Angular Mechanics - Rolling using dynamics
Contents:
•Review
•Linear and angular Qtys
•Tangential Relationships
•Useful Substitutions
•Force causing 
•Rolling | Whiteboard
•Strings and pulleys
•Example | Whiteboard
Angular Mechanics - Angular Quantities
Linear: Angular:
(m) s  - Angle (Radians)
(m/s) u o - Initial angular velocity (Rad/s)
(m/s) v  - Final angular velocity (Rad/s)
(m/s/s) a  - Angular acceleration (Rad/s/s)
(s) t t
- Uh, time (s)
(N) F  - Torque
(kg) m I - Moment of inertia
TOC
Angular Mechanics - Tangential Relationships
Linear: Tangential: (at the edge of the wheel)
(m) s = r
- Displacement
(m/s) v = r
- Velocity
(m/s/s) a = r
- Acceleration*
*Not in data packet
TOC
Angular Mechanics - Useful Substitutions
 = I
 = rF
so F = /r = I/r
s = r, so  = s/r
v = r, so  = v/r
a = r, so  = a/r
TOC
Rolling objects accelerate linearly and angularly:
Force causing 
F
r
 = I
 = rF
so F = /r = I/r
TOC
Rolling:
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
Whiteboards:
Rolling
1|2|3
TOC
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
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 (9.81 m/s2)sin(21o)
a = 2.5111 m/s2 = 2.5 m/s2
2.5 m/s/s
W
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 = 2.5086
m/s2)
v2 = u2 + 2as
v2 = 02 + 2(2.5111 m/s2)(2.75 m)
v = 3.716 m/s = 3.7 m/s
3.7 m/s
W
Angular Mechanics – Pulleys and such
For the cylinder:
r
m1
 = I
rT = (1/2m1r2)(a/r)
(Where T is the tension
in the string)
For the mass:
m2
F = ma
m2g - T = m2a
TOC
Pulleys and such:
r
m1
So now we have two equations:
rT = (1/2m1r2)(a/r) or
T = 1/2m1a
and
m2g - T = m2a
m2
TOC
Angular Mechanics – Pulleys and such
r
m1
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
m2
TOC
Whiteboards:
Pulleys
1|2|3|4
TOC
A string is wrapped around a 12.0 cm
radius 4.52 kg cylinder. A mass of 0.162
kg is hanging from the end of the string.
Set up the dynamics equation for the
hanging mass. (m2)
m2g - T = m2a
r
m1
m2
figure it out for yourself
W
A string is wrapped around a 12.0 cm
radius 4.52 kg thin ring. A mass of 0.162
kg is hanging from the end of the string.
Set up the dynamics equation for the thin
ring (m1)
 = I, I = m1r2,  = a/r,  = rT
rT = (m1r2)(a/r)
T = m1a
r
m1
m2
figure it out for yourself
W
A string is wrapped around a 12.0 cm
radius 4.52 kg thin ring. A mass of 0.162
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
figure it out for yourself
W
A string is wrapped around a 12.0 cm
radius 4.52 kg thin ring. A mass of 0.162
kg is hanging from the end of the string.
Plug the values in to get the acceleration: a
= m2g/(m1 + m2)
a = m2g/(m1 + m2)
a = (0.162 kg)(9.80 N/kg)/(4.52 kg + 0.162 kg)
a = 0.339 m/s/s
r
m1
m2
0.339 m/s/s
W