Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 8 Part 1 Rotational Motion

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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 = Ma cm. You can apply Newton’s 2 nd 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.

Physics 203 – College Physics I Department of Physics – The Citadel Motion of Extended Objects The motion of extended objects or collections of particles is such that the CM obeys Newton’s 2 nd Law.

Physics 203 – College Physics I Department of Physics – The Citadel 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

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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 = 2  rad = 360 o, 180 o =  rad.

Physics 203 – College Physics I Department of Physics – The Citadel 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. x v  

Physics 203 – College Physics I Department of Physics – The Citadel Rotational Kinematics Linear Motion: distance x velocity v acceleration a Rotational Motion: angle  angular velocity  angular acceleration  Greek letters are used for rotational quantities…

Physics 203 – College Physics I Department of Physics – The Citadel Constant (Angular) Acceleration Linear Motion: v = v 0 + at x = x 0 + v 0 t + ½ at 2 v 2 = v a (x – x 0 ) Rotational Motion:  =  0 +  t  0  +  0 t + ½  t 2  2 =   –  0 

Physics 203 – College Physics I Department of Physics – The Citadel 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?

Physics 203 – College Physics I Department of Physics – The Citadel Example (a)  / t. t = 300 s.  f =  (20,000 rev / 60 s) = 2094 rad/s  = (2094 rad/s) / 300 s = 6.98 rad/s 2 (b) Revolutions = f avg t = ½ f max t = 10,000 rpm × 5 min = 50,000 rev. Or, use  = ½  t 2 to get the same result with more effort. 0 t 5 min f 20,000 rpm 10,000 rpm

Physics 203 – College Physics I Department of Physics – The Citadel Rotational Energy Consider a massless rotating disk with small masses m i inserted in it. The total kinetic energy is K =  ½ m i v i 2. v = r  for an object a distance r from the axis. We’d like to express K in terms of .  r4r4 m2m2 m3m3 m4 m4 m5 m5 m1m1 r1r1 r2r2 r3r3 r5r5

Physics 203 – College Physics I Department of Physics – The Citadel Rotational Kinetic Energy In terms of the angular velocity, K =  ½ m i v i 2 = ½  m i r i 2  2 = ½  2  m i r i 2 K = ½ I  2 I is called the moment of inertia moment of inertia. I =  m i r i 2 Units: kg m 2.  r4r4 m2m2 m3m3 m4 m4 m5 m5 m1m1 r1r1 r2r2 r3r3 r5r5

Physics 203 – College Physics I Department of Physics – The Citadel Moment of Inertia For example, the moment of inertia of a hoop of radius R and mass M about an axis through the center is  = MR 2 since all parts are the same distance from the center. R M

Physics 203 – College Physics I Department of Physics – The Citadel Moments of Inertia Thin Hoop Disk Thin Solid Rod Sphere I = MR 2 R R L I = ½ MR 2 I =ML 2 /12 I = 2/5 MR 2

Physics 203 – College Physics I Department of Physics – The Citadel 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 Hollow 2/5 MR 2 2/3 MR 2

Physics 203 – College Physics I Department of Physics – The Citadel Parallel Axis Theorem The moment of inertia about any axis can be found if its value I CM is known about an axis through the CM. I h = I CM + Mh 2 The moment of inertia is always the smallest about an axis through the CM. CM M h

Physics 203 – College Physics I Department of Physics – The Citadel Moment of Inertia of Rod What is the moment of inertia of uniform rod of mass M and length L about the end? I end = I CM + M(L/2) 2 I CM = ML 2 /12 I end = ML 2 /12 + ML 2 /4 = ML 2 /3 CM h = L/2

Physics 203 – College Physics I Department of Physics – The Citadel Rotational Work An object can be given kinetic energy by doing work:  K = 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 = R . F R  d Only the component of F perp. to R does work: W = F R . ┴ FF

Physics 203 – College Physics I Department of Physics – The Citadel Torque The product  F  R is called torque. If  is the angle between R and F,  = RF sin  Linear motion: W = F x Rotational motion: W . R W = F  R . F FF  Power: W = Pt, P = . _

Physics 203 – College Physics I Department of Physics – The Citadel Torque Equivalent expression: If R  is the component of R perpendicular to F, then  F R . R  is the distance of the line of the force from the axis, and is called the lever arm of the force. R  = F  R  RF sin  F RR 

Physics 203 – College Physics I Department of Physics – The Citadel F Torque R   = R F sin . The torque is defined to be the perpendicular component of the force times the distance from the pivot to where it acts:  = R F  FF where F  = F sin . Then Counter-clockwise torque is considered to be positive, as for angles.

Physics 203 – College Physics I Department of Physics – The Citadel F Torque R   = R F sin  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.  = R F  RR  = R  F

Physics 203 – College Physics I Department of Physics – The Citadel 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. The lever arms are the same. A B C D

Physics 203 – College Physics I Department of Physics – The Citadel Rotational Dynamics Linear Motion: mass m force F = ma kinetic energy K t = ½ mv 2 work W t = Fx Rotational Motion: moment of inertia  torque  kinetic energy K r = ½ I  2 work W r =  power P r =  power P t = Fv

Physics 203 – College Physics I Department of Physics – The Citadel 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 M R a

Physics 203 – College Physics I Department of Physics – The Citadel Example Isolate the hanging mass: Newton’s Law: M a = F net = Mg – F T where F T is the tension in the rope. M a mgmg FTFT

Physics 203 – College Physics I Department of Physics – The Citadel Example Isolate the pulley:  with I = ½ m R 2,  = RF T,  = a/R. Then RF T = (½ mR 2 )(a/R). Therefore, F T = ½ ma. m R FTFT a

Physics 203 – College Physics I Department of Physics – The Citadel Example Combine results: M a = F net = Mg – F T = Mg – ½ ma. Then (M + m/2) a = Mg. Result: a = m R FTFT a 1 + m/2M g M

Physics 203 – College Physics I Department of Physics – The Citadel Example Note that the tension does not need to be the same on two sides of a massive pulley. Net torque = R(F 1 – F 2 ) = . m R F1F1 F2F2 R 

Physics 203 – College Physics I Department of Physics – The Citadel Rigid Body Motion The relation  =   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 + K rot with K cm = ½ mv cm 2, K rot = ½  2. “Newton’s Law” F = m a cm (ch. 9),  =  

Physics 203 – College Physics I Department of Physics – The Citadel 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 m m m m F F A B  p = Ft → → → →

Physics 203 – College Physics I Department of Physics – The Citadel Dumbbell A force F is applied for time t to a dumbbell in one of two ways shown. Which gives the greater energy to the dumbbell? (a) A (b) B (c) the same m m m m F A B → F → →

Physics 203 – College Physics I Department of Physics – The Citadel Dumbbell The total kinetic energy is Case A: K = K trans + K rot = ½ mv cm 2 + ½  2 Case B: no rotation: K = ½ mv cm 2 There is more energy in case A. m m m m A B F → F →

Physics 203 – College Physics I Department of Physics – The Citadel 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. m m 2v cm v cm v = 0

Physics 203 – College Physics I Department of Physics – The Citadel Dumbbell Energy The initial kinetic energy is then K total = ½ m (2v cm ) 2 = 2m v cm 2. This is conserved as the dumbbell rotates. K cm = ½ (2m)v cm 2 = mv cm 2 K rot = K cm = ½ K total There is twice as much total energy. m m 2v cm v cm v = 0