Physics 430: Lecture 6 Center of Mass, Angular Momentum

1. Physics 430: Lecture 6 Center of Mass, Angular Momentum Dale E. Gary NJIT Physics Department

2. 3.3 Center of Mass • We are now going to discuss the notion of center of mass, with which you are certainly already familiar. Think of a system of N particles a = 1, …, N, with masses ma and positions ra. The center of mass (or CM) is defined to be the position • Like any vector equation, this represents separate equations for each of the components (X, Y, Z): • You can think of the center of mass as a weighted average of the positions of each mass element, i.e. weighted by the mass of that element, or equivalently it is the vector sum of the ra, each multiplied by the fraction of mass at that location. • To get a feeling for CM, let’s look at the center of mass for a two particle system, which might, for example, represent the Sun and Earth, or two stars in orbit around each other.

3. CM m1 r1 m2 R r2 O Center of Mass and Equation of Motion • In this case, , which can be seen in the figure. • It is easy to show that the distance of the CM from m1 and m2 is in the ratio m2/m1. The figure shows the case where m1 4m2. In particular, if m1 >> m2, then the CM will be very close to m1. • Note that the time derivative of the center of mass for N particles is just the CM velocity so the momentum of an N-particle system is related to its CM by . • Differentiating this expression, we get the very useful relation for the equation of motion: • This says that the CM of a collection of particles moves as if the external forces on all of the individual particles were concentrated at the CM. This is why we can treat extended objects (e.g. a baseball) as a point mass.

4. Calculating the Center of Mass • Although we developed the foregoing for a set of point particles, the result obviously applies to extended objects by replacing the summation with an integral, and treating infinitesimal parts of the object as having mass dm. The CM expression then becomes where the integral extends over the object. • If you have a uniform extended object of total mass M, you may be given the size or volume, from which you can determine the density, or alternatively you may be given the density, from which you determine the volume. In either case, the integral over the mass is replaced by an integral over the volume • Let’s do an example, the CM of a solid cone. You will have a chance to practice this with a solid hemisphere, for homework.

5. z R r=Rz/h h y x Example 3.2: The CM of a Solid Cone • Statement of the problem: • Find the CM position for the uniform solid cone shown in the figure. • Solution: • You should be able to see immediately from the symmetry of the problem that the CM lies on the z axis. This greatly simplifies the problem, since we can now concentrate only on the z component. To find the height Z of the CM, where the density r = M/V can be brought outside the integral because the cone is uniform, and we have replaced dV with the cartesian element of volume dx dy dz. Note that we have to do an integration over x, y and z, despite the fact that this is only the z component. • If there is any trick to this, it is that we can do the x, y integrals in our head—the area at a given height z is a circle of radius r = Rz/h, of area pr2 = pR2z2/h2. The integral then becomes where

6. Example 3.2: Cylindrical Coords • Solution: • Although the text solves the problem as just shown, from the symmetry of the problem it is a more natural choice to use cylindrical coordinates. The cylindrical element of volume is dV = r dr df dz. (Convince yourself this is right.) The integral is then which then leads to the previous result. • If you do not know the volume of a cone, the way to calculate it parallels the above, but without the z:

7. O 3.4 Angular Momentum for a Single Particle • As you know, in addition to the law of conservation of momentum, there is an independent but obviously related law of conservation of angular momentum. • The angular momentum of a single particle is defined as the vector where I am forced to use the over-arrow because I cannot make the script bold. Here r× p is the vector product of the particle’s position vector r, relative to the chosen origin O, and its momentum p, as shown in the figure. • It is important to understand the implications of the statement that the angular momentum is about the origin O. In the figure at left, we can make the angular momentum of the particle disappear by simply shifting our origin. How can we define away a conserved quantity like this? • Just consider that angular momentum has little meaning for a single particle, but when a second body is included, shifting the origin affects that one, too. p=mv r r into page

8. Angular Momentum and Torque • The time derivative of angular momentum is but because , the first term is identically zero (the vector product of a vector with itself is zero). In addition, we can replace the in the second term with the net force F, and we then recognize the torque. The text using the greek capital gamma for torque, and I will, too. Other popular symbols are t and N. • In many two-body problems one should choose the origin O so that the net torque is zero. For example, a planet orbiting the Sun feels a gravitational force F = GmM/r2 from the Sun. A hallmark of such motion is that the force is central, i.e. is directed along the line between the two centers. Choosing the origin at the Sun greatly simplifies the problem because this ensures that there is no torque (r× F = 0), so the angular momentum r× p is constant, from which we can immediately deduce that r and p must remain in a fixed plane through the Sun. Let’s take a closer look at that problem.

9. Q P dA r dA Q P dr = vdt Kepler’s Second Law • Kepler’s second law states that • The situation is shown in the figure below, where we show two segments of the orbit that I will approximate as triangles (the approximation becomes exact in the limit as the width of the triangles goes to zero). Kepler’s 2nd law is equivalent to saying that so long as the elapsed time dt for the planet to go from P to Q is the same as for it to go from P’ to Q’, then the areas of these two triangles must be equal. Equivalently, dA/dt = constant. As each planet moves around the Sun, a line drawn from the planet to the Sun sweeps out equal areas in equal times. • A well-known property of the vector product is that two sides of a triangle are given by vectors a and b, then the area is (see problem 3.24—this is related to area= ½ base×height). Thus, the area of triangle OPQ is . • This can be rearranged to get: which, since the angular momentum implies that Kepler’s law holds.

10. 3.5 Angular Momentum for N Particles • We can extend these ideas to N particles, a = 1, 2, …, N, following very much the same procedure as we did for momentum in lecture 1. Each particle has angular momentum (with all ra measured from the same origin O), so the total angular momentum is • The time derivative of the total angular momentum is where, exactly as before, the net force on particle a is consisting of the inter-particle forces Fab, and the external force Faext. Substituting into the L-dot equation, we have • As before, we use the fact that Fab = -Fba to replace the sum over bawith one over b>a containing matching pairs, to get

11. Conservation of Angular Momentum • In lecture 1, the paired terms canceled directly. This time, the first sum is again zero, but for a different reason. Clearly ra – rb 0, but rather, we assume that the inter-particle forces are central forces, so that the vector ra – rb is aligned with the force Fab, so the cross-product is zero. So finally we are left with the time derivative of total angular moment equal to the externally applied torque • In particular, if there are no applied torques, then L = constant, which leads to • The validity depends on two assumptions: the inter-particle forces are central, and that they obey Newton’s third law. You can imagine a system where the first of these is not true, but for nearly all cases the law holds. Principle of Conservation of Angular Momentum If the net external torque Gext on an N-particle system is zero, the system’s total angular momentum L = S ra× pa is constant.

12. Moment of Inertia • We will deal with (and extend) the idea of moment of inertia in Chapter 10. However, you should be familiar with the basics from your Introductory Physics course. In particular, you should recall that the angular momentum about an axis of rotation (say the z axis) is where I is the moment of inertia about the axis of rotation, and w is the angular velocity. • In your earlier course, you learned that the moment of inertia is known for a few standard bodies (i.e. for a uniform sphere of mass M, radius R, the moment of inertia through the center is I = 2/5 MR2). • In general, for any multiparticle system, I = Smara2, where ra is the distance of mass ma from the axis of rotation. The moment of inertia for an extended object, can be calculated by replacing the sum with an integral.

13. r b q v Location of stuck putty Example 3.3: Collision of a Lump of Putty with a Turntable • Statement of the problem: • A uniform circular turntable (mass M, radius R, center O, moment of inertia about O, ½ MR2) is at rest in the x, y plane and is mounted on a frictionless axle, which lies along the vertical z axis. I throw a lump of putty (mass m) with speed v toward the edge of the turntable so it approaches along a line that passes within a distance b of O, as shown in the figure. When the putty hits the turntable, it sticks to the edge and the two rotate together with angular velocity w. Find w. • Solution: • We use conservation of angular momentum. Since the turntable is not moving initially, the initial angular momentum is that of the putty about O, • After the putty sticks, the turntable starts to turn at some unknown angular velocity, so that the angular momentum is • Equating these two equations and solving for w, we have

14. Special Case: Angular Momentum About CM • The foregoing derivation of was made under the (unstated) assumption that Newton’s second law F = ma holds, but recall that this is only valid in an inertial (non-accelerating) reference frame. We can state that holds for any origin O only for inertial reference frames. • We will see in Chapter 10, but state it now without proof, that the law holds even in accelerating (non-inertial) frames as long as the origin is about the CM of the system, even when the CM is undergoing acceleration. • Stated another way, if Gext(about CM) = 0then L(about CM) is conserved. You can show this yourself with the guidance of Prob. 3.37 of the text, if you are curious. • This result shows the special nature of the CM.

15. y x 2b F Example 3.4: A Sliding and Spinning Barbell • Statement of the problem: • A barbell consisting of two equal masses m mounted on the ends of a rigid massless rod of length 2b is at rest on a frictionless horizontal table, lying on the x axis and centered on the origin, as shown in the figure. At time t = 0, the left mass is given a sharp tap, in the shape of a horizontal force F in the y direction, lasting for a short time Dt. Describe the subsequent motion. • Solution: • The type of force described is called an impulse. When dealing with such a force, we want to focus on the change in momentum due to the force, rather than the force itself. Since , the momentum after the force acts is (this is actually the change in momentum, DP, but the momentum before the impulse is zero). • It is important to recognize that this impulse does two things. It provides momentum to the CM, but at the same time, since it is off-center, it also provides a torque. We need to calculate both. • The CM relation is easy: vCM = FextDt / 2m. Since Fext is in the +y direction, vCM is too.

16. Example 3.4, cont’d • Solution, cont’d: • The rotational motion due to the torque is found using the methods of the chapter. The torque is going to cause a change in angular momentum, with magnitude Gext= Fb. Using the same approach as before to deal with the fact that the force is an impulse, we find the angular momentum after the impulse is • The moment of inertia of the dumbbell is easily calculated (since the rod is massless) as I = 2mb2, since each mass can be considered a point mass, and each contributes mb2 to the moment of inertia. Solving for the angular velocity: • Note that just after the impulse, the velocity of the left mass is vleft = vcm + wb = FDt/m, while the velocity of the right hand mass is vright = vcm – wb = 0. • The subsequent motion is straightforward. The CM moves straight up the y axis while the barbell continues to rotate at angular velocity w. y wb vcm x 2b

17. Example 3.4, Further Remarks • Remarks: • It always bothered me that if the same impulsive force were applied on the bar between the two masses, i.e. at the CM, the barbell would move at the SAME speed we just calculated for the CM, but without rotation. • From energy considerations, it seems that applying the same force in two locations imparts DIFFERENT amounts of energy to the barbell. It always seemed to me that the energy should be the same in the two situations, if the force is the same. Hopefully this gives you the same sense of unease. • After more thought, however, one can understand the difference. Recall that work (energy) is force through a distance. We are given the force, but what about the distance over which it acts? In the case of hitting the barbell at the CM, the force acts for a time Dt on the barbell moving at speed vCM, so the distance is s = vCMDt. In the case of this problem, where we hit the left mass, at a location a distance b from the CM, the left mass moves at speed 2vCM (check it), so the force acts through a distance s = 2vCMDt. It is this difference that accounts for the difference in energy. • If you really want to be confused, however, think about the case where the impulse force is due to a lump of putty that sticks to the barbell. The putty has the same energy in both cases, but the putty plus barbell energy is different depending on where the putty lands. What the… (There is an explanation—can you find it?)