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Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 8 Part 2 Rotational Motion.

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Presentation on theme: "Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 8 Part 2 Rotational Motion."— Presentation transcript:

1 Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 8 Part 2 Rotational Motion

2 Physics 203 – College Physics I Department of Physics – The Citadel Announcements Problem set HW08 is due Thursday. It covers Chapter 8 except angular momentum, rolling, and some torque problems. Today: Rotational Motion, Ch. 10, mostly sec7 – 8. Thursday: chapter 9, sec. 1, 2, and 4. Please read these sections before class. A problem set HW09 on them and the remainder of Ch. 8 will be posted today and due next Tuesday. Deadline for making up Exam 2: Wed. 5 PM.

3 Physics 203 – College Physics I Department of Physics – The Citadel Rotational Analogs of Linear Motion Linear Motion Rotational Motion Kinematics: x, v, a  Dynamics: m, F, K t I, , K r F = ma  K t = ½ mv 2 K r = ½ I  2 W = Fx W =   P = F v P 

4 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.

5 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

6 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

7 Physics 203 – College Physics I Department of Physics – The Citadel Question A force is applied to the rim of two wheels. Assuming the only significant mass is in the rim, what force F 2 will give the wheels identical angular accelerations? (A) 0.25 N (B) 0.5 N (C) 1.0 N (D) 2.0 N (E) 4.0 N I = mR 2

8 Physics 203 – College Physics I Department of Physics – The Citadel Question  = RF = I  = mR 2 . = = → = F 2 = 2 F 1 = 2 N. I = mR 2    R 2 F 2 mR 2 2  F 2 R 2   R 1 F 1 mR 1 2  F 1 R 1

9 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

10 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

11 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

12 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

13 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 

14 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. 7),  =  

15 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 → → → →

16 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 → →

17 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 →

18 Physics 203 – College Physics I Department of Physics – The Citadel Rolling When an object rolls, its circumference moves a distance  r every period, so  and v are related: v = 2  r/T = r  2r2r 2r2r 2r2r2r2r

19 Physics 203 – College Physics I Department of Physics – The Citadel Rolling A solid wheel and a hollow wheel roll down a ramp, starting from rest at the same point. Which gets to the bottom faster?

20 Physics 203 – College Physics I Department of Physics – The Citadel Rolling If an object with mass m and moment of inertia  rolls down an inclined plane of height h and length L, how fast is it rolling when it gets to the bottom? m  Lh

21 Physics 203 – College Physics I Department of Physics – The Citadel Rolling Energy conservation: U i = K trans + K rot mgh = ½ mv 2 + ½  2. Rolling:  = v/R. mgh = ½ (m + I/R 2 ) v 2. √ v = 2gh 1 +  /(mR 2 ) m  Lh

22 Physics 203 – College Physics I Department of Physics – The Citadel Rolling The solid wheel gets to the bottom first, because the object with the smaller moment of inertia relative to its mass and size moves faster.

23 Physics 203 – College Physics I Department of Physics – The Citadel Rotational Analog of Momentum Linear Motion: (one dimension) Momentum: p = mv Impulse:  p = Ft Rotational Motion: (fixed axis) Angular momentum: L   L  t

24 Physics 203 – College Physics I Department of Physics – The Citadel Angular Momentum Units of angular momentum: L =  = [kg m2][s -1 ] = kg. m 2 /s  L  =  t = [mN][s] = Nms = J. s When there is no external torque on a system, angular momentum is conserved. In particular, this applies to collisions between rigid bodies.

25 Physics 203 – College Physics I Department of Physics – The Citadel Figure Skater A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m 2. (a) What was her final moment of inertia?

26 Physics 203 – College Physics I Department of Physics – The Citadel Figure Skater A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m 2.        2  2        kg ∙ m 2 ≈ 1.8 kg ∙ m 2

27 Physics 203 – College Physics I Department of Physics – The Citadel Figure Skater A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/s in 1.5 s. Her initial moment of inertia was 4.6 kg ∙ m 2. (b) What average power did she apply to pull in her arms?

28 Physics 203 – College Physics I Department of Physics – The Citadel Figure Skater P = W/t, W =  K = ½ I  2 2 – ½ I  1 2.    1.0 rev/s (2  rad/rev) = 2.0  rad/s    = 2.5 rev/s (2  rad/rev) = 5.0  rad/s I 1 = 4.6 kg ∙ m 2,    kg ∙ m 2 W = 277 J – 90.8 J ≈ 186 J, t = 1.5 s. P = 124 W.

29 Physics 203 – College Physics I Department of Physics – The Citadel Analog of Inelastic Collision A bar of length 2R is dropped onto a rotating disk of radius R. Suppose M = 2m. If the disk initially rotates at 120 rpm, how fast does it rotate if the stick drops onto it and rotates together with the disk? R  2R2R M m

30 Physics 203 – College Physics I Department of Physics – The Citadel Analog of Inelastic Collision Angular momentum is conserved:        2.    ½ MR 2 = mR 2   and  I 2 = I bar + I disk with  bar = mL 2 /12 = mR 2 /3, I disk = ½ MR 2 = mR 2 I 2 = 4mR 2 /3  2        ¾     ¾  (120 rpm) = 90 rpm  R ff M m

31 Physics 203 – College Physics I Department of Physics – The Citadel Angular Quantities as Vectors If the axis is not fixed, we have to specify a direction for angular displacements and velocities. The convention use a vector pointing along the axis. For fixed axis rotations, the vectors  and  are parallel, but they won’t be if the axis direction changes.    = d  /dt   → → → Right-Hand Rule


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