1 / 30

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 Torque Example | Whiteboard Moment of Inertia D D Example | Whiteboard Torque and Moment of inertia Example | Whiteboard.

iliana
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 • Torque • Example | Whiteboard • Moment of InertiaDD • Example | Whiteboard • Torque and Moment of inertia • 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 - Tangential Relationships Linear: (m) s (m/s) v (m/s/s) a Tangential: (at the edge of the wheel) = r - Displacement = r - Velocity = r - Acceleration* *Not in data packet TOC

  4. 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 Angular:  = /t  = /t*  = o + t  = ot + 1/2t2 2 = o2 + 2  = (o + )t/2*  = I *Not in data packet TOC

  5. Angular Mechanics - Torque Push farther out: F F Push with more force: TorqueA twisting force that can cause an angular acceleration. So how do you increase torque? or TOC

  6. Angular Mechanics - Torque F r TorqueA twisting force that can cause an angular acceleration.  = rxF If r = .5 m, and F = 80 N, then  = (.5m)(80N)  = 40 Nm. Torque is also in foot pounds (Torque Wrenches) TOC

  7. Angular Mechanics - Torque  TorqueA twisting force that can cause an angular acceleration. F  = rxF = rFsin Fsin r Only the perpendicular component gives rise to torque. (That’s why it’s sin) TOC

  8. Example: What’s the torque here?  = 56o F = 16 N r = 24 cm Ok – This is as tricky as it can be: use  =rFsin, r = .24 m, F = 16 N But  is not 56o. You want to use the angle between r and F which is 90 – 56 = 34o Finally,  = (.24 m)(16 N)sin(34o) = 2.1 Nm TOC

  9. Whiteboards: Torque 1 | 2 | 3 TOC

  10. What is the torque when you have 25 N of force perpendicular 75 cm from the center of rotation? •  =rFsin • = (.75 m)(25 N)sin(90o) = 18.75 Nm = 19 Nm W 19 Nm

  11. If you want 52.0 Nm of torque, what force must you exert at an angle of 65.0o to the end of a .340 m long wrench?  =rFsin 52 Nm = (.34 m)(F) sin(65o) F =(52Nm)/((.34m) sin(65o))=168.75 N F = 169 N W 169 N

  12. The axle nut of a Volkswagen requires 200. foot pounds to loosen. How far out (in feet) do you stand on the horizontal wrench if you weigh 145 lbs?  =rFsin 200. ft lbs = (r)(145 lbs) sin(90o) r = (200. ft lbs)/(145 lbs) = 1.3793 feet r = 1.38 feet (1’ 41/2”) W 1.38 feet

  13. Angular Mechanics – Moment of inertia Moment of Inertia - Inertial resistance to angular acceleration. Question - If the blue masses were identical, would both systems respond identically to the same torque applied at the center? Demo TOC

  14. Angular Mechanics – Moment of inertia r F F = ma - We can’t just use “m” for “I”  = I (The position of “m” matters!) • F = ma • rF = mar (Mult. by r) • rF = m(r)r (a = r) • = (mr2) = I ( = rF) So I = mr2 for a point mass, r from the center. TOC

  15. Angular Mechanics – Moment of inertia Some parts are far from the axis Some parts are close to the axis What about a cylinder rotating about its central axis? In this case, I = 1/2mr2 (You need calculus to derive it) TOC

  16. In general, you look them up in your book. Demos: TP, Solids TOC

  17. Three main ones: ½mr2 - Cylinder (solid) mr2 - Hoop (or point mass) 2/5mr2 – Sphere (solid) TOC

  18. Example:Three 40. kg children are sitting 1.2 m from the center of a merry-go-round that is a uniform cylinder with a mass of 240 kg and a radius of 1.5 m. What is its total moment of inertia? The total moment of inertia will just be the total of the parts: Children – use mr2 (assume they are points) MGR – use 1/2mr2 (solid cylinder) I = 3((40kg)(1.2m)2) + ½(240kg )(1.5m)2 I = 442.8 kgm2 = 440 kgm2 TOC

  19. Whiteboards: Moment of Inertia 1 | 2 | 3 TOC

  20. What is the moment of inertia of a 3.5 kg point mass that is 45 cm from the center of rotation? I = mr2 I = (3.5kg)(.45m)2 = .70875 kgm2 I = .71 kgm2 W .71 kgm2

  21. A uniform cylinder has a radius of 1.125 m and a moment of inertia of 572.3 kgm2. What is its mass? I = ½mr2 572.3 kgm2 = ½m(1.125 m)2 m = 2(572.3 kgm2)/(1.125 m)2 m = 904.375 = 904.4 kg W 904.4 kg

  22. A sphere has a mass of 45.2 grams, and a moment of inertia of 5.537 x 10-6 kgm2. What is its radius? (2 Hints) m = .0452 kg I = 2/5mr2 5I/(2m) = r2 r = (5I/(2m))1/2 r = (5(5.537x10-6kgm2)/(2(.0452kg)))1/2 r = .0175 m W .0175 m

  23. Angular Mechanics – Torque and moment of inertia Now let’s put it all together, we can calculate torque and moment of inertia, so let’s relate them: The angular equivalent of F = ma is: F = ma  = I TOC

  24. Example: A string with a tension of 2.1 N is wrapped around a 5.2 kg uniform cylinder with a radius of 12 cm. What is the angular acceleration of the cylinder? • I = ½mr2 = ½(5.2kg)(.12m)2 = 0.03744 kgm2 •  = rF = (2.1N)(.12m) = .252 Nm • And finally, •  = I, •  = /I = (.252 Nm)/(0.03744 kgm2) = 6.7 s-2 TOC

  25. Whiteboards: Torque and Moment of Inertia 1 | 2 | 3 | 4 | 5 TOC

  26. What torque is needed to accelerate a 23.8 kgm2 wheel at a rate of 388 rad/s/s?  = I  = (23.8 kgm2)(388 rad/s/s) I = 9234.4 Nm = 9230 Nm W 9230 Nm

  27. An object has an angular acceleration of 23.1 rad/s/s when you apply 6.34 Nm of torque. What is the object’s moment of inertia?  = I (6.34 Nm) = I(23.1 rad/s/s) I = (6.34 Nm)/(23.1 rad/s/s) = .274 kgm2 W .274 kgm2

  28. If a drill exerts 2.5 Nm of torque on a .075 m radius, 1.75 kg grinding disk, what is the resulting angular acceleration? (1 hint)  = I, I = ½mr2 I = ½mr2 = ½(1.75 kg)(.075 m)2 = .004922 kgm2  = I, 2.5 Nm = (.004922 kgm2)  = /I = (2.5 Nm)/(.004922 kgm2) = 510 s-2 W 510 s-2

  29. What torque would accelerate an object with a moment of inertia of 9.3 kgm2 from 2.3 rad/s to 7.8 rad/s in .12 seconds? (1 hint)  = I,  = o + t 7.8 rad/s = 2.3 rad/s + (.12 s)  = (7.8 rad/s - 2.3 rad/s)/(.12 s) = 45.833s-2  = I = (9.3 kgm2)(45.833s-2) = 426.25 Nm  = 430 Nm W 430 Nm

  30. A merry go round is a uniform solid cylinder of radius 2.0 m. You exert 30. N of force on it tangentially for 5.0 s and it speeds up to 1.35 rad/s. What’s its mass? (1 hint)  = rF, I = 1/2mr2 , = o+t,  = I  = 1.35 s-1/5.0 s = .27 s-2  = rF = 30.N(2.0m) = 60 Nm  = I, I=/ (60Nm)/(.27s-2)=222.2kgm2 I = 1/2mr2, m = 2I/r2 m = 2(222.2kgm2)/(2.0m)2 = 111.1111 kg m = 110 kg W 110 kg

More Related