freely. The filter-regulator of the air supply (separate from the table) should be set for 9 Psi (pounds per square inch) for this experiment; have the instructor adjust this setting if necessary. Fix the lower disk against rotation. 5. Now test the operation of launching a steel ball down the ramp and over to the catcher. There are several variables to consider here. a. The ramp and catcher should be positioned so that the ball arrives perpendicular to the catcher. If the ramp is too far from the catcher, the ball may drop slightly after leaving the ramp, and so not enter the catcher reliably. b. Most such problems, when they occur, result from poor alignment and tightening of the mechanical parts, or an incorrect air flow. Your lab instructor will help you here. In an extreme case, it may be necessary to adjust the height or angle of the ramp by strips of tape under its base. Trial and error, and practice, will achieve the desired results! 6. The apparatus is working correctly when you can launch a ball and catch it on the disk from two positions on the ramp: the full height of the ramp and half this height. You will need to devise a system (pencil mark, tape) that makes these launch points repeatable. 7. Now position the ramp at the edge of the rotation table to make the range measurements that give the launch velocities (see Figure 2 and the procedure leading up to equation (3) in Appendix I). a. Note that there are two velocities, one corresponding to the full ramp height position, and the other to the half-height starting position. 8. To find the length that ball shoots horizontally, we will first need to shoot the ball, to find the rough area of where it will land. Then we place the carbon and white paper sandwich in this area. This paper will mark the landing points for each of the two velocities. a. Sufficient measurements (at least four) should be taken for each mean velocity determination, to give you an estimate of the precision with which the mean has have been measured (the standard deviation of the mean). 9. Return the ramp to the position for launching into the catcher. At least two launches should be made from each of the two starting positions on the ramp. For each starting position, each collision into the catcher should be made at a different radius r!, so the initial angular momentum (L! = m! v! r! ) varies. 3

4 a. Note: when measuring r! you should not just eyeball where the center of the ball is, as it has been caught by the catcher. Instead, to get a more precise measurement, you should measure the distance that each edge of the ball is from disc axis (this is easier to align with the measuring ruler on the catcher). When you have these two measurements, you can find r! by taking the average of the two measurements, as this will be the middle between the two ends of the ball. 10. The final angular momentum L! = I! + I! + I! ω! follows by recording the counter reading, from which the angular velocity ω! is deduced (equation (4) in Appendix I), and multiplying ω! by the sum of moments of inertia of disk, catcher and ball, all rotating about the disk axis. Data The data to be taken as described above is summarized below, along with the quantities to be determined from the measurements. All data and quick check calculations should be done directly in the notebook. You will have four different values of L! and L!, corresponding to two values of r! for each of two values of h. For each of these values of L!, the average of four measurements of v! is used. For each value of L!, the average of five values of N (automatically updated every second) is used to determine ω!. 1. Determine Moments of Inertia I!, I! and I! from measurements of the masses of ball, catcher and disk, r!, and the dimensions of catcher and disk. In this setup, I! is the moment of inertia of the ball at r! rotating about the disk axis, not around its own radius or diameter. Therefore I! depends on r!. 2. Velocity measurements a. Starting the ball from a marked position about halfway up the ramp, make at least four independent measurements of launch velocity v!, determined from l and h measurements by using equation (3) in Appendix I. b. Calculate the mean v! ± Δv! for this launch height. Be sure the paper on which the distance is recorded remains in place during the launch. c. Starting the ball from a second marked position near the top of the ramp, repeat the measurements of part a. F 3. Collision measurements a. Position the ramp so the ball is caught about halfway out of the length of the catcher. Measure this distance r!. Roll the ball from each of the two marked positions on the ramp. Each time the ball is caught record at least 5 successive readings of N as the rotation proceeds, as there can be fluctuations. Calculate N ± ΔN. b. Position the ramp so the ball is caught out near the end of the catcher. Measure this distance r!. Repeat the steps rolling the ball as in part (a). 4

5 Calculations 1. Calculate the initial angular momentum of the system, L! = mv! r! for each combination of v! and r! (at least 4 combinations). 2. For each value of L! include its uncertainty (derived from the uncertainty for v! ). [Reminder if A = constant x then ΔA = constant Δx ] 3. For each of the combinations of v! and r! (at least 4), calculate the final angular momentum of the system, L! = I! + I! + I! ω!. To calculate ω!, use equation (4) in Appendix I with the average of your values of N for that case. For each value of L! include its uncertainty (derived from the uncertainty determined for N). 4. For each case, calculate L! L! /L!. Results and Discussion Compare the initial and final angular momenta in each case. Discuss the results especially with reference to the spread in repeated measurements of v! and the repeated readings of N, assuming the other quantities are constant. Is this assumption valid? Consider the uncertainties in m and r!, and in I; are they relatively small compared to the uncertainties you determined for v! and N? Have you demonstrated the conservation of angular momentum within the limits of your experimental uncertainties? What systematic effects might have caused any differences? How might they have been reduced? You made at least four tests of angular momentum conservation, with two different ball velocities and two different radii of the ball about the axis of rotation. Rank the tests in the order in which you expect the results to be most accurate, and explain the reasons for this ranking. Compare the expected rank with your actual results. References Kestin and Tauc, University Physics, Volume 1, Chapter 8. 5

6 Appendix I: Theory of the collision If the ball of mass m leaves the ramp with a velocity v!, directed perpendicular to a disk radius, and at a distance r! from the disk axis, its angular momentum will be L! = mv! r! (with respect to the disk axis) before collision. With the disk at rest before the collision, this is the total initial angular momentum of the ball-disk-catcher system. Notice that the angular momentum is a well-defined constant quantity at the time of the collision even though no part of the ball-disk system is in rotation before the collision. If we neglect the very slight influence of gravity on the ball as it moves from the ramp to the catcher, the kinetic energy of the ball remains constant until it strikes the catcher. But this quantity changes considerably as the ball collides and lodges in the catcher. When it lodges in the catcher, the entire system of disk, ball and catcher will rotate with an angular velocity ω! that depends on the moments of inertia of the ball, the catcher and the disk itself, with respect to the axis of rotation. Since angular momentum in any system is conserved in the absence of external torques, the system's final angular momentum is equal to its initial angular momentum. L! = L! (1) mv! r! = I! + I! + I! ω! (2) Here I! is the moment of inertia of the ball treated as a point mass rotating about the disk axis at a distance, r!, I! is that of the catcher (treat as a rigid rod fixed at one end) and I! is that of the disk. (Refer to Chapter 9 of the Reference for details of these moments of inertia.) Velocity measurements Two velocities are central to the ball-disk collision. The first is the velocity v! of the ball as it exits from the ramp and moves horizontally toward the disk. The second is ω!, the angular velocity of the disk (with ball and catcher attached) after the ball collides with it. Initial ball velocity If the ball of mass m is released from rest at a height h above the ramp exit it will roll and slide down, acquiring kinetic energy while losing energy to friction (see Figure 2). If it slid down without friction, all the kinetic energy would be translational and you would expect its velocity to be calculable by equating the kinetic energy to the original potential energy 1 2 mv!! = mgh (3) and finding v! accordingly. If, at the other extreme, the ball never slid, but always rolled down the ramp, then part of the potential energy mgh would be expended in rotating the ball faster and faster down the ramp. At the ramp exit, the translational kinetic energy of the ball would be less than that predicted from equation (3) when the ball started from height h. 6

7 Figure 2: Ball falling from ramp exit. In fact, we have no plausible reason to believe that the ball either simply slides or simply rolls down the ramp it probably does both. What we can do is measure the exit speed of the ball from the ramp when it is started from rest at some height h. By doing this several times for a fixed value of h, we obtain a good measure of the velocity, the average velocity bar v! for release height h, and can determine the uncertainty in the velocity from the standard deviation of the measurements. To make the individual measurements of v!, the ramp is oriented on the rotation table so the horizontal exiting ball is directed out from the table to fall to the floor. A sheet of white paper covered with carbon paper records the point of impact (be sure the white paper's position stays fixed). Now the vertical length h from the ramp to the floor and the horizontal distance l from the ramp to the mark made on the paper will yield the velocity v!, as follows: 1. At the point at which the ball leaves the ramp, the horizontal component of its translational velocity is v! and the vertical component of the velocity is zero. The time taken to reach the floor at vertical distance y below the ramp exit under the force of gravity (acceleration g) is obtained from the equation h =!! gt!. The force of gravity has no effect on v!. 2. The ball falls under the influence of gravity for a time t = 2h/g. In this same time, it has moved horizontally a distance l = v! t, so the initial velocity is found as v! = l t = l 2h/g, (3) from measurements of l and h. You may want to refer to Chapter 3, Section 4 of the reference, on projectile motion. 7

8 As mentioned above, repeated measurements are expected to show a spread in values as well as providing a group for averaging. Remember that in addition to the usual uncertainties, there is a good probability that the ball will slide and roll by different amounts from trial to trial, and so a spread in trial values can be expected, beyond the uncertainties in the measurement of l and h. The effect of all these variations will be reflected in the standard deviation of the mean velocity, obtained in the usual way. Angular velocity of the disk This angular velocity is simple to obtain on the rotation tables, because the disks used contain alternate black and white bars completely around the circumference. The table includes an optical detector that produces a pulse each time a black bar passes it. The pulses are counted for a fixed amount of time (one second in our case) and then displayed digitally. Thus, the counter reads the number of black bars per second passing the detector, which can be used to derive the angular velocity of the disk, as discussed below. The digital display is updated each second, when a new count of black bars passing is completed. In the conditions of the current experiment, we would expect only a small change (decrease) in the bar count caused by friction, since no other forces act after collision. To obtain an angular velocity from the bar frequency" reading shown on the counter, we use the fact that the black and white bars are each one millimeter wide. Therefore, one count on the digital display corresponds to two bars (one black and one white) or two mm/sec of disk circumference sweeping past the optical reader. More generally, a bar frequency of N showing on the counter corresponds to the disk circumference sweeping past the detector at a linear speed s, equal to s = 2 mm N 1/sec. But this linear speed of the circumference is simply s = Rω, where R is the radius of the disk and ω is the angular velocity. Therefore ω! = s R = (2 mm)n R radians/sec, (4) where N is the counter reading and R the disk radius. Note that the radius should be measured in millimeters to use the relation in this form (because the 2 in the numerator has millimeter units). 8