Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 8 Part 1 Rotational Motion

2. Announcements • A problem set HW07B was due today. • Today: Rotational Motion, Ch. 10. sec. 1 – 6 • Next: sec. 7 – 8, skipping sec. 9. • Homework set HW08 on sections 1 – 6 will be due next Thursday. • Sections 7 – 8 will be combined with chapter 9, sec. 1, 2, and 4 in a later set, HW09.

3. Motion of the Center of Mass • If an external force F acts on an extended object or collection of objects of mass M, the acceleration of the CM is given by • F= Macm. • You can apply Newton’s 2nd Law as if it were a particle located at the CM, as far as the collective motion is concerned. • This says nothing about the relative motion, rotation, etc., about the CM. That comes up in chapter 8.

4. Motion of Extended Objects • The motion of extended objects or collections of particles is such that the CM obeys Newton’s 2nd Law.

5. Motion of the Center of Mass • The CM of a wrench sliding on a frictionless table will move in a straight line because there is no external force. In this sense, the wrench may be though of as a particle located at the CM. cm motion

6. Motion of the Center of Mass • For example, if a hammer is thrown, its CM follows a parabolic trajectory under the influence of gravity, as a point object would.

7. Motion of the Center of Mass • For example, if a hammer is thrown, its CM follows a parabolic trajectory under the influence of gravity, as a point object would.

8. Rotational Motion • Everything we have done for linear motion has an analog for rotational motion. • We will consider only fixed axis rotations. • Rotational displacements can be expressed in various units: radians are most standard, but degrees and revolutions are common as well. • Conversions: 1 rev = 2prad = 360o, 180o = p rad.

9. Rotational Kinematics • The physics of rotations of a rigid body about a fixed axis is in many ways analogous to the one-dimensional motion of a point particle. v x q w

10. Rotational Kinematics Greek letters are used for rotational quantities… Linear Motion: distance x velocityv acceleration a Rotational Motion: angleq angular velocity w angular acceleration a

11. Constant (Angular) Acceleration Linear Motion: v = v0 + at x = x0 + v0t + ½ at2 v2 = v02 + 2a (x – x0) Rotational Motion: w = w0 + a t q = q0+ w0t + ½ a t2 w2 = w02 + 2a (q–q0)

12. Example • A centrifuge accelerates from rest to 20,000 rpm in 5.0 min. • (a) What is its average angular acceleration? • (b) How many times does it spin during these five minutes?

13. Example • (a) a = w/ t. t = 300 s. • w = 2pf =2p (20,000 rev / 60 s) = 2094 rad/s • a = (2094 rad/s) / 300 s = 6.98 rad/s2 • (b) Revolutions = favg t • = ½ fmaxt = 10,000 rpm × 5 min • = 50,000 rev. • Or, useq= ½at2to get the same result with more effort. 20,000 rpm f 10,000 rpm 0 t 5 min

14. Rotational Energy Consider a massless rotating disk with small masses miinserted in it. The total kinetic energy is K = S ½ mivi2 . v = rwfor an object a distance rfrom the axis. We’d like to express K in terms of w. w m1 m2 r1 r2 r5 r3 m5 r4 m3 m4

15. Rotational Kinetic Energy • In terms of the angular velocity, K = S ½ mivi2 = ½ Smiri2w2 = ½ w2 Smiri2 K = ½ I w2 • I is called the • moment of inertia. I =Smiri2Units: kg m2. w m1 m2 r1 r2 r5 r3 m5 r4 m3 m4

16. Moment of Inertia For example, the moment of inertia of a hoop of radius Rand mass Mabout an axis through the center is I = MR2 since all parts are the same distance from the center. M R

17. Moments of Inertia Thin Hoop Disk R R I = MR2 I = ½ MR2 Thin Solid Rod Sphere L I =ML2 /12 I = 2/5 MR2

18. Solid vs Hollow Sphere • If a solid and hollow sphere have the same mass and size, which has a bigger moment of inertia about an axis through the center? • A: solid • B: hollow • C: the same • The hollow one, since the mass is distributed further from the axis. Solid 2/5 MR2 Hollow 2/3 MR2

19. Parallel Axis Theorem • The moment of inertia about any axis can be found if its value ICM is known about an axis through the CM. • Ih = ICM + Mh2 • The moment of inertia is always the smallest about an axis through the CM. M CM h

20. Moment of Inertia of Rod • What is the moment of inertia of uniform rod of mass Mand length L about the end? • Iend = ICM+ M(L/2)2 • ICM = ML2/12 • Iend= ML2/12 + ML2/4 • = ML2/3 CM h = L/2

21. Rotational Work An object can be given kinetic energy by doing work: DK = W To do work, I can apply a force F at a point given, relative to the axis, by a vector R. The force does work through a distance d = Rq. F R d q F Only the component of Fperp. to R does work: W = FR q. ┴

22. Torque • The product t =FR is called torque. • If f is the angle between R and F, t= RF sin f • Linear motion: W = F x • Rotational motion: W = t q. F W = FR q. f R F _ Power: W = Pt, P = tw.

23. Torque • Equivalent expression: • If R is the component of R perpendicular to F, then • t =FR . • R is the distance of the line of the force from the axis, and is called the lever arm of the force. F t = FR =RF sinf f R R

24. Torque The torque is defined to be the perpendicular component of the force times the distance from the pivot to where it acts: • F F^ R q t= R F^ where F^ = F sin q. Counter-clockwise torque is considered to be positive, as for angles. Then • = R F sin q.

25. Torque The torque can also be expressed in terms of the magnitude of the force and the distance from the axis to the line of the force. The distance R ^ is called the lever arm of the torque. • F R q R^ t= R ^ F t= R F^ • = R F sin q

26. Tightening a Nut with a Wrench • Which use of the wrench is most effective for tightening the nut? • Which is least effective? • Which of A and D is more effective? • Choose E if they are the same. A B C D The lever arms are the same.

27. Rotational Dynamics Linear Motion: mass m forceF = ma kinetic energy Kt = ½ mv2 work Wt = Fx Rotational Motion: moment of inertiaI torque t = I a kinetic energy Kr = ½ Iw2 workWr = tq power Pr = tw powerPt= Fv

28. Example • Mass falling on rope wrapped around a massive pulley. • Assume the pulley is a uniform disk as shown. • What is the acceleration of the hanging mass? m R M a

29. Example • Isolate the hanging mass: • Newton’s Law: • M a = Fnet = Mg – FT • where FTis the tension in the rope. FT M a mg

30. Example • Isolate the pulley: t = I a with I = ½ m R2, t = RFT, a= a/R. • ThenRFT = (½ mR2)(a/R). • Therefore, FT = ½ ma. m R a FT

31. Example • Combine results: • M a = Fnet = Mg – FT • = Mg – ½ ma. • Then (M + m/2) a = Mg. • Result: a = m R g FT 1 + m/2M M a

32. Example • Note that the tension does not need to be the same on two sides of a massive pulley. • Net torque = • R(F1– F2) = Ia. F2 a m R R F1

33. Rigid Body Motion • The relation t = Ia holds for rigid body rotation in any inertial frame. • This always holds in the CM frame of the rigid body, even if it is accelerating. • The energy of a rigid body can be expressed as a sum K = K cm + Krot with • K cm = ½ mvcm2, K rot = ½ Iw2. • “Newton’s Law” F = m acm(ch. 9), t = Ia.

34. m m m m Dumbbell → • A force F is applied for time t to a dumbbell in one of two ways shown. • Which gives the greater speed to the center of mass? • (a)A(b)B • (c) the same → F A → F → → Dp= Ft B

35. m m m m Dumbbell → • A forceF is applied for time tto a dumbbell in one of two ways shown. • Which gives the greater energy to the dumbbell? • (a)A(b)B • (c) the same → F A → F B

36. m m m m Dumbbell → • The total kinetic energy is • Case A: • K= Ktrans+ Krot= ½ mvcm2+ ½ Iw2 • Case B:no rotation: • K = ½ mvcm2 • There is more energy incase A. F A → F B

37. m m Dumbbell Energy • How much more energy does it have if I push at the top? • The bottom mass is initially at rest, so the speed of any point on the dumbbell is initially proportional to its distance from the bottom mass. 2vcm vcm v = 0

38. m m Dumbbell Energy • The initial kinetic energy is then • Ktotal= ½ m (2vcm)2= 2mvcm2. • This is conserved as the dumbbell rotates. • Kcm = ½ (2m)vcm2 = mvcm2 • Krot= Kcm= ½ Ktotal • There is twice as much total energy. 2vcm vcm v = 0