Presentation on theme: "Physics 203 College Physics I Fall 2012"— Presentation transcript:
1 Physics 203 College Physics I Fall 2012 S. A. YostChapter 8 Part 2Rotational Motion
2 AnnouncementsProblem 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 Rotational Analogs of Linear Motion Linear Motion Rotational MotionKinematics: x , v, a q, w, aDynamics: m, F, Kt I, t, KrF = ma t = IaKt = ½ mv Kr = ½ Iw 2W = Fx W = t qP = F v P = tw
4 Torque t = R F^ = R F sin q . The torque is defined to be the perpendicularcomponent of the forcetimes the distance fromthe pivot to where it acts:FF^Rqt = R F^where F^ = F sin q.Counter-clockwise torqueis considered to bepositive, as for angles.Then= R F sin q .
5 Torque t = R ^ F t = R F^ = R F sin q The torque can also be expressed in terms of themagnitude of the force andthe distance from the axisto the line of the force.The distance R ^ is calledthe lever arm of the torque.FRqR^t = R ^ Ft = R F^= R F sin q
6 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 BC DThe lever arms are the same.
7 QuestionA force is applied to the rim of two wheels. Assuming the only significant mass is in the rim, what force F2 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 NI = mR2
8 Question t = RF = I a = mR2 a. = = → = F2 = 2 F1 = 2 N. t R2F mR22 a F R2t R1F mR12 a F R1I = mR2
9 Example a Mass falling on rope wrapped around a massive pulley. R Assume the pulley is a uniform disk as shown.What is the acceleration of the hanging mass?RMa
10 Example FT a mg Isolate the hanging mass: Newton’s Law: M a = Fnet = Mg – FTwhere FT is the tension in the rope.FTMamg
11 ExamplemIsolate the pulley: t = I a with I = ½ m R2, t = RFT , a = a/R.Then RFT = (½ mR2)(a/R).Therefore, FT = ½ ma.RaFT
12 Example FT M a Combine results: R M a = Fnet = Mg – FT = Mg – ½ ma. Then (M + m/2) a = Mg.Result: a =RFTM1 + m/2Mga
13 ExampleF2Note that the tension does not need to be the same on two sides of a massive pulley.Net torque =R(F1 – F2) = Ia.amRRF1
14 Rigid Body MotionThe relation t = I a 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 mvcm2, K rot = ½ Iw 2. “Newton’s Law” F = m acm (ch. 7), t = I a.
15 Dumbbellm→→FA 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 sameAm→F→ →Dp = FtB
16 Dumbbellm→→FA 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 sameAm→FB
17 Dumbbell F F The total kinetic energy is Case A: →FThe total kinetic energy isCase A:K = Ktrans + Krot = ½ mvcm2 + ½ Iw2Case B: no rotation:K = ½ mvcm2There is more energy in case A.Am→FB
18 RollingWhen an object rolls, its circumference moves a distance 2pr every period, so w and v are related:v = 2pr/T = rw2pr2pr2pr2pr
19 RollingA solid wheel and a hollow wheel roll down a ramp, starting from rest at the same point.Which gets to the bottom faster?
20 RollingIf an object with mass m and moment of inertia I rolls down an inclined plane of height h and length L, how fast is it rolling when it gets to the bottom?m,IhL
21 √ Rolling v = Energy conservation: Ui = Ktrans + Krot mgh = ½ mv2 + ½ Iw2.Rolling: w = v/R.mgh = ½ (m + I/R2) v2.√hLm,Iv =2gh1 + I/(mR2)
22 RollingThe 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 Rotational Analog of Momentum Linear Motion:(one dimension)Momentum:p = mvImpulse: Dp = FtRotational Motion:(fixed axis)Angular momentum:L = IwDL = t t
24 Angular Momentum Units of angular momentum: L = I w = [kg m2][s-1] = kg . m2/sDL = t t = [mN][s] = Nms = J.sWhen there is no external torque on a system, angular momentum is conserved.In particular, this applies to collisions between rigid bodies.
25 Figure SkaterA 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 ∙ m2. (a) What was her final moment of inertia?
26 Figure SkaterA 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 ∙ m2. I1w1 = I2 w2 w2 = 2.5 w1 I2 = I1 / 2.5 = 1.84 kg ∙ m2≈ 1.8 kg ∙ m2
27 Figure SkaterA 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 ∙ m2. (b) What average power did she apply to pull in her arms?
28 Figure SkaterP = W/t, W = DK = ½ Iw22 – ½ Iw12. w1 = 1.0 rev/s (2p rad/rev) = 2.0 p rad/s w2 = 2.5 rev/s (2p rad/rev) = 5.0 p rad/s I1 = 4.6 kg ∙ m2, I 2 = 1.84 kg ∙ m2 W = 277 J – 90.8 J ≈ 186 J, t = 1.5 s. P = 124 W.
29 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?2RmMRw0
31 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 q and w are parallel, but they won’t be if the axis direction changes.→q→ →w = d q /dtw→q→ →wRight-Hand Rule