Transcript Thomson Template - UNT College of Arts and Sciences

Physics 1710—Warm-up Quiz
Two 2.0 kg disks, both of radius 0.10 m are sliding
(without friction) and rolling, respectively, down an
incline. Which will reach the bottom first?
1. Rolling disk wins.
2. Sliding disk wins.
3. Tie.
40%
37%
23%
1
2
3
Physics 1710—Chapter 11 Rotating Bodies
Solution:
Kinetic Energy of sliding disk:
K1 = ½ mv2 = mg(h-z); v =√[2g(h-z)]
Kinetic Energy of rolling disk:
K2 = ½ mv2+ ½ I ω2= mg(h-z)
= ½ mv2+ ½ ( ½ mr2) (v/r)2
= 3/4 mv2; v =√[4/3 g(h-z)]
Slider wins!
Physics 1710—Chapter 11 Rotating Bodies
Consider two spindles rolling down a ramp:
c
c
a
b
Which one will win and why?
Physics 1710—Chapter 11 Rotating Bodies
Which one will win and why?
Think!
Peer Instruction Time
No Talking!
Confer!
Physics 1710—Chapter 11 Rotating Bodies
1′ Lecture
•The energy of rotation
K = ½ I ⍵2
• Torque (“twist”) is the vector product of the “moment”
and a force. τ = r x F
• τ = I ⍺ = I d⍵/dt
•Angular momentum L is the vector product of the moment
arm and the linear momentum. L= r x p.
• τ = d L/dt
Physics 1710—Chapter 11 Rotating Bodies
Moment of Inertia—sphere
I = ∭ R 2ρ d V
N.B. : R 2 = r 2 – z 2
R
r
R 2 = r 2 – (r cos θ) 2
=r 2 (1– cos 2 θ)
I = ∫0 dφ∫0 (1– cos
2π
π
2 θ)
a
sin θdθ∫0 r 4 ρdr
= [2π][2- 1/3(2)][1/5 a5] ρ = [4π][2/3][1/5 a5] ρ
=[4π/3][2/5 a5] ρ = 2/5 M a2
Physics 1710—Chapter 11 Rotating Bodies
Kinetic Energy of Rotation:
K = ½ Σi mi vi 2
K = ½ Σi mi (Ri ωi ) 2
K = ½ Σi mi R i 2ωi 2
For rigid body ωi = ω
K = ½ [Σi mi R i 2] ω 2
K=½Iω2
With I =Σi mi R i 2 = the moment of inertia.
Physics 1710—Chapter 11 Rotating Bodies
Why do round bodies roll down slopes?
F = m a = mr dω/dt
2dω/dt
rF
=
rF
sinθ
=
mr
g
Fsinθ
Torque = r x F = I α
θ
Fg
θ
τ = r x F, |τ | = rFsinθ
The torque is the “twist.”
Physics 1710—Chapter 11 Rotating Bodies
Torque and the Right Hand Rule:
r
X
F
Physics 1710—Chapter 11 Rotating Bodies
Vector Product:
C=AxB
Cx = Ay Bz – Az By
Cyclically permute: (xyz), (yzx), (zxy)
|C| =√[Cx2 + Cy2 + Cz2 ]
= AB sin θ
Directed by RH Rule.
Physics 1710—Chapter 11 Rotating Bodies
Vector Product:
AxB=-BxA
Ax(B+C)=AxB+AxC
d/dt ( A x B ) = d A /dt x B + A x d B/dt
ixi=jxj=kxk=0
ixj=-jxi=k
jxk =-kx j=i
kxi=-ixk=j
Physics 1710—Chapter 11 Rotating Bodies
Torque Bar:
τ
τ=r xF
r
F
Physics 1710—Chapter 11 Rotating Bodies
Teeter-totter:
τ=r xF
F1
A
B
C
F2
Where should the fulcrum be place to
balance the teeter-totter?
Physics 1710—Chapter 11 Rotating Bodies
Where should the fulcrum be place to balance
the teeter-totter?
A.
B.
C.
86%
6%
A
8%
B
C
Physics 1710—Chapter 11 Rotating Bodies
Torque Ladder
?
Which way will the
torque ladder move?
Physics 1710—Chapter 11 Rotating Bodies
Which way will the torque ladder move?
70%
A. Clockwise
B. Counterclockwise
C. Will stay balanced
27%
3%
A
B
C
Physics 1710—Chapter 11 Rotating Bodies
Torque Ladder
?
r sin θ
r
Which way will the
torque ladder move?
Physics 1710—Chapter 11 Rotating Bodies
Second Law of Motion
F=ma
Or F = dp/dt
Then:
r x F = d (r x p)/dt
Torque = τ = d L/dt
L = r x p is the “angular momentum.”
Physics 1710—Chapter 11 Rotating Bodies
Angular Momentum:
L=rxp
The angular momentum is the vector product of
the moment arm and the linear momentum.
∑ T = d L/dt
The net torque is equal to the time rate of change
in the angular momentum.
Physics 1710—Chapter 11 Rotating Bodies
Angular Momentum:
Proof:
And
∑ T = r x ∑F = r x d p/dt
d L/dt = d( r x p) /dt
= d r/dt x p + r x d p/dt.
But p = m d r/dt , therefore d r/dt x p = 0
And thus
d L/dt = r x d p/dt
∑ T = d L/dt.
Physics 1710—Chapter 11 Rotating Bodies
Second Law of Motion
Torque = τ = d L/dt
If τ = 0, then
L is a constant.
L = constant means angular momentum si
conserved.
Physics 1710—Chapter 11 Rotating Bodies
Rotating Platform Demonstration
Physics 1710—Chapter 11 Rotating Bodies
Analysis:
•Why does an ice skater increase her angular velocity
without the benefit of a torque?
L=rxp
= r x ( m v)
= r x ( m r x ⍵)
Li = mi ri 2 ⍵
Lz = (∑i mi ri 2 ) ⍵
Lz = I ⍵; & ⍵ = Lz / I
Therefore, a decrease in I ( by reducing r) will result in an
increase in ⍵.
Physics 1710—Chapter 11 Rotating Bodies
Summary:
•The total Kinetic energy of a rotating system is the sum
of the rotational energy about the Center of Mass and
the translational KE of the CM.
K = ½ ICM ⍵ 2 + ½ MR 2 ⍵ 2
τ=rxF
Physics 1710—Chapter 11 Rotating Bodies
Summary:
•Angular momentum L is the vector product of the
moment arm and the linear momentum.
L=rxp
• The net externally applied torque is equal to the time
rate of change in the angular momentum.
∑ τz = d Lz /dt = Iz ⍺