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r. Angular Momentum. FYI: For all you brainiacs, the genuine official definitions are. FYI: The direction of L is the same as the direction for , and given by the right hand rule. Topic 2.2 Extended H – Angular momentum. v.
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r Angular Momentum FYI: For all you brainiacs, the genuine official definitions are... FYI: The direction of L is the same as the direction for , and given by the right hand rule. Topic 2.2 ExtendedH – Angular momentum v If you recall, linear momentump was the mass m times the velocity v. Just as torque was the rotational equivalent of force, and defined as ( = rF) = rFsin we define angular momentum L as the rotational equivalent of linear momentum: L = rpsin (L = rp) L = rmvsin 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 Since vt = r we have L = rmvt. L = rm(r) so that L = mr2. L = I
Newton's 2nd (linear momentum) Newton's 2nd (angular momentum) Topic 2.2 ExtendedH – Angular momentum Now we can derive the following: L = I Given L = I Provided I is constant L t t Why? = I L t = I Why? L t = Compare with Newton's 2nd law (linear momentum): P t = F
0 0 Conservation of Angular Momentum FYI: You may recall doing the same thing when we looked at linear momentum... P t = Fint + Fext Topic 2.2 ExtendedH – Angular momentum CONSERVATION OF ANGULAR MOMENTUM We can divide the torque into internal and external torques, so that L t = int + ext 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 t = ext Then if all of the external torques sum to zero, we have L t = 0 which implies that L = a constant, or L0 = Lf I00 = Iff Provided ext = 0
L ωi m m m m m m ωf Topic 2.2 ExtendedH – Angular momentum CONSERVATION OF ANGULAR MOMENTUM M 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 ωias shown: If an internal force pulls each point mass to half its original radius, what is the new angular speed ωf?
L Ii If ωf = ωi ωi m m m m m m 1 12 Irod,i = Irod,f = ML2 ωf × 2 × 2 L2 4 L2 2 Ipoint,i = m Ipoint,f = m 1 2 1 8 1 12 1 12 mL2 2 mL2 8 Ii = ML2 + mL2 If = ML2 + mL2 = = Topic 2.2 ExtendedH – Angular momentum CONSERVATION OF ANGULAR MOMENTUM M Since there are no external torques, angular momentum is conserved: Li = Lf Iiωi = Ifωf The total rotational inertia is given by the sum of the individual ones:
L ωf = ωi ωi 2M + 12m m m m m m m ωf = ωi 2M + 3m ωf 1 2 1 8 1 12 1 12 2·3 + 12·2 ML2 + mL2 ML2 + mL2 ωf = 10 2·3 + 3·2 Topic 2.2 ExtendedH – Angular momentum CONSERVATION OF ANGULAR MOMENTUM M Our final results: Which reduces to 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? = 25 rad/s
Topic 2.2 ExtendedH – 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.
Δθcraft Δt Icraft = Ifly Δθfly Δt Icraft Ifly Δθfly = Δθcraft Topic 2.2 ExtendedH – Angular momentum CONSERVATION OF ANGULAR MOMENTUM Here’s howMethod (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 Why are the Δtsunsubscripted? Icraft Δθcraft = Ifly Δθfly
3.75 0.45 Δθfly = ·90° Icraft Ifly Δθfly = Δθcraft Rotational Motion and Equilibrium8-5 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 = 750° Question: Why can we use degrees if we want?