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Angular Mechanics - Torque and moment of inertia Contents: Review Linear and angular Qtys

Angular Mechanics - Torque and moment of inertia Contents: Review Linear and angular Qtys Tangential Relationships Angular Kinematics Rotational KE Example | Whiteboard Rolling Problems Example | Whiteboard. Angular Mechanics - Angular Quantities. Linear: (m) s (m/s) u

celestyn-leighton
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Angular Mechanics - Torque and moment of inertia Contents: Review Linear and angular Qtys

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  1. Angular Mechanics - Torque and moment of inertia • Contents: • Review • Linear and angular Qtys • Tangential Relationships • Angular Kinematics • Rotational KE • Example | Whiteboard • Rolling Problems • Example | Whiteboard

  2. Angular Mechanics - Angular Quantities Linear: (m) s (m/s) u (m/s) v (m/s/s) a (s) t (N) F (kg) m Angular:  - Angle (Radians) o - Initial angular velocity (Rad/s)  - Final angular velocity (Rad/s)  - Angular acceleration (Rad/s/s) t - Uh, time (s)  - Torque I - Moment of inertia TOC

  3. Angular Mechanics - Angular kinematics Linear: s/t = v v/t = a u + at = v ut + 1/2at2 = s u2 + 2as = v2 (u + v)t/2 = s ma = F 1/2mv2 = Ekin Fs = W Angular:  = /t  = /t*  = o + t  = ot + 1/2t2 2 = o2 + 2  = (o + )t/2*  = I Ek rot = 1/2I2 W = * TOC *Not in data packet

  4. Angular Mechanics - Useful Substitutions  = I  = rF so F = /r = I/r s = r, so  = s/r v = r, so  = v/r a = r, so  = a/r TOC

  5. Angular Mechanics - Rotational Ke Two types of kinetic energy: Of course a rolling object has both IP Demo Rolling.ip Translational: Ekin = 1/2mv2 Rotational: Ek rot = 1/2I2 TOC

  6. Example: What Energy does it take to speed up a 23.7 kg 45 cm radius cylinder from rest to 1200 RPM? TOC

  7. Whiteboards: Rotational KE 1 | 2 | 3 TOC

  8. What is the rotational kinetic energy of an object with an angular velocity of 12 rad/s, and a moment of inertia of 56 kgm2? Ek rot = 1/2I2 Ek rot = 1/2(56 kgm2)(12 rad/s)2 Ek rot = 4032 J = 4.0 x 103 J W 4.0 x 103 J

  9. What must be the angular velocity of a flywheel that is a 22.4 kg, 54 cm radius cylinder to store 10,000. J of energy? (hint) Ek rot = 1/2I2, I = 1/2mr2 Ek rot = 1/2(1/2mr2)2 = 1/4mr22 2 = 4(Ek rot)/mr2 =(4(Ek rot)/mr2)1/2=(4(10000J)/(22.4kg)(.54m)2)1/2  = 78.25 rad/s = 78 rad/s W 78 rad/s

  10. What is the total kinetic energy of a 2.5 cmdiameter 405 g sphere rolling at 3.5 m/s? (hint) I=2/5mr2, = v/r, Ek rot=1/2I2 , Ekin=1/2mv2 Ek total= 1/2mv2 +1/2I2 Ek total= 1/2mv2 +1/2(2/5mr2)(v/r)2 Ek total= 1/2mv2 +2/10mv2 = 7/10mv2 Ek total= 7/10mv2 = 7/10(.405 kg)(3.5m/s)2 Ek total= 3.473 J = 3.5 J W 3.5 J

  11. Angular Mechanics – Rolling with energy h m r - cylinder I = 1/2mr2  = v/r mgh = 1/2mv2 + 1/2I2 mgh = 1/2mv2 + 1/2(1/2mr2)(v/r)2 mgh = 1/2mv2 + 1/4mv2 = 3/4mv2 4/3gh = v2 v = (4/3gh)1/2 TOC

  12. Whiteboards: Rolling with Energy 1 | 2 | 3 TOC

  13. A 4.5 kg ball with a radius of .12 m rolls down a 2.78 m long ramp that loses .345 m of elevation. Set up the energy equation without plugging any of the knowns into it. Make substitutions for I and , but don’t simplify. I = 2/5mr2,  = v/r mgh = 1/2mv2 + 1/2I2 mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2 W mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2

  14. Solve this equation for v: mgh = 1/2mv2 + 1/2(2/5mr2)(v/r)2 mgh = 1/2mv2 + 2/10mr2v2/r2 mgh = 1/2mv2 + 2/10mv2 = 7/10mv2 10/7gh = v2 v = (10/7gh)1/2 W v = (10/7gh)1/2

  15. A 4.5 kg ball with a radius of .12 m rolls down a 2.78 m long ramp that loses .345 m of elevation. What is the ball’s velocity at the bottom? (v = (10/7gh)1/2) v = (10/7(9.8m/s/s)(.345 m))1/2 = 2.1977 m/s v = 2.20 m/s W 2.20 m/s

  16. A 4.5 kg ball with a radius of .12 m rolls down a 2.78 m long ramp that loses .345 m of elevation. What was the rotational velocity of the ball at the bottom? (v = 2.1977 m/s) • = v/r = (2.1977 m/s)/(.12 m) = 18.3 s-1 • = 18 s-1 W 18 rad/s

  17. A 4.5 kg ball with a radius of .12 m rolls down a 2.78 m long ramp that loses .345 m of elevation. What was the linear acceleration of the ball down the ramp? (v = 2.1977 m/s) v2 = u2 + 2as v2/(2s) = a (2.1977 m/s)2/(2(2.78 m)) = .869 m/s/s W .869 m/s/s

  18. In General: I tend to solve all rotational dynamics problems using energy. • Set up the energy equation • (Make up a height) • Substitute linear for angular: •  = v/r • I = ?mr2 • Solve for v • Go back and solve for accelerations TOC

  19. Angular Mechanics – Pulleys and such r h m1 m2 Find velocity of impact, and acceleration of system r = 12.5 cm m1 = 15.7 kg m2 = .543 kg h = .195 m TOC

  20. Angular Mechanics – Pulleys and such r h = (made up) m1 m2 Find acceleration of system r = 46 cm m1 = 55 kg m2 = 15 kg m3 = 12 kg h= 1.0 m m3 TOC

  21. Angular Mechanics – yo yo ma h= 1.0 m Find acceleration of system (assume it is a cylinder) r1 = 6.720 cm r2 = .210 cm m= 273 g r1 r2 TOC

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