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

2. Angular Mechanics - Radians
Full circle:
360o = 2 Radians
 = s/r
Radians = m/m = ? r s 
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3. 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)
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4. 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)
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5. Whiteboards:
Conversions
1 | 2 | 3 | 4
TOC

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

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 W
27.7 rev

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

9. 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

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*
*Not in data packet
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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)
 = s-1
r = .011m/2 = m
v = r = ( s-1)(.0055 m)
v = 5.8 m/s
(show ‘em!)
(pitching machines)
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12. Tangential relationships
Whiteboards:
Tangential relationships
1 | 2 | 3 | 4 | 5 | 6
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 W
8.8 m/s

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 W
95 rad/s

15. A. 450 m radius marking wheel rolls a distance of 123. 2 m
A .450 m radius marking wheel rolls a distance of 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

16. (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

17. (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) = Rad/s/s
 = 8.3 Rad/s/s W
8.3 Rad/s/s

18. (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 ( )
(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 = m
s = r, m = (.36 m)
 = (121.5 m)/(.36 m) = Rad
 = 340 Rad W
340 Rad

19. 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
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20. 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)
 = s-1
 = o + t
s-1 = 0 + (.78s)
 = ( s-1)/(.78s) =1342.6=1300 rad/s/s
(u + v)t/2 = s ( = (o + )t/2)
( s-1)(.78s)/2 = = 410 rad
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21. Whiteboards:
Angular Kinematics
1 | 2 | 3 | 4 | 5 | 6 | 7
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 W
8.2 rad/s

23.  = (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

24. (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

25.  = (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) = rad
2 = o2 + 2
 = (2 - o2)/(2)
 = (02 - (120 rad/s)2)/(2(18.85 rad))
 = = -380 rad/s/s W
-380 rad/s/s

26. 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?
 = = -380 rad/s/s
v/t = a,
t = v/a = (120 rad/s)/t
= (120 s-1)/( s-2) = .31 sec W
.31 s

27.  = 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) =
 = = 91 rad/s W
91 rad/s

28. (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
 = = 250 rad W
250 rad