1. Angular Mechanics - Rolling using dynamics Contents:
Review
Linear and angular Qtys
Tangential Relationships
Useful Substitutions
Force causing 
Rolling | Whiteboard
Strings and pulleys
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 - Useful Substitutions
 = 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 - Force causing 
 = rF
so F = /r = I/r
TOC

6. Angular Mechanics - Rolling
mgsin
I = 1/2mr2 m
r - cylinder 
F = ma + /r
mgsin = ma + I/r ( = I)
mgsin = ma + (1/2mr2)(a/r)/r ( =a/r)
mgsin = ma + 1/2ma = 3/2ma
gsin = 3/2a
a = 2/3gsin
TOC

7. Whiteboards:
Rolling
1 | 2 | 3
TOC

8. mgsin = ma + I/r, I = 2/5mr2,  = a/r mgsin = ma + (2/5mr2)(a/r)/r
A marble (a solid sphere) has a mass of 23.5 g, a radius of 1.2 cm, and rolls 2.75 m down a 21o incline. Solve for a in terms of g and 
mgsin = ma + I/r, I = 2/5mr2,  = a/r
mgsin = ma + (2/5mr2)(a/r)/r
mgsin = ma + 2/5ma = 7/5ma
gsin = 7/5a
a = 5/7gsin
5/7gsin W

9. a = 5/7gsin = 5/7 (9.8m/s2)sin(21o) a = 2.5086 m/s2 = 2.5 m/s2
A marble (a solid sphere) has a mass of 23.5 g, a radius of 1.2 cm, and rolls 2.75 m down a 21o incline. Plug in and get the actual acceleration. (a = 5/7gsin)
a = 5/7gsin = 5/7 (9.8m/s2)sin(21o)
a = m/s2 = 2.5 m/s2 W
2.5 m/s/s

10. A marble (a solid sphere) has a mass of 23. 5 g, a radius of 1
A marble (a solid sphere) has a mass of 23.5 g, a radius of 1.2 cm, and rolls 2.75 m down a 21o incline. What is its velocity at the bottom of the plane if it started at rest? (a = m/s2)
v2 = u2 + 2as
v2 = ( m/s2)(2.75 m)
v = m/s = 3.7 m/s W
3.7 m/s

11. Angular Mechanics – Pulleys and such
For the cylinder:
 = I
rT = (1/2m1r2)(a/r)
(Where T is the tension in the string) r m1 m2
For the mass:
F = ma
m2g - T = m2a
TOC

12. Angular Mechanics – Pulleys and such
So now we have two equations:
rT = (1/2m1r2)(a/r) or
T = 1/2m1a
and
m2g - T = m2a
TOC

13. Angular Mechanics – Pulleys and such
T = 1/2m1a
m2g - T = m2a
Substituting:
m2g - 1/2m1a = m2a
Solving for a:
m2g = 1/2m1a + m2a
m2g = (1/2m1 + m2)a
m2g/(1/2m1 + m2) = a
TOC

14. Whiteboards:
Pulleys
1 | 2 | 3 | 4
TOC

15. A string is wrapped around a 12 cm radius 4. 52 kg cylinder. A mass of
A string is wrapped around a 12 cm radius 4.52 kg cylinder. A mass of .162 kg is hanging from the end of the string. Set up the dynamics equation for the hanging mass. (m2) r m1 m2
m2g - T = m2a W
figure it out for yourself

16. A string is wrapped around a 12 cm radius 4.52 kg thin ring. A mass of .162 kg is hanging from the end of the string. Set up the dynamics equation for the thin ring (m1) r m1 m2
 = I, I = m1r2,  = a/r,  = rT
rT = (m1r2)(a/r)
T = m1a W
figure it out for yourself

17. A string is wrapped around a 12 cm radius 4.52 kg thin ring. A mass of .162 kg is hanging from the end of the string. Solve these equations for a:
m2g - T = m2a
T = m1a
m2g - m1a = m2a
m2g = m1a + m2a
m2g = a(m1 + m2)
a = m2g/(m1 + m2) r m1 m2 W
figure it out for yourself

18. A string is wrapped around a 12 cm radius 4.52 kg thin ring. A mass of .162 kg is hanging from the end of the string. Plug the values in to get the acceleration: a = m2g/(m1 + m2) r m1 m2
a = m2g/(m1 + m2)
a = (.162 kg)(9.80 N/kg)/(4.52 kg kg)
a = .339 m/s/s W
.339 m/s/s