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Chapters 11 & 12: Angular Momentum/Static Equilibrium Chapters 11 & 12 Goals: • we’ll do these chapters out of order • to specialize, as in lab, to the case of no net force nor no net torque: engineering statics •To relax the assumption of a fixed rotation axis, and explore the implications on how a net torque affects changes in angular momentum • to show how the translational and rotational versions of N2 allow us to completely understand the motion of a body, at least qualitatively • to briefly understand the free symmetric top: letting the axis do its own thing: gyroscopic motion Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Idea of the Center of Gravity (CG) • a system (may or may not be rigid) is acted on by gravity because all of its masses have weight • these forces act at different places, obviously, and we regard them as ‘external’ • of course there may be forces beside gravity Nforces… Fg mi g gM • the total weight force is clearly i 1 • this result assumes that g has the same value at all points • what about torque due to weight force? Use ri = rCM + r’i N N N N i 1 i 1 i 1 i 1 τ g ri mi g mi rCM r 'i g mi rCM g mi r 'i g N N mi rCM g mi r 'i g MrCM g 0 [CM is at 0 as seen from it] i 1 i 1 τ g rCM Mg • for gravity torque, gravity acts at CMCG is same as CM Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Science of Statics • system feels no net force: Fext,net = 0 system momentum is constant CM does not accelerate, and we assume that it is not translating either • system feel no net force moment : text,net = 0 system angular momentum is constant, and we assume no rotational motion either • This state of affairs is called static equilibrium • upside: the right sides of the two N2 laws are zero • downside: choice of origin is important, and one must count equations carefully so that system is is not ‘overdetermined’ • example of an overdetermined system: 4-legged table!! • a three-legged stool is stable… A four-legged chair will always not quite be sure which 3 legs to use Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example: the massive lever 250 N 100 N 50 kg d L=4m S W FL FR Two forces act as shown. The lever itself has a mass of 50 kg, and has a length of 4 m. The whole system is supported by the fulcrum, a distance d from the left end. Find support force and d. • Take + to be UP and IN • put all forces where they act (c.o.g.) • Forces: – FL – W – FR + S = 0 • S = FL + W + FR = 850 N • Torques: put origin at (say) left end • 0∙250 – d∙850 + 2∙500 + 4∙100 = 0 • d = 1400/850 = 1.65 m • a massless lever FL/FR = dR/dL •‘The lever equation’ Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A heavy sign (20 kg) hangs from a massless bar and a support wire. Find the Example: the building sign tension, and find both components of the support force S. Some kid has thrown 3 Solution bar a pair of 3 kg shoes out at the end 0 q • take the body to be the bar above the sign c • take + to be UP, to RIGHT, and IN Fargo m • put all forces where they act (including CG) • 3:4:5q 37°; sin q = .60; cos q = .80 • forces UP: S cos f – Fg + T sin q – fg = 0 40 • forces RIGHT: S sin f – T cos q = 0 cm • put origin at left so S does not appear in equation!! S • torques IN: 0S + .20∙200 – .4T sin 143° – .4 ∙30 = 0 T 40 12 28 T 117 N f .4 sin 143 .4.6 q S x S cosf T cos q 133 N Fg fg S y S sin f Fg T sin q f g 200 100 30 130 N S S S (133) 130 2 x 2 y 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 2 Sy 186 N; f Tan 44.3 Sx -1 Not in book: the idea of stability in gravity • as can tips, CG rises: stable • once CG is directly above pivot: tipping point! • from then on, object tips over and c.o.g. descends Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular momentum in terms of moment of inertia • angular velocity vector w points along the axis by RHR • However, L is parallel to that vector ONLY if there is no external force moment on the body • consider the ‘dumbbell’ comprising two masses m and a massless stick of length L O • axis through the c.o.m. perpendicular to the stick; origin at c.o.m. 2 mL IC 2 4 2 mL 2 • L = L1 + L2 and both masses contribute an identical angular momentum; total magnitude is mL2w/2 and it points up, exactly along w -- and it is a constant • Therefore, there is no need for a force moment on system • we have that L = ICM w in this case: L is parallel to w Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens if we alter the axis? • put origin not at CM, but distance O D from it ┴ • angular velocity vector w points along the axis by RHR I A I CM 1 MD ML2 MD 2 2 2 • we still have L = IA w and L is constant • note now that because the CM is accelerating, there must be a net horizontal force on the system… the thing is wobbly… the force at this instant points to R since CM is accelerating centripetally in that direction • but there is still no need for a net torque since L is constant .. so L is parallel to the angular velocity Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens if we alter the axis differently? • construct L = r x p for left mass • p is out, so L is up and to left • get same result for right mass • clearly L CHANGES as τdumbell dL dt whirls around and L is NOT parallel to w: L ≠ ICM w • furthermore, L is changing as the thing flies around so by virtue of r L O p net,ext τ net,ext dL dt • we need a torque in this case!! Fnet,ext is however zero • thus we need a torque couple.. how do we push at this instant? Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example IA O An object is made from a thin stick that is 75 cm in length and of mass 2.4 kg. At one end there is a solid sphere of radius 5.0 cm and mass 4.0 kg. The object turns about its center at 8 rad/sec. Find the c.o.m. and find the contribution to I and K for each part. Where is center of mass of system? Put origin at left end: 1 4(0) 2.4(37.5) 90 kg - cm XC mi xi 14.1 cm M 4 2.4 6.4 kg What is I for left mass? Use parallel axis theorem and IC for ball: I AL I CL m mL R mL 2 2 5 2 L 2 2 5 4.05 4.375 2 2 .0040 .563 .567 kg - m 2 2 2 2 I I m L 2 . 4 . 75 . 113 kg m What is I for stick? AS CS 12 S S 12 1 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 1 Example IB O I A I AL I AS .57 .11 .68 kg - m 2 Question: can we regard this K as the K of c.o.m. + K referred to c.o.m.? KA 1 2 I w 2 A .68(8) 2 21.76 J 2 KA ? ? 12 MVC2 12 I Cw 2 2 6 . 4 ( 8 ) .234 1 1 2 2 2 MVC 2 Mw 11.21 J 2 2 2 2 RL 1 64 L2 2 2 2 I w 2 mL 5 .141 mS 12 .234 2 C 324.0010 .0199 2.4.0469 .0548 2 32.084 .244 10.50 J Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 11.21 J + 10,50 J = 21.71 J YES!! Example II A mass of 5 kg hangs from a string that is wrapped several times around a hoop (radius 60 cm) with a horizontal frictionless fixed axle. The mass has a = 4 m/s2 5 Find the angular atan 4 m/s 2 2 acceleration of the atan R R .60 m 6.7 rad/s hoop, and its mass. Make FBDs for both objects. Put T origin at axle. Choose UP & IN + S O Force on m : T mg ma T F g F T m( g a) 5(10 4) 30 N g T 30 Moment on M : RT I C MR M 7.5 kg R .6(6.7) 2 Force on M : T S Mg 0 S Mg T 105 N Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A taste of angular momentum conservation By Example We model an ice skater as a cylinder (head/torso/legs) of mass 50 kg and radius 25 cm) along with two cylindrical arms, each of mass 4 kg, radius 5 cm and length 80 cm) a) Find moment of inertia with arms DOWN and with arms OUT b) Argue that angular momentum should be conserved when a skater spins, and then assuming an initial rotation rate of 2 rotations/sec with arms OUT, find the final rotation rate with arms DOWN. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. I C,axial 2 MR 2 1 example II I A,normal@end 4 MR 2 3 ML2 1 1 .05 m 2 2 I C, DOWN 50 kg .25 m 24 kg 2 .30 m 1 2 2 1.56 kg - m 2 .73 kg - m 2 2.29 kg - m 2 • first term is torso axial c.o.m. moment of inertia • factor of 2 x mass is for two arms of mass 4 kg • first term in big parenthesis is arm’s axial c.o.m. IC • second term in big parenthesis is parallel-axis part 2 2 I C,OUT 1 2 50 kg .25 m 2 24 kg .054m .80 m 3 .25 m 2 1.56 kg - m 2 2.21 kg - m 2 3.77 kg - m 2 • L is conserved because the external force moment is zero!! Linit Lfin I initwinit I finwfin wfin winit 2 rev 2 s rad rev 4 rad wfin s I init w init I fin 3.77 rad rad 4 6.6 2.29 s rev Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How does torque affect angular momentum, in general? • consider a wheel (a hoop, say), that is rotating with M, R d L a large angular velocity (and hence a large angular O momentum • the wheel is supported at one point only at the end of its axis a distance d from wheel’s CM • put an origin at the support point • the net external torque tnet,ext on the wheel is therefore of magnitude Mgd, and it points which way? This is the same as the (instantaneous) direction of dL/dt, of course!! • right! into the page!! So L’s change is into the page, which means that L moves around on a big circle • The wheel does NOT fall over.. It precesses dL dw dw gd Mgdφˆ MR 2 Mgd 2 rate of precession dt dt dt R Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. τ net,ext