LECTURE 25. Rotational Dynamics and Angular Momentum

Transcription

1 LECTURE 5 Rotational Dynamics and Angular Momentum Introduction To introduce the subject of this lecture, consider this spinning wheel suspended by a rope tied to one end of its axle But rather, when the wheel is spinning the axle turns about the suspension point so that the plane of rotation of the wheel remains vertical. Suspension cord Rotating Bicycle wheel Suspension cord Weight of bicycle wheeel The curious thing about this wheel is that it doesn t fall so that its axis becomes vertical as, it would if it weren t spinning. This despite the fact that the weight of the wheel, which acts at the center of the wheel, gives a torque which should do just that. Suspension cord Weight of bicycle wheeel Weight of bicycle wheeel Some of you may recognize this to be an example of the gyroscope, a central piece of equipment for navigating in space and often a toy sold in science shops. But did you actually ever figure out how it works? Why it does what it does? In this lecture I hope to lead to some understanding of that and to show the importance of this understanding in biomechanics. The Equations of Circular Motion The basic kinematic equations of circular motion were introduced earlier in the course. These are the angular displacement expressed in radians and the angular velocity expressed in radians per second. These have their direct analogs in linear displacement, s, and linear velocity, v.

2 Physics 0A - Physics for the life sciences y ω r x Thus the equations relating linear displacement, velocity and accelerations have their direct analogs in angular motion. Continuing from a study of kinematics to mechanics, the concept of a force in linear mechanics has its analog, torque, in rotational mechanics. Linear s Angular Displacement Displacement Linear Velocity v Angular Velocity ω If the velocities are constant s= vt = ωt Although I have no yet emphasized it in this course, angular acceleration α is defined as the rate of change of angular velocity, just as linear acceleration is defined as the rate of change of linear velocity. Linear Displacement s Angular Displacement Linear Velocity v Angular Velocity ω Linear a Angular α Acceleration Acceleration Force F Torque Τ That torque is the direct analog of force is shown by the work involved in pushing an object in a circle. F r Distance travelled = r α r The distance the force moves the object is r. The work the force does is then F r. Rearranging this gives W = Fr = T Linear s Angular Displacement Displacement Linear Velocity v Angular Velocity ω Linear Acceleration a Angular Acceleration If the accelerations are constant s= vot+ at = ωot+ αt α Linear s Angular Displacement Displacement Linear Velocity v Angular Velocity ω Linear a Angular α Acceleration Acceleration Force F Torque Τ Work sf Work Τ Power vf Power ωt

3 Lecture 5 - Rotational Dynamics and Angular Momentum 3 Continuing on to dynamics, the ubiquitous F = ma has a direct analog in rotational dynamics. This is shown by again considering the case of an object being forced to move in a circle. α F r a a form analogous to F = ma by representing the product mr by the symbol I. T = Iα This symbol can also be used to simplify the equation for the kinetic energy of rotational motion. mv m r mr ω Kinetic Energy = = ( ω) = Kinetic Energy = Iω So our table of analogies has become even more extensive: As an example, suppose the object of mass 6 kg was moved in a circle of radius 0. m by a force of 600N. The tangential acceleration of the object, that is the acceleration along the circumference of the circle, is calculated from the equation F = mat at = 600 / 6 = 00m/s This tangential acceleration is related to the angular acceleration is the same way that the linear displacement and velocity are related to the angular displacement and velocity: s= r v = rω at = rα This results in F = mrα Multiplying both sides of this equation by the radius r results in Fr = mr α The left hand side of this equation is now the torque. The equation can therefore be made into Linear Displacement s Angular Displacement Linear v Angular ω Velocity Velocity Linear a Angular α Acceleration Acceleration Force F Torque Τ Work sf Work Τ Power vf Power ωt Dynamics F=ma Dynamics Τ=Ια Kinetic Energy mv Kinetic Energy Iω The product mr is so important in dynamics that it is given its own name; The moment of inertia. Again the word moment here means importance. It implies that the: moment of inertia is the importance of the inertia. Essentially, the importance of the mass of an object is proportional to the square of its distance from the center of rotation. The only item left in linear dynamics for which there should be an analog in rotational dynamics is then momentum. Is there a rotational concept analogous to linear momentum? Indeed there is, and for reasons I will try to make clear later, it is the most important concept there is regarding rotational motion. Following the sequence above it would seem that if linear momentum is the product mv, then angular momentum should be Iω. But is this reasonable? The momentum of an object is changed by a force action on it. In fact, the rate of change of

4 Physics 0A - Physics for the life sciences 4 the momentum of an object is equal to the unbalanced force on the object. F = dp dt For the analogy with rotational motion to be complete then the rate of change of angular momentum would have to be equal to the unbalanced torque on the object. In equation form: Rate of change of angular momentum = Rate of change of Iω If I is a fixed quantity, such as when a mass m moves at a constant radius from a center of rotation, then the only thing that can change is ω. Since the rate of change of ω is just α, then Rate of change of angular momentum = Iα This is, of course, just the torque on the object and the table of equivalences is complete. Linear Displacement s Angular Displacement Linear v Angular ω Velocity Velocity Linear a Angular α Acceleration Acceleration Force F Torque Τ Work sf Τ Power vf ωt Dynamics F=ma Τ=Ια Kinetic Energy mv Iω Linear mv Angular Iω momentum momentum Angular momentum is such an important quantity that by International agreement it is designated as L. An Application of the Equations of Circular Motion At first it might seem that the equations of angular motion are just another way of expressing motion that could be just as easily expressed, perhaps even more so, by the more easily understood linear equations. To show how they can be used to actually simplify the analysis of a motion that can indeed be described by linear equations consider again the pendulum. You may recall, with a feeling that is certainly not nostalgia, how the analysis of this device was carried out in the linear equations. (If you like self-flagellation you might go back to the notes for Lecture.) Now proceed with the analysis using rotational dynamics. l Wsin m W = mg The torque of the weight of the object is the component of the weight that is perpendicular to the string times the string length: T = mglsin Here the sign is negative because the torque is clockwise. In the small angle approximation the sine is equal to the angle in radians and so T = mglsin But the torque is just the moment of inertia multiplied by the angular acceleration, and the moment of inertia is just the mass times the length of the pendulum squared T = Iα = ml α so that mgl = ml α Flipping the equation left to right, doing a little manipulation and noting that α is the second time derivative of gives

5 Lecture 5 - Rotational Dynamics and Angular Momentum 5 d g = dt l Since the equation for simple harmonic motion is d = ω dt the equation for the angular frequency of the pendulum is g ω = ; ω = l This is considerably more straightforward than the treatment by linear dynamics. Of course, one of the reasons this is so is that the motion of the mass on the string is in a circle. Power and the Equations of Circular Motion As an example of the use of the rotational equations of motion in power calculations, consider a set of specifications for a North American automobile engine; peak torque of 0 ft-lbs at 3600 rpm and 00 HP at 5000 rpm. Why is the torque peak at 3600 while the peak Horsepower (it must the peak horsepower, otherwise why would they advertise it) at 5000 rpm? First it is necessary to wade into the thicket of units that gives engineering in North America a certain mystic. Converting the information into the units of the World (outside the US), a ft-lb (that is a foot-pound in real English) becomes 0.305m times kg or kg-m. This is converted to the scientific N-m by multiplying by 9.8 to get.36 N-m. Thus 0 ft-lb is 98.9 or, to the accuracy of these car specifications, 300 N m. The power of this torque at 3600 rpm is Tω. This is 300 π 3600/60 = 3 kwatt. Since HP is 746 Watt, then the power of the engine at 3600 rpm is 5 HP. How come the peak engine power was not at this peak torque? Because while the torque is less at higher engine speeds, the higher speed gives more horsepower even at this reduced torque. This can be checked by calculating the torque that comes with the peak HP at 5000 rpm. Here the calculation is the reverse of that we just did. 00 HP is = 49. kw rpm is 5000/60 π = 53.6 rad/s and so 49. kw is achieved with only 4900/53.6 = 85 N-m, or 85/.36 = 09 ft-lb of torque. g l There are many such examples where knowledge of the rotational equations of motion is necessary to understand the motion. However, the most important aspect of rotational motion that I want you to concentrate on here is that of angular momentum. Is it, like its linear counterpart, conserved? The Conservation of Angular Momentum As a demonstration that introduces the conservation of angular momentum, consider the case of me spinning in a chair with a very good bearing that allows me to spin for several minutes. I now take two 4-kg masses, one in each hand, and, holding them as far from my body as I can, I use my foot against the support structure to give myself a gentle spin, of about seconds per revolution. s per revolution

6 Physics 0A - Physics for the life sciences 6 I now pull the masses in as close as I can to my body. Less than a second per revolution How to Calculate Moments of Inertia As an introduction to the calculation of moments of inertia, consider a body made up of a number of masses, each small enough to be considered a point but distributed as in the diagram below. 0.5 kg m It is easy to see that my rotation has sped up. You can estimate it to be now less than a second per revolution. But how could this have happened? I applied no torque to my body, I just pulled the masses in toward the center. Any torque to my system (me, the chair and the masses) would have to be applied from outside, such as the torque I had to apply to get myself going at the beginning, which came from my foot being pressed against the support structure. The solution to this conundrum comes in considering the moment of inertia of the complete system before and after pulling in the weights. Pulling in the weights, and my arms brought more of the mass toward the center, thereby decreasing the overall moment of inertia of the system. Since angular momentum is the product of the moment of inertia times the angular velocity and the moment of inertia is reduced then for the angular momentum to be preserved the angular velocity has to increase: Lbefore = Lafter ( Iω) = ( Iω) before after I ω = I ω before before after after But do the numbers add up? To get an estimate of the actual moment of inertia of my system we have to consider how moments of inertia are calculated. CM kg.5 m 0.5 m m kg kg The total moment of inertia of the structure about its center of mass is the sum of the products of each mass multiplied by the square of its distance from the center. Going clockwise from the upper mass this is I total = = = 35. kg m Note that the 0.5 kg mass has 4 times the moment of inertia of the much larger kg mass because of its greater distance from the mass center. This shows the meaning of moment in moment of inertia. Masses that are far from the center are the ones whose inertia has the greatest moment, i.e. the greatest importance. This sort of calculation can be carried out with any structure of masses. If the structure is a shape that can be described by a mathematical function then the moment of inertia can be calculated using integral calculus (which essentially sums up to contributions of an infinite number of elements of the system). This is not a subject for this course, and is really only a subject

7 Lecture 5 - Rotational Dynamics and Angular Momentum 7 of importance in mechanical engineering, and even there modern computer techniques have made calculus of less and less importance even in that subject. However, whatever the origin it is useful for what follows to know a few of the simple formulas. First consider a ring: R RING R I=MR I= MR Mass = M All of the mass of a ring is at the same radius from its center. The contribution of each part of this mass to the overall moment of inertia of the ring is therefore the mass of that part times the square of the ring radius. The total moment of inertia of the hoop is therefore its total mass, M, multiplied by this square. I = MR ring ring Now consider a disk spinning about its center. DISK R Mass = M I= MR This disk can be considered to be made up of many rings, all very thin and all contributing to the total moment of inertia in proportion to the square of their radii. Since the inner rings have a smaller radius a ring of the same mass will contribute less to the moment of inertia. The actual contribution can be calculated using integral calculus but again that is not a subject for this course. What is important here is the formula that results: Idisk = MRdisk Of course this formula applies to a long rod rotating about its axis, even though it does not look like a disk. (It can be thought of as a whole stack of disks.) Mass = M Formulae for the moment of inertia for almost any object that has a regular geometry can be found in handbooks of Physics, Chemistry and Engineering. For example the moment of inertia of a sphere rotating about its center is /5 MR. For the discussion that follows, two particular formulae are of importance. First, the moment of inertia of a long thin rod rotating about its center. Mass = M Its moment of inertia is I = rod Ml l I= Ml

8 Physics 0A - Physics for the life sciences 8 And finally, the moment of inertia of a rod rotating about one of its ends: Mass = M Its moment of inertia is l I= Ml 3 I = rod Ml 3 Note that the moment of inertia of an object depends on the choice of axis about which it rotates. In general, the more of the mass that is distributed away from the center of rotation, the greater is the moment of inertia. In summary, the total moment of inertia of an object is the sum of the moments of inertia of its component parts. In most of the examples above the moment of inertia has been calculated for rotation about the center of mass. However, there is a very useful rule, that can be derived from integral calculus, that the moment of inertia for rotation of an object about any center is the moment of inertia for rotation about the center of mass plus the mass of the object multiplied by the square of the distance of the center of mass from the center of rotation, R CM Mass = M The Moment of Inertia of Humans You can use the rules just given for calculating moments of inertia to estimate the moment of inertia of human beings in various activities, such as figure skating. First consider the range of possible values for an average human. Consider a person of mass 75 kg and height.8 m. Modeling that person as basically a cylinder of diameter 0.5 m, the moment of inertia about a vertical axis through the center of mass gives a moment of inertia I = Mr = = 0. 6 kg-m Now consider that same person stretched out horizontal and rotating about a vertical axis through the center of mass. The moment of inertia is now I = Ml = = 0 kg-m These are the two extremes of the moment of inertia of the person rotating about the person s center of mass. They cover a range of about 30. In between will be the situation where the person is rolled into a ball. Using the equation for the moment of inertia for a sphere and noting that the radius of a 75 kg sphere of body mass would be about 0.5 m gives I = Mr = = kg-m 5 5 Now consider a typical figure skater. Again assume the skater weighs 75 kg and is.8 m tall. A simplified mathematical model of such a skater preparing for a spin is shown below..8 m Center of rotation I = ICM + MR Object You can check this formula by comparing the equations for the rotation of a rod about its center of mass and about one of its ends. For the moment of inertia about its end we merely have to add the mass of the rod times the square of half its length: Arms 0 kg 0.3 m Body 50 kg Upright leg 5 kg I = Ml M l + end Ml Ml = + 4 = 3 Mechanical model of figure skater (starting position for a spin)

9 Lecture 5 - Rotational Dynamics and Angular Momentum 9 The moment of inertia of the skater in this position is approximated by a vertical cylinder for the upright leg with the skate that the skater is spinning on, a horizontal cylinder representing the upper body and the other leg, and another transverse cylinder representing the outstretched arms. The radius of the upright leg cylinder will be about 0. m. The moments of inertia of the upright leg and the horizontal body are: I = Mr + Ml = = = 3. 6 kg-m It is seen that the moment of inertia of the upright leg is insignificant. To the above figure we must add the moment of inertia of the outstretched arms:.6 m Arms 0 kg moment of inertia calculated earlier for a person rotating about an axis along the body length, you can see that it is about 30 times greater. This implies that if the skater executes this maneuver his or her angular velocity must increase by 30 times if angular momentum is to be preserved. Starting at a slow turn of about 3 seconds per revolution will therefore result in an incredible 0 revolutions per second at the finish. Of course, this would imply no loss of angular momentum due to friction of the skate on the ice but a good figure skater can indeed achieve incredibly high angular velocities by simply changing their moment of inertia during a spin. (You might notice that due to the tremendous centrifugal forces in a typical spin the skater has great difficulty in bringing the elbows and the free skate inward toward the center of the body. The ratio of final angular velocity to initial angular velocity is therefore probably more like 0: Return now to my very tame rotation in a comfortable chair on a rotary bearing. A crude mathematical model of my system is shown below. Mechanical model of figure skater (view from above) This is the moment of inertia of the arms about the center of mass of the arms plus the mass of the arms times the square of the distance of the center of mass of the arms from the center of rotation. I = Ml + Mr = = = 30. kg - m The total moment of inertia of the skater is then about 7 kg-m. Comparing this to the Upper body 30 kg (R = 0. m) Rotating platform 5 kg Arms 0 kg 0. m 0.6 m 0. m Thighs 5 kg.0 m Lower legs 0 kg (R= 0.05 m) Mechanical model of me on chair (chair mass is negligible) Here my upper body in a sitting position is represented as an upright cylinder of radius 0. m and mass 30 kg at a distance of 0. m from the center of rotation, my thighs as a 0.6 m long

10 Physics 0A - Physics for the life sciences 0 horizontal cylinder of mass 5 kg with its center of mass at the center of rotation and my lower legs are represented by a vertical cylinder of radius 0.05 m and of mass 0 kg at a distance of 0. m from the center of rotation. In addition, my two arms are represented by a cylinder.6 m long and of mass 0 kg with its center of mass at 0.m from the center of rotation. In addition there are, of course, the two masses of 4 kg each at a distance of 0.8 m from the center of rotation..6 m Arms 0 kg 4 kg 4 kg Mechanical model of me on chair (view from above) The moment of inertia of my upper body is that of a cylinder rotating about an axis that is 0. m from its own axis. That is = = 35. kg-m The moment of inertia of my thighs is that of a cylinder rotating transverse to its axis. That is = 0. kg-m The moment of inertia of my lower legs is that of a cylinder rotating about an axis that is 0. m from its own axis. That is = = 0. 4 kg-m The moment of inertia of the platform is that of a disk rotating about its center of mass. It is = kg-m The total moment of my system, not counting my arms and the 4 kg masses, is then.5 kg-m. For my initial moment of inertia I must add the moments of inertia of my arms and the masses. For my arms, as for the figure skater, it is the moment of inertia for a rotation about their center of mass plus their mass times the square of the distance of their center of mass from the center of rotation. Ml + Mr = = = 5. kg - m For the masses the moment of inertia is simply the sum of the two masses times the square of their distance from the center of rotation. Mr = = 5. kg-m The chair is of plastic and its mass and moment of inertia are negligible compared to the remainder of the system. The total moment of inertia of my system with my arms stretched out is then I initial = = 0. = 0 kg m (to the accuracy of the model) The vertical view of when I brought the weights in toward my body is shown below. r = 0. m Arms 4 kg.0 m 0 kg 4 kg Mechanical model of me on chair (arms pulled in)

11 Lecture 5 - Rotational Dynamics and Angular Momentum My arms now formed a cylinder of only about meter length. Their moment of inertia is then Ml + Mr = = =. kg - m The moment of inertia of the masses is now Mr = 8 0. = 03. kg-m The total moment of inertia of my system is now = 4.0 = 4 kg-m. This moment of inertia is only 40% of what it was before I brought the weights inward. Conservation of angular momentum therefore implies that my rotational velocity after I brought in the weights must have been.5 times what it was before. This indeed is compatible with what was observed. Relationship of The Law of Conservation of Angular Momentum to the Other Conservation Laws. We now have three conservation laws: The Conservation of Energy The Conservation of Linear Momentum The Conservation of Angular Momentum Do we really need all three? Can t at least this new intruder be shown to be just a result of other two? No it cannot. And to show this consider a case of two masses of, say, kg each rotating at an angular velocity of 5 rad per second at a distance of meter from the center of rotation. Now suppose these masses moved inward to 0.5 m from the center of rotation. By the law of conservation of angular momentum the moment of inertia has decreased by a factor of 4 from 4 kg-m to kg-m and the angular velocity would then have increased by a factor of 4 to 0 radians per second. But this means that the velocity of each mass has actually increased, from initially 5 m/s to finally 0 m/s. Regarding momentum as a scalar would mean that the momentum has actually doubled! We only save the Law of Conservation of Momentum by noting that momentum is a vector and since the two masses are moving in opposite directions the total momentum both before and after was zero. So the Law of Conservation of Angular Momentum is completely independent of the Law of Conservation of Linear Momentum. But what about the Law of Conservation of Energy? Since the velocities of the masses has actually doubled, their kinetic energy has quadrupled. Where did this energy come from? Go back to the case of me pulling in the weights to get them closer to my body. To do this I did have to pull against them. This was so as to create the radial acceleration that comes with circular motion. This means that I had to do work on the weights in order to pull them in. It will not be part of this course but if you evaluated the area under the force displacement graph for me pulling in the weight you would find that this work is just equal to the kinetic energy the weights gained as they were being pulled in. So indeed, all three laws are independent and all three are applicable. The General Motion of Bodies in Space. I will not pretend to derive the general theorems about the motion of bodies in space, but rather state them, hoping that by now you will see them to be reasonable. They are:. The center of mass of a body will be accelerated according to the equation F = ma where F is the vector sum of all the forces acting on the body.. The angular acceleration of the body about its center of mass will be according to the equation T = Iα where T is the sum of all the torques acting acting on the body. An example of the consequences of these laws is that of a diver. After leaving the board the only force of any importance on the diver s body is his or her weight, which acts on the center of mass. If you follow this center of mass through all the contortions that the diver may make you

12 Physics 0A - Physics for the life sciences will see that it follows the smooth parabola of a simple mass thrown into the air to fall into the pool. Also, once the diver has left the board there is no net torque on his or her body. Therefore, no matter what contortions the diver may undergo, his or her angular momentum remains constant. In the picture to the left you can see that when the diver is curled into the tuck position, where her moment of inertia could be modeled to be that of a disk of about radius 0. m, she has a moment of inertia, if her mass is 60 kg, of about. kg-m. When stretched out as she enters the water her moment of inertia, if she is.6 m tall, would be about 3 kg-m, or about 0 times greater. Her angular velocity will therefore be reduced by 0 times from her tuck position to her dive position. This is very important in that it allows her to complete her somersaults in a very short time while enabling her to almost stop her rotation just before she enters the water. The sequence of pictures to the left are approximately 0. second apart and so you can get a rough estimate of her rotational velocity at various points in her trajectory. The conservation of angular momentum takes a particularly bizarre twist (pun intended I must admit) in the well-known case of dropping a cat from a height of about a meter when it is held by its paws upside down. (I assure you that there has never been a cat that has been injured by this exercise.) While it is in this short fall the cat will execute a maneuver in which it lands on its feet. If angular momentum is preserved, and if you allowed the cat to have no angular momentum when you released it, how could it have achieved the angular rotation required for it to get its feet under itself? In fact, time-lapse photographs of this maneuver show that the cat achieves it by rotating part of its body one way and the rest the other way, thereby keeping its total angular momentum to be zero. As a result of this twisting it can get its legs, which have very little moment of inertia compared to the rest of its body, to rotate the required 80 degrees while the remainder of its body rotates only a few degrees the other way. And the cat does this without knowing anything about the Law of Conservation of Angular Momentum. It is just that the Law of Conservation of Angular Momentum requires that it do it that way. The same with humans. Very few Olympic divers know anything about the Law of Conservation of Angular Momentum. However, if you are to understand anything about the dynamics of such maneuvers and the strains

13 Lecture 5 - Rotational Dynamics and Angular Momentum 3 they can put on body parts, then you should know something about that Law. Angular Momentum as a Vector! Up to now you may have been taking solace in the appearance that while angular momentum was somewhat abstract it was at least not a vector. Sorry but not true. Consider another demonstration. While sitting on the magic chair I have this heavy bicycle wheel in my hand. I then put it into a spin in a horizontal plane, i.e. with its axis vertical. I then lift my feet from the floor and, while I have no connection to the floor, I invert the axis of the wheel. You can see that the result is that me and the chair are now spinning. Where did I get this spin? It represents angular momentum; angular momentum that I did not have before I turned over the wheel. The only way I can recover the Law of Conservation of Angular Momentum is to say that by turning over the wheel I changed the sign of its angular momentum. Before I turned the wheel it had an angular momentum that, say, was positive. By turning it over I made this angular momentum negative. Therefore I changed its angular momentum by an amount that was twice its initial value. If, for example, it was kg-m - rad/sec at the start then it would have been kg m -rad/sec at the end. In order for the angular momentum of the system to be preserved, I would need to have 4 kg-m -rad/sec of angular momentum to make the new total the same as before, i.e. kg-m -rad/sec. But what if I only turned the wheel half-way so that it was spinning in a vertical plane? You can see my spinning is now only about half what it was when I turned the wheel completely. And if I only turned the wheel a little bit, I only turned a little bit. By now you are probably convinced that there is a vector lurking in the spin of this wheel. And indeed there is. Angular momentum has a magnitude. All it needs to be a vector is a direction. That direction is along the axis of rotation. By convention, the angular momentum is said to be positive in the direction of so-called right-handed rotation. That is a rotation which, if you look along the axis results in a counterclockwise motion. Another way of thinking about it is to consider the thumb and forefingers of your right hand. By curling your fingers your thumb will point in the direction of the angular momentum vector for a circular motion in the direction of your forefingers. So if the bicycle wheel was initially rotating counterclockwise, looking down, then its angular momentum vector was pointing up. When I turned it over it was spinning clockwise looking down and its angular momentum vector was down. This negative change in its angular momentum vector required a positive change in my angular momentum vector to counteract it. That means that I turned clockwise looking down. Finally you may be in a position to understand the hanging wheel at the beginning of the lecture. This wheel was spinning in a vertical plane so that its angular momentum vector was horizontal.

14 Physics 0A - Physics for the life sciences 4 T Suspension cord Weight of bicycle wheeel New L L Change in L produced by T Now if torque is the rate of change of angular momentum then the torque on this system will be changing this angular momentum vector. This torque is the weight of the wheel times the distance of its center from the point of support. Suspension cord diagram. The action of the weight of the wheel is therefore to turn it in a horizontal circle. The Origin of Conservation of Angular Momentum Like the Conservation of Energy and the Conservation of Linear Momentum, the Conservation of Angular Momentum can be derived from a basic symmetry principle. And in this case it is deceptively simple; the laws of the Universe do not change when you look at the Universe from a different angle. This may seem to be such an obvious assumption that it should not have a consequence as great as a fundamental Law. However, it does. And indeed the consequences of the assumption go even beyond the conservation of angular momentum. This is because of the simple fact that when you turn a complete circle the universe is exactly back to what it was before you started. The logic of this is very abstract but is central to modern quantum mechanics. Whatever, the end result is that angular momentum cannot be continuous but must come in discrete units, just as matter is made of atoms. The unit of angular momentum is now referred to as Plank s Constant, after Plank who discovered the discrete nature of atomic interactions, and is the must fundamental quantized unit (more fundamental then the quantized unit of matter; the atom). By International agreement it is given the symbol h and has the value h = Joule - seconds s W T = Ws But what is the direction of this torque? Like angular momentum, torque is also a vector and points in the direction of the axis of rotation. In the diagram for the rotating wheel this is perpendicular to the axis of the wheel and into the plane of the above drawing, as shown above. Therefore the torque tries to change the angular momentum vector by adding to it in the direction of the torque. This means that the angular momentum vector must be forced to pivot about its origin in the manner shown in the (It may seem strange to give the units of h as J-s, instead of the kg-m -rad/s that comes from Iω. However, note that kg-m/s is ma, of which the unit of that of force, i.e.the N. Since kg-m -rad/s can be expressed as (kg-m/s ) times (m-s-rad) and since radians are unitless, the units become N-m-s. Since N-m are Joules the units of angular momentum become simply J-s.) Relevant material in Hecht Chap 8 - Section 8.. Also figure 8.8 and associated text. Also recommended are the Discussion Questions.