Continue Chapter 8
Analogies
Torque and Angular Acceleration
Newton’s Second Law for a Rotating Object
Στ = Iα
linear-----rotation
analogous to
x becomes θ
m becomes I
v becomes ω
a becomes α
F becomes
I = moment of inertia
2
2
For Uniform Ring I = Σmi ri = MR
τ
Other Moments of Inertia
∑F = ma
moment of inertia depends on
quantity of matter
and its distribution
and location of axis of rotation
Newton’s Second Law for Rotation-Example
Draw free body diagrams of
each object
Only the cylinder is rotating,
so apply Στ = I α
The bucket is falling, but not
rotating, so apply ΣF = ma
Remember a = α r and
solve the resulting equations
Torque and Angular Acceleration
Angular Momentum
L=Iω
Just like p = mv
L always conserved no matter what!
Στ = Iα
∑F = ma
T - mg = ma
− 12 Ma − mg = ma
a=
− mg
M
+m
2
ur
ur
R T = 12 MR 2 α
ur
Rα = a
− R/ T = 12 MR/ a
Look at a as m→0 or ∞ or M→0 or ∞
Impulse
∆L
∆t
ur
∆p
Just like ∑ F =
∆t
Στ =
Conservation of Angular Momentum
Στ = 0, Li = Lf or Iiω i = If ω f
Isolated system
Applying Conservation Rules
If the net torque is zero, the
angular momentum remains
constant
Conservation of Angular
Momentum states: The angular
momentum of a system is
conserved when the net external
torque acting on the systems is
zero.
In an isolated system, the following
three quantities are conserved:
Mechanical energy
Linear momentum
Angular momentum
Στ = 0, Li = Lf or Iiω i = If ω f
Applying Conservation Rules
In an isolated system, the
following
three quantities are
conserved:
Mechanical energy
Linear momentum
Angular momentum
Conservation of Angular
Momentum-- Example
How does a skater
spin faster in the
air?
L is conserved
As arms come in-I decreases
w increases
Rotational Kinetic Energy
1
KE = Iω 2
2
1 2
Equivalent to KE = mv
2
Ball rolling down an incline
How fast does it leave the
bottom of the incline?
h
Conservation of Mechanical Energy
(KEt + KEr + PEg + PE s )i = (KEt + KEr + PE g + PE s )f
Work-Energy Theorem instead of Conservation of
Energy: Wnc = ∆KEt + ∆KER + ∆PE
Conservation of Mechanical Energy
(KEt + KE r + PE g + PE s )i = (KEt + KE r + PE g + PE s )f
+ mgh + 0
= mv2/2+Iω2/2 + 0 + 0
2
I=2mr /5 and rω=v so Iω2/2= mv2/5
2
Giving: mgh = mv /2+ mv2/5 = 7 mv2/10
0
+
0
Note m cancels out as usual
or
v= 10gh/7