Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 8 Part 3 Chapter 9 Angular Momentum, Rotational Equilibrium

Physics 203 – College Physics I Department of Physics – The Citadel 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

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

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?

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

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

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.

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

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.

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?

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

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?

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.

Physics 203 – College Physics I Department of Physics – The Citadel 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.

Physics 203 – College Physics I Department of Physics – The Citadel 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. See-Saw 1kg m1gm1gm1gm1g F m2gm2gm2gm2g m2m2m2m2

Physics 203 – College Physics I Department of Physics – The Citadel Torque about pivot: x 2 m 2 g + x 1 m 1 g = 0 x 1 = –3 x 2 m 2 – 3 m 1 = 0 m 2 = 3 kg. See-Saw 1kg m1gm1gm1gm1g F m2gm2gm2gm2g m2m2m2m2 x 1 x 2

Physics 203 – College Physics I Department of Physics – The Citadel 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? 1kg mg A 0.25 kg B 0.5 kg C 1kg D 2 kg E 4 kg

Physics 203 – College Physics I Department of Physics – The Citadel 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. 1kg AB mg mg FAFAFAFA FBFBFBFB

Physics 203 – College Physics I Department of Physics – The Citadel Balancing Torques Torques about B: (1m)mg + (0.5m)mg – (0.5 m) F A = F A = 1.5 mg 1kg AB mg mg FAFAFAFA FBFBFBFB F A = 3mg =29.4N

Physics 203 – College Physics I Department of Physics – The Citadel Balancing Torques Torques about A: (0.5 m) mg – (0.5 m) F B = 0 F B = mg 1kg AB mg mg FAFAFAFA FBFBFBFB F B = 9.8 N 3mg mg mg The forces also balance.

Physics 203 – College Physics I Department of Physics – The Citadel Sign Problem L =6.0 ft h L/2 L/4 L/4 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? SavannahRiverSeafood

Physics 203 – College Physics I Department of Physics – The Citadel Sign Problem Show all forces on the bar. T W 1 W 2 F L = 6.0 ft T = 100 lb W 1 = 25 lb W 2 = 75 lb h Weights act at the center of gravity. L/2 L/4 L/4 Isolate the bar – don’t consider forces on anything else.

Physics 203 – College Physics I Department of Physics – The Citadel Sign Problem y F T  h W 1 W 2 Torques about left end of beam: F does not contribute:  ½ L W 1 (cw)  (3/4) L W 2 (cw) L T sin  (ccw) L/2 L/4 L/4 Balance torques: ½ LW 1 + (3/4)LW 2 = LT sin  25 lb 75 lb 0.5 W W 2 = T si n  = 100 sin   = 43.4 o h = L tan  = 5.68 ft. x

Physics 203 – College Physics I Department of Physics – The Citadel Sign Problem y F T  h W 1 W 2 What is the force F of the wall on the beam? Balancing forces: F x =  T x = T cos  = = 72.7 lb. Balance torques about the end: LF y – ½ LW 1 – ¼ L W 2 = 0 25 lb 75 lb F y = ½ W 1 + ¼ W 2 = lb F = 79.1 lb  = 23.3 o  L/2 L/4 L/4 x

Physics 203 – College Physics I Department of Physics – The Citadel Question Which of the boxes shown will tip over if the center of gravity is at the point shown?

Physics 203 – College Physics I Department of Physics – The Citadel Hint The contact force F c must be equal and opposite the weight W. When the box is about to tip, there is only one contact point, the lower corner. FcFc W

Physics 203 – College Physics I Department of Physics – The Citadel Question In which case(s) is impossible for the CG to be above a point on the box where the contact force could act?

Physics 203 – College Physics I Department of Physics – The Citadel Hanging A Sign Kitty Klub The sign is 1.20 m wide.          How far from the left edge is the center of mass?    

Physics 203 – College Physics I Department of Physics – The Citadel Hanging A Sign Kitty Klub If three non-parallel forces act on an object in equilibrium, their lines must all meet at a point.     T 1 T 2 mgmgmgmg → → →

Physics 203 – College Physics I Department of Physics – The Citadel Hanging A Sign Kitty Klub If three non-parallel forces act on an object in equilibrium, their lines must all meet at a point.     T 1 T 2 mgmgmgmg → → →

Physics 203 – College Physics I Department of Physics – The Citadel Hanging A Sign Kitty Klub x tan  1 = (L – x) tan  2     x L – x L = 1.20 m         x / √3 = (L – x) √3 x = 3(L – x) 4 x = 3 L x = 3L/4 = 0.90 m

Physics 203 – College Physics I Department of Physics – The Citadel Pushing a Box Suppose I want to push the box shown across a floor at constant speed, and it has coefficient of friction  k = 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? w h y F w = 18 cm h = 84 cm

Physics 203 – College Physics I Department of Physics – The Citadel Pushing a Box Forces balance: F = F fr =  k F N, F N = mg. F =  k mg Gravity acts at the center. When about to tip, the contact forces act at the leading edge. w h y F w = 18 cm h = 84 cm  k = 0.36 FNFNFNFN F fr mg

Physics 203 – College Physics I Department of Physics – The Citadel Pushing a Box F =  k mg. Balance torque about the front edge: yF = mg w/2.  k mg y = mg w/2 y = w/(2  k ) = (18 cm)/ (0.72) = 25 cm. w h y F w = 18 cm h = 84 cm  k = 0.36 FNFNFNFN F fr mg