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

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

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

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

Whiteboards: Conversions 1 | 2 | 3 | 4 TOC

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

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

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

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

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

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

Tangential relationships Whiteboards: Tangential relationships 1 | 2 | 3 | 4 | 5 | 6 TOC

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 W 8.8 m/s

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 W 95 rad/s

A. 450 m radius marking wheel rolls a distance of 123. 2 m 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 W 274 rad

(a) What is the linear acceleration? 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 W 3.0 m/s/s

(b) What is the tire’s angular acceleration? 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 W 8.3 Rad/s/s

(c) What angle do the tires go through? 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 W 340 Rad

Linear: s/t = v v/t = a u + at = v ut + 1/2at2 = s u2 + 2as = v2 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 Angular:  = /t  = /t*  = o + t  = ot + 1/2t2 2 = o2 + 2  = (o + )t/2* *Not in data packet TOC

Example: My gyro spinner speeds up to 10,000 RPM, in. 78 sec 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

Whiteboards: Angular Kinematics 1 | 2 | 3 | 4 | 5 | 6 | 7 TOC

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 W 8.2 rad/s

 = (89 rad/s - 34 rad/s)/(2.5 sec) = 22 s-2 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 W 22 rad/s/s

(34 rad/s + 89 rad/s)(2.5 s)/2 = 150 rad 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 W 150 rad

 = (02 - (120 rad/s)2)/(2(18.85 rad))  = -381.97 = -380 rad/s/s A wheel stops from 120 rad/s in 3.0 revolutions. (a) What is the angular acceleration?  = (3.0)(2) = 18.85 rad 2 = o2 + 2  = (2 - o2)/(2)  = (02 - (120 rad/s)2)/(2(18.85 rad))  = -381.97 = -380 rad/s/s W -380 rad/s/s

A wheel stops from 120 rad/s in 3. 0 revolutions 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 W .31 s

 = 45.0 rad/s + (12.4 rad/s/s)(3.7 s) =  = 90.88 = 91 rad/s 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 W 91 rad/s

(b) What angle does it go through? 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?  = ot + 1/2t2  = (45.0s-1)(3.7s) + 1/2 (12.4s-2)(3.7s)2  = 251.378 = 250 rad W 250 rad