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Chapter 7: Rigid Body Dynamics

Chapter 7: Rigid Body Dynamics. Chapter 7 Goals:. To define the rigid body as a system of point masses Through use of forces and force moments, to introduce the physics of static equilibrium To distinguish the force moment from the special case of the torque

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Chapter 7: Rigid Body Dynamics

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  1. Chapter 7: Rigid Body Dynamics Chapter 7 Goals: • To define the rigid body as a system of point masses • Through use of forces and force moments, to introduce the physics of static equilibrium • To distinguish the force moment from the special case of the torque • To learn about the moment of inertia of a rigid body about a given axis, and to consider various situations • To apply N2 and its rotational equivalent to the rotation of a rigid body about a fixed axis • To discuss how conservation laws for energy and angular momentum may be applied fruitfully

  2. What is a rigid body? • extended in space, has mass M, and has a c.o.m. • conceive of it as a collection of very large number of infinitesimal point masses dm • thedistance between any pair of point masses is fixedit is rigid (for N masses, there are therefore N(N–1)/2 kinematic constraints on system • Its motion may be ‘broken down’ into pure translation of the c.o.m. combined with pure rotation about the c.o.m. • many of the summations we have been considering, as sums over all the bodies from i=1 to N, are going to become, at worst, volume integrals • YIKES!!!

  3. A couple of those integrals • in practice, the multiple integrals can be handled pretty easily once one identifies how to express dm • the key tasks are to think about the mass density of the stuff, and whether the object is one-dimensional (like a thin stick, or an arc), or 2d (a plate or a shell) or 3d (a brick; a ball; a cylinder)

  4. 1d Example Ia A piece of wire is bent into the arc of a circle of radius R which subtends an angle q in radians. Find the mass and the c.o.m. of the object. The linear mass density of the material is l, in units of kg/m. y • Put origin at the center of the arc • Let the wire subtend an angle 2q0, so that is lies between –q0 and +q0 • The length of the wire is the arc length: L = 2 Rq0, where angles are in radians • Consider a very small arc length ds, which contains mass dm, and subtends an angle dq0 (see next slide) x

  5. 1d Example Ib • Infinitesimal mass dm ‘lives’ at angle q • It subtends an infinitesimal angle dq • Its infinitesimal arc length is ds= R dq • Since in 1d any mass is “density x length”  dm = lR dq[key step in the solution!] y dq dm q x • Obviously YC = 0 by ‘up-down’ symmetry’ • dm is located at (x,y) = (R cosq, R sinq)

  6. 1d Example IIa A stick of length L is fabricated from a material with a variable linear mass density, that increases from the value lL on the left end to the value lR on the right end. Find the mass and locate the c.o.m. l(x) vs. x lR lL L x • Using slope-intercept ideas • dm = l dx for the infinitesimal bit of mass located at x

  7. 1d Example IIb l(x) vs. x lR lL x L This result is a bit hard to interpret, but if we take lL = 0 as a special case, we get M = lRL/2 and so XC = lRL2/3M = 2L/3

  8. Review of system ‘momentous’ quantities • assume N3 and central forces: double sum zero!!

  9. The idea of the ‘center of gravity’ • 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’ • the total external force is clearly • what about moment of force? Use ri = RC + r’i • for gravity force moment, gravity acts at c.om c.o.g. IS the c.o.m.

  10. The science of statics • system feels no net force: FEXT = 0  system momentum is constant  c.o.m does not accelerate, and we assume that it is not translating either • system feel no net force moment : MEXT = 0  system angular momentum is constant, and we 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!!

  11. 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. Example: the massive lever 250 N 100 N 50 kg S d 4 m • 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 • Moments: 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’ L W FR FL

  12. 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 Example: the building sign 30 cm q • Take + to be UP, to RIGHT, and IN • put all forces where they act (c.o.g.) • 3:4:5q = 37°; sin q = .60; cosq = .80 • Forces UP: Scosf– W + T sin q = 0 • Forces RIGHT: S sin f– Tcosq = 0 • put origin at left so S does not appear in equation – clever!! • Moments IN: 0S + .20∙200 – .4T sin 143° = 0 Fargo 40 cm f T S q W

  13. Not in book: the idea of stability in gravity • as can tips, c.o.g. rises: stable • once c.o.g. is directly above pivot: tipping point! • from then on, object tips over and c.o.g. descends

  14. Torque as distinct from/related to force moment • a system of point masses (not necessarily a rigid body) with positions relative to O written ri feeling external forcesFi,ext(again, we assume internal forces are central and cancel in pairs) • R is position vector relative to O of a second origin O’ • positions of the masses relative to O’ : r’i:= ri– R • express the force moments using ri = r’i– R • so, if there is no external force on the system, the external force moment is origin-independent!! • if this is the case, we call it the torque T

  15. Review of rigid body dynamics • now let the second origin be at the c.o.m.  RC • The system external force causes the c.o.m. to accelerate – this is also manifest in the moment equaton • the external force moments cause the body to change its angular moment um with respect to (‘about’) the c.o.m. • if the system external force is zero, the c.o.m. does not accelerate, and all that happens is rotational acceleration about the c.o m. We therefore have, in this case,

  16. Kinetic energy in terms of moment of inertia • a rigid body is fixed to an axis, not necessarily through its c.o.m. and is currently turning at angular speed w • treat it as a collection of point masses: with the origin on the axis (normal to the screen), vi= wsiwhere si is the distance from the axis to mass i • IA is called the moment of inertia or rotational inertia • it is the rotational analog to the mass – it depends on the masses; also quadratically on the distances from the axis • It is crucial to bear in mind that the axis must be described clearly – and it may or may not even penetrate the body

  17. How to calculate the moment of inertia • the moment of inertia captures the resistance of the body to being angularly accelerated about that axis • it is the rotational analog to the mass – it depends on the masses; also quadratically on the distances from the axis Example of calculation of the moment of inertia Find the moment of inertia for a long thin bar of mass M and length L about an axis through its c.o.m. and perpendicular to the bar x O

  18. Example of calculation of the moment of inertia Find the moment of inertia for a hoop of mass M and radius R about an axis through its c.o.m. and along the hoop’s axis This one is trivial since all of the matter is at the same distance from the axis: hoops are special!! Example of calculation of the moment of inertia Find the moment of inertia for a solid cylinder of mass M and radius R about an axis through its c.o.m. and along the axis

  19. Kinetic energy of c.o.m. and referred to c.o.m.: the parallel-axis theorem • a second axis through c.o.m. point, parallel to the first • let l be the displacement vector from first to c.o.m. axis (perpendicular to both axes, as always in geometry) • re-express origin-to-axis vector si = l + ri where ri is the perpendicular vector from c.o.m. axis to mass i

  20. Parallel-axisand perpendicular-axis theorems • interpret this by comparing to the previous expression for K in terms of IA and w • there is also a perpendicular-axis theorem • for a thin plate-like object lying in xy plane, let Izz be the moment of inertia for an axis through a point along z z y • for an axis along x or y, the distances are trivial: just y2 or x2 respectively x

  21. 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 d, O • axis through the c.o.m. perpendicular to the stick; origin at c.o.m. • L = L1 + L2 and both masses contribute an identical angular momentum; total magnitude is md2w/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 = ICw in this case

  22. What happens if we alter the axis? • put axis at x, not d/2; still ┴ • angular velocity vector w points along the axis by RHR O • note now that because the c.o.m. is accelerating, there must be a net horizontal force on the system… the thing is wobbly… but there is still no need for a net force moment • construct L = r x p for left mass • get same result for right mass • clearly L CHANGES as dumbell whirls around and L is NOT parallel to w: L ≠ ICw L O r p • we need a torque in this case!! FEXT is however zero

  23. 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. What is I for stick? Where is center of mass of system? Put origin at left end: What is I for left mass? Use parallel axis theorem and IC for ball:

  24. Example IB O Question: can we regard this K as the Kofc.o.m. + Kreferred toc.o.m.? 11.21 J + 10,50 J = 21.71 J YES!!

  25. Example II with a shocking result?!! O FH A heavy door (1.25 m x 2.0 m) is hung by hinges at one end. Its mass is 40 kg. A force of 20 N is applied on the outer edge in a direction normal to the door. Find the angular acceleration of the door and the force of the hinges on the door during the acceleration. Hint: put an origin at the hinges b F What is net moment on the door as seen from the origin? the c.o.m translates due to net force: a = ab/2 = .75 m/s2 THE HINGES ALSO PUSH TO THE RIGHT ON THE DOOR!

  26. Example III 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 acceleration of the hoop, and its mass. Make FBDs for both objects. Put origin at axle. Choose UP & IN  + S T O T W W

  27. Kinematics of rolling motion I • Any object of circular cross-section • c.o.m. is at the center for smooth rolling • no sliding (so if a ≠ 0, need static friction) • let v = center of mass speed w(t) • in one rotation, c.o.m. andpoint of contact (p.o.c.) move a distance of one circumference 2pR • time is rotation period: T = distance/speed = 2pR/v •  v = 2pR/T = wR: same formula as for linear speed as related to angular speed for a turning object, where R is distance from axis!! • p.o.cappears to move… but that bit of roller can’t slide

  28. Kinematics of rolling motion II w(t) • c.o.m. moves at speed v = wR, as established • p.o.c. cannot move unless slippage: v = 0 • thus p.o.c. acts as an instantaneous axis of rotation, that slides along at speed v • top of roller is distance 2R from the instantaneous axis, so it moves forward at speed 2v (!!) • rotation (left) • translation (mid) • both t+r (right)

  29. Kinematics of rolling motion III w(t) • c.o.m. moves forward at v in steady motion • a dot on the edge moves on a path cycloid • for an instant at the very bottom, it stops • at the top, it is moving forward at 2v • top of roller is distance 2R from the instantaneous axis, so it moves forward at speed 2v (!!) • what does a flanged wheel do??

  30. Energetics of rolling motion I • a solid ball [IC = (2/5)MR2] rolls down a ramp of height h • as it speeds up, it is both translating and rotating • find c.o.m. speed at bottom • two approaches A and B give same result: different origins • A: origin at p.o.c.  pure rotation • by parallel-axis thm, IA = (2/5)MR2 +MR2 = (7/5)MR2 • B: origin at c.o.m.  translation + rotation

  31. Energetics of rolling motion II • The ball rolls down in gravitational field • Two forces act, N and W Only W does work • W has a potential energy: U = Mgy = Mgh • Since E is conserved, in the process DK = –DU • assume ball released from rest, so • thus, in a race, a fricitionless slider beats a roller • hoops, having the largest I, fare the worst • static friction is necessary to provide the net external moment that allows angular acceleration {show Active Figure 1-_26}

  32. Example III revisited A mass of 5 kg hangs from a string that is wrapped several times around a hoop (radius 60 cm) of mass 7.5 kg with a horizontal frictionless fixed axle. Find the speed when the mass has fallen 2.0 m by conserving energy. 5 If it were a different object--say, a solid cylinder--the answer would be different. In that case M would appear as M /2 – do you see why?

  33. conservation of energy is assumed • falling mass is m, spool radius is r, time of fall of m is t, distance of fall of m is h • final velocity of falling mass is v • final angular velocity of object is w = v/r The lab equation example

  34. A taste of angular momentum conservation 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) • 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.

  35. 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. IC • second term in big parenthesis is parallel-axis part • L is conserved because the external force moment is zero!!

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