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Chapter 10: Rotation of a Rigid Body about a Fixed Axis Chapter 10 Goals: • to reintroduce ourselves to the rotational analogues of position, velocity and acceleration • to understand the limitations inherent in the fixed axis assumption • to remind ourselves of the connections between the rotation of a body, and the translational kinematics of a point of the body • to define the kinetic energy of rotation and thereby introduce the rotational analogue of the mass: the moment of inertia • to utilize the parallel axis theorem • to define the (axis-dependent) torque associated with a force and to understand the relationship between torque and angular acceleration • to appreciate the kinematics of rolling motion Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rotational Kinematics about a Fixed Axis • here is a body which we want to be turning about an axis • rotation axis is a line “normal” to page through the origin O • we overlay an xy coordinate system • a spot (fiducial point) on the body moves from A to B • the line from O to the fiducial point will move through an angular displacement Dq : = qf – qi in a time Dt = tf – ti Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rotational Kinematics about a Fixed Axis 3 Warnings (1) if the axis’s direction is fixed in space but the axis moves (e.g. rolling motion), then the body is translating as well as rotating!! (2) in fact, if the axis is fixed, only points ON the axis are not translating (3) if the axis’s direction is not fixed in space then one cannot even DEFINE Dq as a simple number!! It’s even more strange than a vector!! It is a tensor and you need nine numbers in 3d to express it!! r arc length s • The radian: q : s circle radius r q • To convert: p rad = 180 ° Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Angular Velocity and Acceleration • given Dq : = qf – qi in a time Dt = tf – ti • define average angular velocity: angular displaceme nt Dq rad sec 1 avg : time interval Dt sec • of course, this is the slope of the line on a graph of q (t) that connects (qi , ti) to (qf , tf ) • now we define instantaneous angular velocity: Dq dq : lim : slope of the tangent t o q (t) graph Dt 0 Dt dt D d d 2q • Angular avg : and : 2 Dt dt dt accelerations: • obviously, these are slopes of the (t) graph • units: [||] = rad/sec2 = sec–2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Analogy Is Pretty Darn Good • with a fixed axis, it is as if it is a 1d kinematics problem • quantities can be positive or negative, of course • a useful alternative to radians as the angle unit is the cycle, or revolution: 2p radians = 1 revolution • 1 cycle per second := 1 Hz [Hertz] • 1 Hz = 60 revolutions per minute = 60 rpm • angular speed := magnitude of angular velocity = || • with the period T as the time for one cycle, =2p/T The Case of Constant Angular Acceleration • needless to say, we recover the ‘big five’: constant q q i i t 12 t 2 q q i 12 i t i t 2 i2 2 q q i Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Connection between Linear and Angular Velocities for a Fiducial Point on a Turning Object • If a point is a distance r from the v 2d axis line, the point moves in a circle of that radius r • and since the object rotates at (t), the average speed of the point is vavg circumfere nce 2pr avg r time to go around once T • notice how the rad unit sort of disappears… • this is also true on an instantaneous basis, even if is a function of time: v(t) = (t) r • velocity vector is tangent to the circular path in space • we can say that v r tˆ or vt r and vr 0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Connection between Linear and Angular Accelerations for a Fiducial Point on a Turning Object • velocity vector is tangent to the path v • how does it change with time? In two at ways!! Its length may be changing, and its direction is obviously changing too!! • the acceleration of that point has both ar components, tangential and radial: length of v changes changing speed ( angular accelerati on) dv d r d tangential accelerati on is at : r : r dt dt dt direction of v changes centripeta l accelerati on radial accelerati on is ar 2 r Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinetic Energy of Rotation • here is a rotating body, turning about a fixed axis (along z) at angular speed • the axis does not have to pass through the body, nor must it be along z!! • if the axis passes through the CM, that’s fun • think of the body as being made up of a collection of point masses.. a system • ri is the distance of mi from the AXIS, not from the origin (‘tho figure implies otherwise) • vi is the speed of mi so vi = ri rotational kinetic energy is K R 12 mi vi2 12 mi 2 ri 2 12 m r 2 2 ii Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Moment of Inertia rotational kinetic energy is K R 12 mi vi2 12 mi 2 ri 2 12 m r 2 2 ii that summation is called the moment of inertia... I : m r 2 ii so we can write KR 12 I 2 in the form like 12 mv2 • • • • for a collection of point masses one can do the sum pretty easily remember: r is the distance from the 1d axis, not from an origin!! if the system is an extended body (an object) then we have to think! break the body up into a very large number of very tiny masslets I : m r 2 ii 2 dm r where the big challenge is to write dm body • To express dm we need to define some kind of ‘mass density’ which may be 1d, 2d or 3d • examples will clarify the ideas… Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Parallel-Axis Theorem • imagine a second axis, parallel to the first, through the CM, separated by a ‘distance vector’ D (a ‘line-to-parallel-line’distance) • let the vectors {ri} be expressed as ri = r’i + D where r’I is measured from the CM axis to mi I m r 2 m r 'i D r 'i D m r '2 D 2 m 2D m r '1 ii i i i i i I CM MD 2 2D the first moment of the mass as seen from CM I CM MD 2 0 the parallel - axis theorem • one implication is that the CM moment of inertia is the smallest possible one for a given axis direction • Another cool fact is that we can ‘split up’ KR: K R 12 I 2 12 I CM MD 2 2 12 I CM 2 12 MD 2 2 2 but here vCM D K R 12 I CM 2 12 MvCM Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How to Calculate a Moment of Inertia Find the moment of inertia for a stick of length L and mass M for an axis normal to the stick and through the CM Solution Place the stick on the x-axis with origin at CM The stick has a linear mass density l = M/L and so a thin ‘slice’ of stick of infinitesimal thickness dx has an infinitesimal mass dm = l dx The thin slice’s distance from the axis is x, so I CM : 2 dm r body L/2 2 l x dx L/2 M L L/2 x 2 dx L/2 3 L/2 Mx 3L O dx ML2 23 M L/2 1 L2 12 8 Suppose the axis is normal to the stick but at one end… Then D = L/2 and by P.A.T we get I I CM MD 2 1 12 ML M 2 L 2 2 ML2 1 1 12 4 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 1 3 ML2 x 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 I CM 2 r dm body 2 2 R dm R body 2 dm MR body • This one is trivial since all of the matter is at the same distance from the axis: hoops are special and have the greatest possible ICM!! • Sometimes, even though the body is (say) 3d, you can get away with a 1d (or 2d) integral, by exploiting simpler results Example of calculation of the moment of inertia Find the moment of inertia for a solid cylinder of mass M and radius R and length L, about an axis through its CM and along the cylinder’s symmetry axis. Here, we take dm to be a very thin cylindrical shell of radius r and thickness dr and length L. The volume mass density of the stuff is r = Mass/Volume = M/(pR2L), and the infinitesimal mass has a volume 2prL so dm = 2prLr dr which becomes dm = 2Mr dr/R2 R 2M 3 2Mr I CM : dm r 2 r dr 2 R 4 R body 0 2 4 R 12 MR 2 0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. New Concepts: the Line of Action, and the Moment Arm, of a Vector • an origin-dependent quantity: given a vector, we have the line of action of the vector in space • the distance d from the origin to the line of action of the vector quantity is called the moment arm of the vector • we are here applying the notion to the force • we will see soon enough how to apply it to the momentum • it might be ‘obvious’ that if you want to turn something about an axis through O, you’ll succeed better for a given force F if you maximize the associated d. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. We Want to Maximize the ‘Force of Twist’ by Using the Wrench as Wisely as Possible • r is the position vector from the origin O to the point of application of the force F • the line of action of F is also shown • any component of the force ║ to r will do nothing to help turn the nut • this component is F cos f, where f is the angle between F and r tailtail • the component of the force ┴ to r is what really does the job • this component is F sin f • notice that the moment arm is d, where d = r sin f • conclusion: the ‘force of twist’ is force x the moment arm, or force x distance x sin f Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A Definition: the Cross Product of Two Vectors • the vector product of a vector with a vector: the cross(vector(outer(wedge))) product A × B • the cross product of two vectors is itself a vector!! magnitude of A B is A B A B sin q direction of A B is normal to plane of A and B • which normal? Use Right Hand Rule: swing A into B through the smaller tail-tail angle, according to how your right hand operates; its thumb points along A x B •so we have a completely new ‘thing’, in a completely new direction, derived from A and B, which couldn’t be captured by the dot product • A x B = sort of ‘signed’ area of parallelogram spanned by A and B; a measure of their ‘perpendicularness’ Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. More about the cross product of two vectors A |A|sin q q B • magnitude of cross product is magnitude of B times ‘moment’ of A [ |A| sin q ] from perpendicular projection of A along line of B Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Definition of the Torque of a Force F d = |r| sin q r r d = |r| sin q q tip-to-tail, where F lives q F tail-to-tail, in standard position • the moment arm is d = |r| sin q, with units of length • t = r × F [|t|] = N-m [NOT Joules! USA: foot-pounds] • called the torque ; greek letter tau • also called the moment of the force • note: the moment of the force is perpendicular to the force • the magnitude is t = (force)(moment arm) = Fd • direction is given by R.H.R. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. O Net Torque about an Origin for a Fixed Axis • the net torque about an origin is the moment of the net force • tnet = S ri × Fi in an obvious defintion • what we care about here is the component of the vector net torque along the fixed axis, which is associated with the tangential component of the net force (the radial component of the forces {Fi} contribute no torque since they are parallel to ri) • for now focus on a single point mass ... • so since at = r and Fnet,t = mat, (recall: r = distance from axis to point that is accelerating tangentially), we get for the component of the vector torque along the axis as follows: t net rFnet,t rmat r 2m : I • looks for all the world like Fnet = ma, doesn’t it? • if the axis is fixed (or not changing direction, more generally), the analogy is pretty darned good.. • so net torques cause angular accelerations about a fixed axiss!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What if the Net Torque about an Origin for a Fixed Axis is Zero? F 13 mm F • these are the forces applied to a bolt head by a wrench • note that the net force is zero: a so-called force couple • there are other forces applied to the bolt by the thing that the bolt is being tightened into • suppose the specification says to ‘torque’ the 13 mm bolt to 70 N-m • the 13 mm bolt has a ‘face dimension’ of 13mm(1/√3) ≈ 7.5 mm • put the origin at the center of the bolt’s face • each force therefore has a moment arm of half of that: 3.75 mm • for the torque to be 70 N-m, we get t 70 N - m t 2 Fs F 9300 N -3 2s (2)(3.75 x 10 m) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example A heavy bank vault door of mass M requires a force applied at its edge of 200 N to accelerate it so that it slams shut in 4 sec after turning F through an angle of 90°. The door is w = 1.5 r meters wide and h = 2.5 meters tall. a) Find the needed angular acceleration b) Find the moment of inertia of the door about an axis through the hinge in terms of the unknown mass Solution c) Find that unknown mass hint: put the axis at the hinge line!! [Why??] Note that F ┴ r F 2q p rad q q i i t t q 0 0 t 2 t 16 s 2 1 2 1 2 2 P.A.T. I I CM MD 2 • • • • 1 12 2 Mw M 2 w 2 2 13 Mw2 Now, what are the forces on the door as seen from the top? F is the push on the edge, with w = moment arm from the axis There is also a horizontal force at the hinge! It turns out to not be equal to F (because the CM of the door has to accelerate too) But the torque associated with the hinge force is zero!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Solution Continued t net t net Fw sin 90 1 I I 3 Mw 2 3Fw 3(200 N) (600 N)(16 s 2 ) / M 2040 kg 2/ 2 p 16 rad/s 1.5 m p rad 1.5 m w • • • • • • can we find the size of the other force, applied at the hinge? yes! The CM accelerates at at = r, where r = w/2 at = (p/16)(rad/s2)(.75 m) = .147 m/s2 therefore, since Fnet = MaCM = 300 N = 200 N + Fhinge amazingly, the hinge pushes on the door in the SAME direction as F, with a force of 100 N (!!) This actually makes some sense since the attempt to close the door in part makes the door want to rotate about its CM which means the hinge edge of the door pushes against the hinges, so by N3 the hinges push back on the door Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Examples that Synthesize All This Stuff: 10.8&10.11 Fp F Fg r A bar of length L and mass M is initially horizontal and supported at its left end by a frictionless pivot. The end of the bar is released from rest. Here is the FBD. a) Find the initial angular acceleration of the bar, and the initial linear acceleration of the CM, and the initial angular acceleration of the right tip of the bar. b) When the bar is vertical, what is its angular velocity, and what is the velocity of its CM? Solution to (a) Put the origin/axis at the pivot, so the pivot force has a zero moment arm The weight force’s moment arm is L/2 t net I mg aCM L mgL mgL/ 3g 2 2 2I 2L 2 mL 3 L 3 g 2 4 but aTIP 3 L g !!! 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Solution to (b) Bar is initially motionless and its height is zero, so initial energy is Emech = U + K = 0 vCM At bottom, bar is turning at and its CM has speed vCM, and its CM height is yCM = − L/2 Emech but vCM MgL 1 1 MgL 1 2 MvCM I CM 2 I PIVOT 2 2 2 2 2 2 L 2v MgL MgL 1 1 2v or CM Emech ML2 CM 2 L 2 2 23 L Emech MgL 2 3 3 2 2 MvCM 0 vCM gL vCM gL 2 3 4 2 A third question: what point (distance d from the pivot) on the bar is moving at √(2gL), which is the speed of a point mass that has fallen a distance L? 2v so since CM 3g / L L 2 therefore we must have 2 gL d 3g / L d L .816 L 3 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 2 Example: the Atwood Machine I m1 Solution Based on Energy Methods m2 • Assume system starts from rest and denote the heights of rising/falling masses to be zero initially • obviously, the pulley does not rise or fall • so initially Emech,i = Ki + Ui = 0 • now release masses.. 1 drops and 2 climbs a distance h, while system picks up linear/angular speed too mechanical energy is now Emech, f m2 m1 gh 1 m1 m2 v 2 1 I 2 2 2 with pulley mass M and radius R, we have I MR 2 and v / R 1 M 2 2 mechanical energy conservati on 0 m2 m1 gh 2 m1 m2 v 2 v 2m1 m2 gh 2 a m1 m2 gh v m1 m2 m2 m1 gh v m1 m2 M 2 m m M 2 1 2m2 M • one can also analyze using Newtonian methods T1 m1 g m m M • put an origin at the axis of the pulley and note = a/R 2 1 • there are two tensions, T1 and T2 now, as well 2m1 M T m g 2 • three FBDs three equations relating T1 and T2 and a 2 m1 m2 M 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 1 Work Considerations in Rotation I • we can work by analogy, almost.. work = force × distance so this smells like work = torque × angular displacement • more generally, dW = F·ds where ds is an infinitesimal displacement • assume the body is pivoting about a fixed axis, so ds = r dq [where r is distance to point of application of F from axis, again, and dq is an infinitesimal rotation angle of the body. ds r F dq • let the angle between F and ds (tail-tail) be dW = F cos ds by the usual usage of the dot product • since ds ┴ r the angle between r and F and r is f = 90° − Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work Considerations in Rotation II • but sin(90° − ) = cos F • so dW = F ds sin(90° − ) = ds F r sin(90° − ) dq 90° − • so dW = F (moment arm) dq r dq • dW = t dq • the analogy looks good and we can push it much further W t dq but only for a fixed axis! ! • if the torque is the net external torque then we can say more! • so, summing over the system, tnet = S text = I I d/dt Wnet t net dq I d dq 1 dq I d I d D I 2 K f K i dt dt 2 • thus we have the rotational analog of the work-energy theorem Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinematics of rolling motion I • Any object of circular cross-section (t) • CM is at the center for smooth rolling • no sliding (so if ≠ 0, need static friction) • let v = CM speed • in one rotation, CM and point of contact (p.o.c.) move a distance of one circumference 2pR • time is rotation period: T = distance/speed = 2pR/v • v = 2pR/T = R: same formula as for linear speed as related to angular speed for a turning object, where R is distance from axis!! • p.o.c appears to move… but that bit of roller can’t slide so instead we realize that there is no relative motion between the two surfaces (if there were, there’d be a lot of grinding and smoke!!) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. (t) Kinematics of rolling motion II • CM moves at speed v = R, as established • p.o.c. cannot move unless slippage: vbot = 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) + = Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. • translation (mid) • both t+r (right) (t) Kinematics of rolling motion III • 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?? Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Does it matter where the origin is placed? • a trivial example: a single force F acts on a body • let t and t’ be referred to origins O and O’ r’ F O’ r τ r F τ : t ( F )( moment arm) ( F )(b) Fb b O τ ' r 'F τ ' : t ' ( F )( moment arm) ( F )(0) 0 • clearly here the torque depends on the choice of origin • But… suppose the net force on the body is zero.. so there MUST be at least two forces acting on the body if there is any torque at all F1 • So take the case of a pair of forces that satisfy F1 = − F2 , and let D be the displacement vector from the point of application of F1 to that of F2 • The net force on the body is zero aCM = 0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. D F2 What Controls whether Torque is Origin-independent •let t1 and t2 be referred to origin O τ net,ext r1 F1 r2 F2 F1 r1 τ1 τ 2 r1 F1 r1 D F1 D r2 F2 r1 F1 r1 F1 D F1 D F1 and that is origin - independen t!! O • The force couple seems to contain only one force, but the other one is lurking in the pair’s torque • the torque is independent of the origin the net force on the body is zero the CM obeys N1 so body must be rotating about the CM, while the CM may be in steady translational motion (or may be stationary) • if the net force on the body is not zero, then we know how the CM moves, and we know how it rotates about the CM: dP where P p i dt dL τ net,ext ri Fi ,ext where L ri p i dt as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing Fnet,ext Fi ,ext