both the absolute and relative angular momentum are identical. As we shall see, this will simplify our nalysis significantly. In general, the absolute and relative angular momentum with respect to an arbitrary not the We can now go back to equation 9 and consider the time variation of HG HG=∑(xm(v+)+∑(rxm1)=0+∑(rxF) In the above equation, the term ri x miri is clearly zero, and iai(ri x miUG)=-UG x Camiri= Ug(liar miTi)/dt=0. Thus, we have that HG=MG (13) Here, MG=airX Fi), is the total moment, about G, of the applied external forces The above expression is very powerful and allows us to solve, with great simplicity, a large class of problems in rigid body dynamics. Its power lies in the fact that it is applicable in very general situations In the derivation of equation 13, we have made no assumptions about the motion of the center of mass G. That is, equation 13 is valid even when G is accelerated We have implicitly assumed that the reference frame used to describe r in equation 10 is non-rotating with respect to the fixed frame xyz(otherwise, we would have written ri=v+wxr, with w, the angular velocity of the frame considered). It is not difficult to show that equation 13 is still valid if the reference frame rotates, provided the angular velocity is constant. If the reference frame rotates with a constant angular velocity, the angular momentum will differ from that of equations, 9 and 11 by a constant, but equation 13 still will be valid (see approach 2 in example below) The angular momentum of the particle system about the center of mass, G, can be evaluated using two Iternative representations, 9 and 11. In particular, equation 1l allows us to determine HG without making any reference to the absolute velocities, vi, or position vectors, r Finally, by combining equations 7 and 9, the angular momentum about a fixed point, O, can be expressed as a function of the angular momentum about the center of mass. as Ho=rc×mva+H Example Rotating dumbbell [ 1] We consider a steadily spinning turntable with angular velocity S, with a symmetric dumbbell mounted at a distance R from the center. The dumbbell is free to spin about its midpoint. We assume that the rod connecting the two masses is massless. Therefore, we can model the dumbbell as a system consisting of two particles. We want to determine the relative motion of the dumbbell. We will consider three different approachesboth the absolute and relative angular momentum are identical. As we shall see, this will simplify our analysis significantly. In general, the absolute and relative angular momentum with respect to an arbitrary point are not the same. We can now go back to equation 9 and consider the time variation of HG, H˙ G = Xn i=1 (r˙ ′ i × mi(vG + r˙ ′ i )) +Xn i=1 (r ′ i × miv˙ i) = 0 + Xn i=1 (r ′ i × Fi) . (12) In the above equation, the term r˙ ′ i × mir˙ ′ i is clearly zero, and Pn i=1(r˙ ′ i × mivG) = −vG × Pn i=1 mir˙ ′ i = −vG × d( Pn i=1 mir ′ i )/dt = 0. Thus, we have that H˙ G = MG . (13) Here, MG = Pn i=1(r ′ i × Fi), is the total moment, about G, of the applied external forces. The above expression is very powerful and allows us to solve, with great simplicity, a large class of problems in rigid body dynamics. Its power lies in the fact that it is applicable in very general situations: • In the derivation of equation 13, we have made no assumptions about the motion of the center of mass, G. That is, equation 13 is valid even when G is accelerated. • We have implicitly assumed that the reference frame used to describe r ′ i in equation 10 is non-rotating with respect to the fixed frame xyz (otherwise, we would have written r˙ ′ i = v ′ i + ω ′ × r ′ i , with ω ′ , the angular velocity of the frame considered). It is not difficult to show that equation 13 is still valid if the reference frame rotates, provided the angular velocity is constant. If the reference frame rotates with a constant angular velocity, the angular momentum will differ from that of equations, 9 and 11 by a constant, but equation 13 still will be valid (see approach 2 in example below). • The angular momentum of the particle system about the center of mass, G, can be evaluated using two alternative representations, 9 and 11. In particular, equation 11 allows us to determine HG without making any reference to the absolute velocities, vi , or position vectors, ri . Finally, by combining equations 7 and 9, the angular momentum about a fixed point, O, can be expressed as a function of the angular momentum about the center of mass, as, HO = rG × mvG + HG. Example Rotating dumbbell [1] We consider a steadily spinning turntable with angular velocity Ω, with a symmetric dumbbell mounted at a distance R from the center. The dumbbell is free to spin about its midpoint. We assume that the rod connecting the two masses is massless. Therefore, we can model the dumbbell as a system consisting of two particles. We want to determine the relative motion of the dumbbell. We will consider three different approaches: 4