1 / 21

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

Dive into the world of Angular Mechanics with this extensive guide covering torque, moment of inertia, kinetic energy, useful substitutions, and real-world examples explained on whiteboards.

francesgarrett
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related