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Angular Mechanics - Kinematics Contents: Radians, Angles and Circles Linear and angular Qtys Conversions | WhiteboardConversionsWhiteboard Tangential Relationships.

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Presentation on theme: "Angular Mechanics - Kinematics Contents: Radians, Angles and Circles Linear and angular Qtys Conversions | WhiteboardConversionsWhiteboard Tangential Relationships."— Presentation transcript:

1 Angular Mechanics - Kinematics Contents: Radians, Angles and Circles Linear and angular Qtys Conversions | WhiteboardConversionsWhiteboard Tangential Relationships Example | WhiteboardExampleWhiteboard Angular Kinematics Example | WhiteboardExampleWhiteboard © Microsoft Encarta

2 Angular Mechanics - Radians r  s Full circle: 360 o = 2  Radians  = s/r Radians = m/m = ? TOC

3 Angular Mechanics - Angular Quantities Linear: (m) s (m/s) u (m/s) v (m/s/s) a (s) t Angular:  - Angle (Radians)  i - Initial angular velocity (Rad/s)  f - Final angular velocity (Rad/s)  - Angular acceleration (Rad/s/s) t - Uh, time (s) TOC

4 Conversions TOC Radians Revolutions Rad/s Rev/min (RPM) = rev(2  ) = rad/(2  ) = (rev/min)(2  rad/rev)(min/60s) = (rev/s)(2  rad/rev) = (rad/s)(60 s/min)(rev/2  rad)

5 Whiteboards: Conversions 11 | 2 | 3 | 4234 TOC

6 How many radians in 3.16 revolutions? rad = rev(2  ) rad = (3.16 rev)(2  ) = 19.9 rad 19.9 rad W

7 If a drill goes through 174 radians, how many revolutions does it go through? rev = rad/(2  ) rev = (174 rad)/(2  ) = 27.7 rev 27.7 rev W

8 Convert 33 RPM to rad/s rad/s = (rev/min)(2  rad/rev)(min/60s) = (33rev/min)(2  rad/rev)(min/60s) rad/s = 3.5 rad/s 3.5 rad/s W

9 Convert 12 rev/s to rad/s rad/s = (rev/s)(2  rad/rev) rad/s = (12 rev/s)(2  rad/rev) rad/s = 75 rad/s 75 rad/s W

10 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* TOC *Not in data packet

11 Example: s =  r, v =  r, a =  r A certain gyro spinner has an angular velocity of 10,000 RPM, and a diameter of 1.1 cm. What is the tangential velocity at its edge?  = (10,000rev/min)(2  rad/rev)(1 min/60 sec)  = 1047.19 s -1 r =.011m/2 =.0055 m v =  r = ( 1047.19 s -1 )(.0055 m ) v = 5.8 m/s (show ‘em!) (pitching machines) TOC

12 Whiteboards: Tangential relationships 11 | 2 | 3 | 4 | 5 | 623456 TOC

13 What is the tangential velocity of a 13 cm diameter grinding wheel spinning at 135 rad/s? v =  r, r =.13/2 =.065 m v = (135 rad/s)(.065 m) = 8.8 m/s 8.8 m/s W

14 What is the angular velocity of a 57 cm diameter car tire rolling at 27 m/s? v =  r, r =.57/2 =.285 m 27 m/s =  (.285 m)  = (27 m/s)/ (.285 m) = 95 rad/s 95 rad/s W

15 A.450 m radius marking wheel rolls a distance of 123.2 m. What angle does the wheel rotate through? s =  r 123.2 m =  (.450 m)  = (123.2 m)/(.450 m) = 274 rad 274 rad W

16 A car with.36 m radius tires speeds up from 0 to 27 m/s in 9.0 seconds. (a) What is the linear acceleration? v = u + at 27 m/s = 0 + a(9.0s) a = (27 m/s)/(9.0s) = 3.0 m/s/s 3.0 m/s/s W

17 A car with.36 m radius tires speeds up from 0 to 27 m/s in 9.0 seconds. (a) a = 3.0 m/s/s (b) What is the tire’s angular acceleration? a =  r (3.0 m/s/s) =  (.36 m)  = (3.0 m/s/s)/(.36 m) = 8.3333 Rad/s/s  = 8.3 Rad/s/s 8.3 Rad/s/s W

18 A car with.36 m radius tires speeds up from 0 to 27 m/s in 9.0 seconds. (a) a = 3.0 m/s/s (b)  = 8.3 Rad/s/s (8.33333333) (c) What angle do the tires go through? s =  r, s = (u + v)t/2, r =.36 m s = (27 m/s + 0)(9.0 s)/ 2 = 121.5 m s =  r, 121.5 m =  (.36 m)  = ( 121.5 m )/(.36 m) = 337.5 Rad  = 340 Rad 340 Rad W

19 Angular Mechanics - Angular kinematics Linear:  s/  t = v  v/  t = a u + at = v ut + 1 / 2 at 2 = s u 2 + 2as = v 2 (u + v)t/2 = s Angular:  =  /  t  =  /  t*  =  o +  t  =  o t + 1 / 2  t 2  2 =  o 2 + 2   = (  o +  )t/2* *Not in data packet TOC

20 Example: My gyro spinner speeds up to 10,000 RPM, in.78 sec. What is its angular accel., and what angle does it go through?  = ?,  o = 0, t =.78 s  = (10,000rev/min)(2  rad/rev)(1 min/60 sec)  = 1047.19 s -1  =  o +  t 1047.19 s -1 = 0 +  (.78s)  = (1047.19 s -1 )/(.78s) =1342.6=1300 rad/s/s (u + v)t/2 = s (  = (  o +  )t/2) (0 + 1047.19 s -1 )(.78s)/2 = 408.4 = 410 rad TOC

21 Whiteboards: Angular Kinematics 11 | 2 | 3 | 4 | 5 | 6 | 7234567 TOC

22 Use the formula  =  /  t to convert the angular velocity 78 RPM to rad/s. Hint: t = 60 sec,  = 78(2  )  =  /  t  = (78(2  ))/(60 sec) = 8.2 rad/s 8.2 rad/s W

23 A turbine speeds up from 34 rad/s to 89 rad/s in 2.5 seconds. What is the angular acceleration?  =  o +  t 89 rad/s = 34 rad/s +  (2.5 sec)  = (89 rad/s - 34 rad/s)/(2.5 sec) = 22 s -2 22 rad/s/s W

24 A turbine speeds up from 34 rad/s to 89 rad/s in 2.5 seconds. What is the angular acceleration? (b) What angle does it go through? (u + v)t/2 = s (34 rad/s + 89 rad/s)(2.5 s)/2 = 150 rad 150 rad W

25 A wheel stops from 120 rad/s in 3.0 revolutions. (a) What is the angular acceleration?  = (3.0)(2  ) = 18.85 rad  2 =  o 2 + 2   = (  2 -  o 2 )/(2  )  = (0 2 - (120 rad/s) 2 )/(2(18.85 rad))  = -381.97 = -380 rad/s/s -380 rad/s/s W

26 A wheel stops from 120 rad/s in 3.0 revolutions. (a) What is the angular acceleration? (b) What time did it take?  = 381.97 = -380 rad/s/s  v/  t = a,  t =  v/a = (120 rad/s)/  t = (120 s -1 )/(381.97 s -2 ) =.31 sec.31 s W

27 A motor going 45.0 rad/s has an angular acceleration of 12.4 rad/s/s for 3.7 seconds. (a) What is the final velocity?  =  o +  t  = 45.0 rad/s + (12.4 rad/s/s)(3.7 s) =  = 90.88 = 91 rad/s 91 rad/s W

28 A motor going 45.0 rad/s has an angular acceleration of 12.4 rad/s/s for 3.7 seconds. (a) What is the final velocity? (b) What angle does it go through?  =  o t + 1 / 2  t 2  = (45.0s -1 )(3.7s) + 1 / 2 (12.4s -2 )(3.7s) 2  = 251.378 = 250 rad 250 rad W


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