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Topic 2.2 Extended FYI: For all you brainiacs, the genuine official definitions are... Angular momentum FYI: The direction H of L–is the same as the direction for , and given by the right hand rule. v If you recall, linear momentum p was the mass m times the velocity v. Just as torque was the rotational r equivalent of force, and defined as ( = rF) = rF sin we define angular momentum L as the rotational equivalent of linear momentum: L = rp sin L = rmv sin (L = rp) The effect of the "sin " term of course is to take into account the fact that only the perpendicular component of v can cause the mass to move in the circle. If = 90° then v = vt, the tangential velocity. Thus L = rmvt. Since vt = r we have L = rm(r) so that L = mr2. L = I Angular Momentum Topic 2.2 Extended H – Angular momentum Now we can derive the following: L = I L = I Given Provided I is constant L = I Why? t t L = I Why? t L = Newton's 2nd t (angular momentum) Compare with Newton's 2nd law (linear momentum): P = F t Newton's 2nd (linear momentum) FYI: You may recall Topic doing the same when we looked at linear 2.2thing Extended 0 momentum... HP– =Angular Fint + Fext momentum t CONSERVATION OF ANGULAR MOMENTUM We can divide the torque into internal and external torques, so that L = 0+ int ext t From Newton's 3rd, for every force there is an equal and opposite reaction. Thus all the internal forces sum to zero: In an analogous way, all the internal torques also sum to zero: ...Thus L = ext t Then if all of the external torques sum to zero, we have L = 0 t which implies that L = a constant, or Conservation of L0 = Lf Angular Momentum I00 = Iff Provided ext = 0 Topic 2.2 Extended H – Angular momentum CONSERVATION OF ANGULAR MOMENTUM A thin rod of mass M and length L having two point masses m on its ends is spinning about its central axis at an angular speed ωi as shown: M m m L m m ωi If an internal force pulls each point mass to half its original radius, what is the new angular speed ωf? m ωf m Topic 2.2 Extended H – Angular momentum CONSERVATION OF ANGULAR MOMENTUM Since there are no external torques, angular momentum is conserved: Li = Lf Iiωi = Ifωf Ii ωf = ω If i M m L m Ipoint,i = Ii = m L 2 1 ML2 12 2 + × 2 1 2 mL2 mL2 = 2 m ωi The total rotational inertia is given by the sum of the individual ones: Irod,i = Irod,f = m m m ωf 1 ML2 12 Ipoint,f = If = m L 4 1 ML2 12 2 + × 2 1 8 mL2 mL2 = 8 Topic 2.2 Extended H – Angular momentum CONSERVATION OF ANGULAR MOMENTUM Our final results: ωf = 1 ML2 12 + 1 2 mL2 + 1 8 mL2 1 2 ML 12 m ωi Which reduces to ωf = M 2M + 12m ωi 2M + 3m m L m m ωi m ωf m Note: The numerator is bigger than the denominator showing that the angular speed increases as the masses approach the center. If M = 3 kg, m = 2 kg, and ωi = 10 rad/s, what is the final angular speed? 2·3 + 12·2 ωf = 10 2·3 + 3·2 = 25 rad/s Topic 2.2 Extended H – Angular momentum CONSERVATION OF ANGULAR MOMENTUM Suppose we send a satellite into space whose orientation has to be fine-tuned after being placed in orbit. There are basically two ways to do this: (1)Fire small external rockets. (2)Rotate an internal flywheel. Method (1) is very hard to manage because you also have to fire rockets to stop the motion. Method (2) is very easy to manage because you can simply run an electric motor until the craft is aimed the right way. Topic 2.2 Extended H – Angular momentum CONSERVATION OF ANGULAR MOMENTUM Here’s how Method (2) works: If you want to turn the craft clockwise, you rotate the flywheel counterclockwise: Conservation of angular momentum does the rest. L0 = Lf 0 = Icraftωcraft - Iflyωfly Icraftωcraft = Iflyωfly Δθfly Δθcraft Icraft = Ifly Δt Δt Why are the Δts unsubscripted? Icraft Δθcraft = Ifly Δθfly Δθfly Icraft = Δθcraft Ifly Rotational Motion and Equilibrium 8-5 Angular Momentum CONSERVATION OF ANGULAR MOMENTUM Suppose Icraft is 3.75 kg·m2, and Ifly is .45 kg·m2. If you want to turn the craft clockwise through 90° then we must turn the flywheel Δθfly = Icraft Ifly Δθfly = 3.75 0.45 Δθcraft ·90° = 750° Question: Why can we use degrees if we want?