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**Physics 203 College Physics I Fall 2012**

S. A. Yost Chapter 8 Part 2 Rotational Motion

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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.

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**Rotational Analogs of Linear Motion**

Linear Motion Rotational Motion Kinematics: x , v, a q, w, a Dynamics: m, F, Kt I, t, Kr F = ma t = Ia Kt = ½ mv Kr = ½ Iw 2 W = Fx W = t q P = F v P = tw

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**Torque t = R F^ = R F sin q . 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 .

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**Torque t = R ^ F t = R F^ = R F sin q 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

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**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.

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Question A 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 N I = mR2

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**Question t = RF = I a = mR2 a. = = → = F2 = 2 F1 = 2 N.**

t R2F mR22 a F R2 t R1F mR12 a F R1 I = mR2

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**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? R M a

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**Example FT a mg Isolate the hanging mass: Newton’s Law:**

M a = Fnet = Mg – FT where FT is the tension in the rope. FT M a mg

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Example m Isolate 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. R a FT

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**Example FT M a Combine results: R M a = Fnet = Mg – FT = Mg – ½ ma.**

Then (M + m/2) a = Mg. Result: a = R FT M 1 + m/2M g a

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

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Rigid Body Motion The 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.

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Dumbbell m → → F 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 A m → F → → Dp = Ft B

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Dumbbell m → → F 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 A m → F B

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**Dumbbell F F The total kinetic energy is Case A:**

→ F 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 in case A. A m → F B

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Rolling When an object rolls, its circumference moves a distance 2pr every period, so w and v are related: v = 2pr/T = rw 2pr 2pr 2pr 2pr

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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?

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Rolling If 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,I h L

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**√ Rolling v = Energy conservation: Ui = Ktrans + Krot**

mgh = ½ mv2 + ½ Iw2. Rolling: w = v/R. mgh = ½ (m + I/R2) v2. √ h L m,I v = 2gh 1 + I/(mR2)

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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.

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**Rotational Analog of Momentum**

Linear Motion: (one dimension) Momentum: p = mv Impulse: Dp = Ft Rotational Motion: (fixed axis) Angular momentum: L = Iw DL = t t

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**Angular Momentum Units of angular momentum:**

L = I w = [kg m2][s-1] = kg . m2/s DL = t 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.

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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 ∙ m2. (a) What was her final moment of inertia?

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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 ∙ m2. I1w1 = I2 w2 w2 = 2.5 w1 I2 = I1 / 2.5 = 1.84 kg ∙ m2 ≈ 1.8 kg ∙ m2

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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 ∙ m2. (b) What average power did she apply to pull in her arms?

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Figure Skater P = 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.

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**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? 2R m M R w0

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**Analog of Inelastic Collision**

Angular momentum is conserved: I1w1 = I2 w2 . I1 =½ MR2 = mR2 and I2 = Ibar + Idisk with Ibar = mL2/12 = mR2/3, Idisk = ½ MR2 = mR2 I2 = 4mR2/3 w2= ( I1 / I2) w1 = ¾ w1 = ¾ (120 rpm) = 90 rpm. m R wf M

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**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 /dt w → q → → w Right-Hand Rule

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