Chapter 16 Three Dimensional Rotational Dynamics and Angular Momentum 16 1 Angular Momentum for the Rotational and Translation of a Rigid Body We now return to our translating and rotating rod Figure 16 1 which is also Figure 14 1 that we first considered when we began our discussion of rigid bodies We noted that the rod could both translate with a center of mass velocity v cm and rotate about the center of mass Figure 16 1 The motion of a thrown rigid rod We shall now show that the total angular momentum of a body about a point S can be decomposed into two vector parts the angular momentum of the center of mass motion about the point and the angular momentum of the rotational motion about the center of mass Consider a system of N particles located at the points labeled i 1 2 N The total angular momentum about the point S is N r Ltotal LS S i 1 N i i 1 S i mi v i 16 1 1 where rS i is the vector from the point S to the ith particle See Figure 15 5 and Figure 16 2 below For a given coordinate system the ith particle has position vector denoted ri Let ri be the vector from the center of mass to the ith particle and let rS cm be the vector from the point S to the center of mass Figure 16 2 illustrates these vectors 7 25 2008 1 Figure 16 2 Coordinate system for the center of mass From Figure 16 2 we see that rS i rS cm ri 16 1 2 We can differentiate Equation 16 1 2 with respect to time to find the law of addition of velocities v i v cm v i 16 1 3 We can use Equations 16 1 2 and 16 1 3 in Equation 16 1 1 to calculate the angular momentum about S Ltotal rS i mi v i rS cm ri mi v cm v i S N N i 1 i 1 16 1 4 When we expand the expression in Equation 16 1 4 we have four terms N N Ltotal rS cm mi v cm rS cm mi v i S i 1 i 1 N N i 1 i 1 ri mi v cm ri mi v i 16 1 5 The vector rS cm is a constant vector that depends only on the location of the center of mass and not on the location of the ith particle Therefore in the first term in the above equation rS cm can be taken outside the summation Similarly the velocity of the center of mass v cm is the same for each term in the summation and may be taken outside the summation The first term in Equation 16 1 5 becomes 7 25 2008 2 N total r m v r S cm i cm S cm mi v cm rS cm m v cm i 1 i 1 rS cm p total N 16 1 6 where p total m total v cm is the total linear momentum of the rigid body Recall that total momentum in the center of mass reference frame is zero N m v 0 i 1 i i 16 1 7 therefore the second term in Equation 16 1 5 vanishes N N i 1 i 1 rS cm mi v i rS cm mi v i 0 16 1 8 The center of mass is located at the origin in the center of mass reference frame so rS cm 1 N miri 0 m total i 1 16 1 9 therefore the third term in Equation 16 1 5 also vanishes N total r m v i i cm miri v cm m rS cm v cm 0 i 1 i 1 N 16 1 10 where as in the first term the velocity of the center of mass v cm has been taken outside of the summation Substituting Equations 16 1 6 16 1 8 and 16 1 10 into Equation 16 1 5 yields N Ltotal rS cm p total ri mi v i S 16 1 11 i 1 Consider the first term in Equation 16 1 11 rS cm p total the vector rS cm represents any translational displacement of the center of mass If we treat the rigid body as a point like body located at the center of mass then the momentum of this point like body is p total Thus the angular momentum of this point like body about the point S is the first term which is the orbital angular momentum about S Lorbital rS cm p total for this rigid body S 7 25 2008 3 In the second term in Equation 16 1 11 the vector ri is the vector from the center of mass to each ith particle and so the quantity inside the summation ri mi v i represents the angular momentum about the center of mass for each ith particle Summing over all the particles gives the total angular momentum about the center of mass N Lspin cm ri mi v i 16 1 12 i 1 which is the spin angular momentum Thus we see that the total angular momentum about the point S is the sum of these two terms Ltotal Lorbital Lspin S S cm 16 1 13 Remember that the angular momentum is a vector at the point S that is now thought of as the sum of two vectors The orbital angular momentum depends on the point S while the spin angular momentum is independent of the point S and both must be added to define the total angular momentum For fixed axis rotation about the center of mass the spin angular momentum is Lspin cm I cm spin 16 1 14 where spin and hence Lspin cm are directed along the fixed axis 16 1 1 Example Earth s motion around the sun The earth of mass me 5 97 1024 kg and mean radius Re 6 38 106 m moves in a nearly circular orbit of radius rs e 1 50 1011 m around the sun with a period Torbit 365 25 days and spins about its axis in a period Tspin 23 hr 56 min the axis inclined to the normal to the plane of its orbit around the sun by 23 5 Figure 16 10 Figure 16 3 Orbital and spin motion of the earth 7 25 2008 4 If we approximate the earth as a uniform sphere then the moment of inertia of the earth about its center of mass is 2 I cm me Re2 5 16 1 15 a derivation of Equation 16 1 15 is given in the Appendix to this chapter If we choose the point S to be at the center of the sun and assume the orbit is circular then the orbital angular momentum is Lorbital rS cm p total rs e r mevcm rs e mevcm k S 16 1 16 The velocity of the center of mass of the earth about the …
View Full Document
Unlocking...