Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 8 Part 3 Chapter 9 Angular Momentum, Rotational Equilibrium

2. Announcements • Today: chapter 8, sec. 7,8 • chapter 9, sec. 1, 2, and 4 . • Tuesday: Election day, no class • Next Thursday: Problem set HW09 Due • Read Ch. 10, sec. 1 – 7 (a lot of these are short) • Fluids: Density, Pressure, and Buoyancy

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

4. 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?

5. Rolling • If an object with mass m and moment of inertiaI 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

6. Rolling • Energy conservation: • Ui= Ktrans+ Krot • mgh= ½ mv2 + ½Iw2. • Rolling: w= v/R. • mgh= ½ (m + I/R2) v2. • √ h L m,I v = 2gh 1 + I/(mR2)

7. 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.

8. 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

9. 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.

10. 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?

11. Figure Skater • A figure skater increases her rotational rate from 1.0 rev/s to 2.5 rev/sin 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

12. 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?

13. Figure Skater • P = W/t, W = DK = ½ Iw22 – ½ Iw12. • w1 = 1.0 rev/s (2prad/rev) = 2.0 prad/s • w2= 2.5 rev/s (2prad/rev) = 5.0 prad/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.

14. Equilibrium • An object in equilibrium can be at rest, moving at constant velocity, and/or rotating with constant angular velocity. • The condition for equilibrium is that the sum of all forces must be zero, and the sum of all torques due to the forces must also be zero. • You can calculate the sum of torques about any axis.

15. See-Saw • Suppose a see-saw has negligible mass, but two rocks are on the ends. What is the mass of the rock on the right? The length on the left is 3 times that on the right. m2 F 1kg m2g m1g

16. See-Saw • Torque about pivot: • x2 m2g + x1 m1g = 0 • x1 = –3 x2 • m2 – 3 m1 = 0 • m2 = 3 kg. m2 F 1kg m2g x1 x2 m1g

17. Balancing Torques • Suppose a 1 kg rock hangs on a meter-long rod that balances 25 cm from one end. • What is the mass of the rod? A 0.25 kg B 0.5 kg C1kg D 2 kg E 4 kg mg 1kg

18. Balancing Torques • What is the force of each support on the meter stick, assuming the stick is firmly attached to each and its mass is 1 kg. FA A B FB mg mg 1kg

19. Balancing Torques • Torques about B: • (1m)mg + (0.5m)mg – (0.5 m) FA = 0 • 0.5 FA = 1.5 mg FA = 3mg =29.4N FA A B FB mg mg 1kg

20. Balancing Torques • Torques about A: • (0.5 m) mg – (0.5 m) FB = 0 • FB = mg FB = 9.8 N 3mg FA A mg B FB mg mg 1kg The forces also balance.

21. Sign Problem A 75 lb sign is supported by a 6-foot long, 25 lb beam and a wire. How high must the wire be attached if its tension cannot be more than 100 lb? h L/2 L/4 L/4 Savannah River Seafood L =6.0 ft

22. Sign Problem Show all forces on the bar. T h F Weights act at the center of gravity. L/2 L/4 L/4 W1W2 L = 6.0 ft T = 100 lb W1 = 25 lb W2 = 75 lb Isolate the bar – don’t consider forces on anything else.

23. Sign Problem What is the force F of the wall on the beam? Balancing forces: Fx = -Tx = Tcosq = = 72.7 lb. y h F T f q x L/2 L/4 L/4 W1W2 25 lb 75 lb Balance torques about the end: LFy – ½ LW1 – ¼ L W2 = 0 F = 79.1 lb f= 23.3o • Fy = ½ W1 + ¼ W2 = 31.25 lb

24. Question • Which of the boxes shown will tip over if the center of gravity is at the point shown?

25. Hint • The contact force Fcmust be equal and opposite the weight W. • When the box is about to tip, there is only one contact point, the lower corner. Fc W

26. Question • In which case(s) is impossible for the CG to be above a point on the box where the contact force could act?

27. Hanging A Sign q1 q2 Kitty Klub The sign is 1.20 m wide. q1 = 30o, q2 = 60o. How far from the left edge is the center of mass?

28. Hanging A Sign → → T1 T2 q1 q2 Kitty Klub → mg If three non-parallel forces act on an object in equilibrium, their lines must all meet at a point.

29. Hanging A Sign → → T1 T2 q1 q2 Kitty Klub → mg If three non-parallel forces act on an object in equilibrium, their lines must all meet at a point.

30. Hanging A Sign x L – x q1 q2 Kitty Klub L = 1.20 m q1 = 30o q2= 60o x = 3(L – x) 4 x = 3 L x = 3L/4 = 0.90 m x tan q1 = (L – x) tan q2 x / √3 = (L – x) √3

31. Pushing a Box • Suppose I want to push the box shown across a floor at constant speed, and it has coefficient of friction mk= 0.36. The box’s center of mass is at its center. What is the highest point I can push the box horizontally to accomplish this without tipping it over? F w w = 18 cm h = 84 cm h y

32. Pushing a Box • Forces balance: F = Ffr = mkFN, FN = mg . • F = mkmg • Gravity acts at the center. • When about to tip, the contact forces act at the leading edge. F w w = 18 cm h = 84 cm h y mg FN Ffr mk= 0.36

33. Pushing a Box • F = mkmg . • Balance torque about the front edge: yF = mg w/2. • mkmg y = mg w/2 • y = w/(2mk)= (18 cm)/ (0.72) = 25 cm. F w w = 18 cm h = 84 cm h y mg FN Ffr mk= 0.36