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Topic 2.2 Extended H – Angular momentum

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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Topic 2.2 Extended H – Angular momentum

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  1. 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 ( = rF)  = rFsin we define angular momentum L as the rotational equivalent of linear momentum: L = rpsin (L = rp) 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

  2. 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

  3. 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 I00 = Iff Provided ext = 0

  4. 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?

  5. 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:

  6. 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

  7. 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.

  8. 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

  9. 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?

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