Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 Chapters 11 & 12: Angular Momentum/Static Equilibrium Chapters 11 & 12 Goals:

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 forces… this result assumes that g has the same value at all points what about torque due to weight force? Use r i = r CM + r’ i the total weight force is clearly for gravity torque, gravity acts at CM  CG is same as CM

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Science of Statics system feels no net force: F ext,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 :  ext,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 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: – F L – W – F R + S = 0  S = F L + W + F R = 850 N Torques: put origin at (say) left end 0∙250 – d∙ ∙ ∙100 = 0  d = 1400/850 = 1.65 m a massless lever F L /F R = d R /d L ‘The lever equation’ 50 kg 100 N 250 N FLFL FRFR W S d L = 4 m

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example: the building sign A heavy sign (20 kg) hangs from a massless bar and a support wire. Find the tension, and find both components of the bar support force S. Some kid has thrown a pair of 3 kg shoes out at the end take the body to be the bar above the sign take + to be UP, to RIGHT, and IN put all forces where they act (including CG) 3:4:5   37°; sin  =.60; cos  =.80 forces UP: S cos  – F g + T sin  – f g = 0 forces RIGHT: S sin  – T cos  = 0 put origin at left so S does not appear in equation!! torques IN: 0S +.20∙200 –.4T sin 143° –.4 ∙30 = 0 40 cm Fargo 30cm30cm  Solution S T FgFg   fgfg

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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  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 L = L 1 + L 2 and both masses contribute an identical angular momentum; total magnitude is mL 2  /2 and it points up, exactly along  -- and it is a constant Therefore, there is no need for a force moment on system we have that L = I CM  in this case: L is parallel to  axis through the c.o.m. perpendicular to the stick; origin at c.o.m. O

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens if we alter the axis? we still have L = I A  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 put origin not at CM, but distance D from it ┴ angular velocity vector  points along the axis by RHR O

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 O r p L we need a torque in this case!! F net,ext is however zero thus we need a torque couple.. how do we push at this instant? p is out, so L is up and to left get same result for right mass clearly L CHANGES as dumbell whirls around and L is NOT parallel to  : L ≠ I CM   furthermore, L is changing as the thing flies around so by virtue of

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example IA 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. O Where is center of mass of system? Put origin at left end: What is I for left mass? Use parallel axis theorem and I C for ball: What is I for stick?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example IB O J + 10,50 J = J YES!! Question: can we regard this K as the K of c.o.m. + K referred to c.o.m.?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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/s 2 Find the angular acceleration of the hoop, and its mass. Make FBDs for both objects. Put origin at axle. Choose UP & IN  + 5 T FgFg S O T FgFg

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A taste of angular momentum conservation 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) By Example 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. example II 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. I C second term in big parenthesis is parallel-axis part L is conserved because the external force moment is zero!!

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How does torque affect angular momentum, consider a wheel (a hoop, say), that is rotating with a large angular velocity (and hence a large angular momentum in general? L M, R d 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 O the net external torque  net,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