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UB, Phy101: Chapter 9, Pg 1 Physics 101: Chapter 9 l Today’s lecture will cover Textbook Sections 9.1 - 9.6.

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Presentation on theme: "UB, Phy101: Chapter 9, Pg 1 Physics 101: Chapter 9 l Today’s lecture will cover Textbook Sections 9.1 - 9.6."— Presentation transcript:

1 UB, Phy101: Chapter 9, Pg 1 Physics 101: Chapter 9 l Today’s lecture will cover Textbook Sections

2 UB, Phy101: Chapter 9, Pg 2 Rotation Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R See text: chapter 8 See Table 8.1

3 UB, Phy101: Chapter 9, Pg 3 New concept: Torque See text: chapter 9 Rotational analog of force Torque = (magnitude of force) x (lever arm)  = F l

4 UB, Phy101: Chapter 9, Pg 4 Comment on axes and sign (i.e. what is positive and negative) Whenever we talk about rotation, it is implied that there is a rotation “axis”. This is usually called the “z” axis (we usually omit the z subscript for simplicity). Counter-clockwise (increasing  ) is usually called positive. Clockwise (decreasing  ) is usually called negative. z 

5 UB, Phy101: Chapter 9, Pg 5 Chapter 9, Preflight The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which of the cases is the torque on the nut the biggest? 1. Case 1 2. Case 2 3. Case 3 CORRECT

6 UB, Phy101: Chapter 9, Pg 6 Chapter 9, Preflight The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which of the cases is the torque on the nut the smallest? 1. Case 1 2. Case 2 3. Case 3 CORRECT

7 UB, Phy101: Chapter 9, Pg 7

8 UB, Phy101: Chapter 9, Pg 8

9 UB, Phy101: Chapter 9, Pg 9 Static Equilibrium A system is in static equilibrium if and only if: l a cm = 0  F ext = 0 l  = 0   ext = 0 (about any axis) torque about pivot due to gravity:  g = mgd (gravity acts at center of mass) Center of mass pivot d W=mg This object is NOT in static equilibrium

10 UB, Phy101: Chapter 9, Pg 10 Center of mass pivot d W=mg Torque about pivot  0 Center of mass pivot Torque about pivot = 0 Not in equilibrium Equilibrium

11 UB, Phy101: Chapter 9, Pg 11 Homework Hints l Painter is standing to the right of the support B. FAFA FBFB Mg mg l What is the maximum distance the painter can move to the right without tipping the board off?

12 UB, Phy101: Chapter 9, Pg 12 Homework Hints l If its just balancing on “B”, then F A = 0 è the only forces on the beam are: FBFB Mg mg Using F TOT = 0: F B = Mg + mgThis does not tell us x x

13 UB, Phy101: Chapter 9, Pg 13 Homework Hints l Find net torque around pivot B: (or any other place) FBFB Mg mg  (F B ) = 0 since lever arm is 0  (Mg ) = Mgd 1 d1d1 d2d2  (mg ) = -mgd 2 Total torque = 0 = Mgd 1 -mgd 2 So d 2 = Md 1 /m and you can use d 1 to find x

14 UB, Phy101: Chapter 9, Pg 14 Homework Hints l Painter standing at the support B. FAFA FBFB Mg mg Find total torque about this axis D d  (F A ) = - F A D  (Mg) = Mgd  (F B ) = 0 (since distance is 0)  (mg) = 0 (since distance is 0) Total torque = 0 = Mgd -F A D So F A = Mgd /D

15 UB, Phy101: Chapter 9, Pg 15 l MORE EXAMPLES (bar and weights suspended by the string): Find net torque around this (or any other) place  (m 1 g) = 0 since lever arm is 0 x T Mg m2gm2g m1gm1g

16 UB, Phy101: Chapter 9, Pg 16  (Mg ) = -Mg L/2  (m 1 g) = 0 since lever arm is 0 L/2 T Mg m2gm2g m1gm1g

17 UB, Phy101: Chapter 9, Pg 17  (Mg ) = -Mg L/2  (m 1 g) = 0 since lever arm is 0 x T Mg m2gm2g m1gm1g  (T ) = T x

18 UB, Phy101: Chapter 9, Pg 18  (Mg ) = -Mg L/2  (m 1 g) = 0 since lever arm is 0 L T Mg m2gm2g m1gm1g  (T ) = T x  (m 2 g ) = -m 2 g L All torques sum to 0: Tx = MgL/2 + m 2 gL So x = (MgL/2 + m 2 gL) / T

19 UB, Phy101: Chapter 9, Pg 19

20 UB, Phy101: Chapter 9, Pg 20

21 UB, Phy101: Chapter 9, Pg 21

22 UB, Phy101: Chapter 9, Pg 22 Moment of Inertia & Rotational KE Textbook Sections :

23 UB, Phy101: Chapter 9, Pg 23 Torque and Stability Center of mass outside of base: --> unstable Center of mass over base: --> stable

24 UB, Phy101: Chapter 9, Pg 24

25 UB, Phy101: Chapter 9, Pg 25

26 UB, Phy101: Chapter 9, Pg 26 Moments of Inertia of Common Objects Hollow cylinder or hoop about central axis I = MR 2 Solid cylinder or disk about central axis I = MR 2 /2 Solid sphere about center I = 2MR 2 /5 Uniform rod about center I = ML 2 /12 Uniform rod about end I = ML 2 /3

27 UB, Phy101: Chapter 9, Pg 27

28 UB, Phy101: Chapter 9, Pg 28

29 UB, Phy101: Chapter 9, Pg 29 Chapter 9, Preflight The picture below shows two different dumbbell shaped objects. Object A has two balls of mass m separated by a distance 2L, and object B has two balls of mass 2m separated by a distance L. Which of the objects has the largest moment of inertia for rotations around the x-axis? 1. A 2. B 3. Same CORRECT x 2L L m m 2m A B I = mL 2 + mL 2 = 2mL 2 I = 2m(L/2) 2 + 2m(L/2) 2 = mL 2

30 UB, Phy101: Chapter 9, Pg 30 Rotational Kinetic Energy Translational kinetic energy: KE trnas = 1/2 MV 2 cm Rotational kinetic energy: KE rot = 1/2 I  2 Rotation plus translation: KE total = KE trans + KE rot = 1/2 MV 2 cm + 1/2 I  2

31 UB, Phy101: Chapter 9, Pg 31 Angular Momentum l Textbook Section 9.6

32 UB, Phy101: Chapter 9, Pg 32 See text: chapters 8-9 See Table 8.1 Define Angular Momentum MomentumAngular Momentum p = mV L = I  conserved if  F ext = 0conserved if  ext =0 VectorVector! units: kg-m/sunits: kg-m2/s

33 UB, Phy101: Chapter 9, Pg 33 Chapter 9, Pre-flights You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.

34 UB, Phy101: Chapter 9, Pg 34 Chapter 9, Preflight What happens to your angular momentum as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same L1L1 L2L2 This is like the spinning skater example in the book. Since the net external torque is zero (the movement of the arms and hands involve internal torques), the angular momentum does not change. CORRECT

35 UB, Phy101: Chapter 9, Pg 35 Chapter 9, Preflight 11 22 I2I2 I1I1 L L What happens to your angular velocity as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same as with the skater example given in the book....as you pull your arms in toward the rotational axis, the moment of inertia decreases, and the angular velocity increases. CORRECT My friends and I spent a good half hour doing this once, and I can say...based on a great deal of nausea, that the angular velocity does increase.

36 UB, Phy101: Chapter 9, Pg 36 Chapter 9, Preflight What happens to your kinetic energy as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same CORRECT Your angular velocity increases and moment of inertia decreases, but angular velocity is squared, so KE will increase with increasing angular velocity 11 22 I2I2 I1I1 L L (using L = I  )

37 UB, Phy101: Chapter 9, Pg 37 l Two different spinning disks have the same angular momentum, but disk 2 has a larger moment of inertia than disk 1. è Which one has the biggest kinetic energy ? (a) disk 1 (b) disk 2 Spinning disks

38 UB, Phy101: Chapter 9, Pg 38 If they have the same L, the one with the smallest I will have the biggest kinetic energy.  disk 2  disk 1 I 1 < I 2 (using L = I  )

39 UB, Phy101: Chapter 9, Pg 39 Preflights: Turning the bike wheel A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens? 1. She starts to spin CCW. 2. She starts to spin CW. 3. Nothing CORRECT

40 UB, Phy101: Chapter 9, Pg 40 Turning the bike wheel... l Since there is no net external torque acting on the student- stool system, angular momentum is conserved. è Remenber, L has a direction as well as a magnitude! LL Initially: L INI = L W,I LLL Finally: L FIN = L W,F + L S L L W,F LLSLLS L L W,I LLL L W,I = L W,F + L S

41 UB, Phy101: Chapter 9, Pg 41 Rotation Summary (with comparison to 1-d linear motion) AngularLinear See text: chapters 8-9 See Table 8.1


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