Announcements
1. Physic Colloquium today --The Physics and Analysis of
Non-invasive Optical Imaging
2. Today’s lecture –
Brief review of momentum & collisions
Example HW problems
Introduction to rotations
Definition of angular variables
Moment of inertia
Energy associated with rotations
10/9/2003
PHY 113 -- Lecture 11
1
From HW 9 –
6. HRW6 9.P.031. A space vehicle is traveling at v0 = 5700 km/h
relative to the Earth when the exhausted rocket motor is
disengaged and sent backward with a speed of vR = 85 km/h
relative to the command module. The mass of the motor is four
times the mass of the module. What is the speed of the command
module relative to Earth after the separation?
[
] km/h
4M
4M
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M
vM − vR
v0
M
5Mv0 = 4 M (vM − vR ) + MvM
vM
PHY 113 -- Lecture 11
⇒ solve for vM
2
From HW 9 –
7. HRW6 9.P.035. A certain radioactive nucleus can transform to another nucleus by
emitting an electron and a neutrino. (The neutrino is one of the fundamental particles of
physics.) Suppose that in such a transformation, the initial nucleus is stationary, the
electron and neutrino are emitted along perpendicular paths, and the magnitudes of the
linear momenta are 1.4×10-22 kg · m/s for the electron and 6.1×10-23 kg · m/s for the
neutrino. As a result of the emissions, the new nucleus moves (recoils).
(a) What is the magnitude of its linear momentum?
[ ] kg · m/s
(b) What is the angle between its path and the path of the electron?
[ ]°
(c) What is the angle between its path and the path of the neutrino?
[ ]°
(d) What is its kinetic energy if its mass is 5.7 ×10-26 kg?
[ ]J
pe
pN
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θe
θν
pν
PHY 113 -- Lecture 11
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From HW 10 –
6. HRW6 10.P.046. Two 2.0 kg masses, A and B, collide.
The velocities before the collision are vA = 12i + 30j and
vB = -5i + 15.0j. After the collision, v'A = -6.0i + 22j. All
speeds are given in meters per second.
(a) What is the final velocity of B? [ ] m/s i + [ ] m/s j
(b) How much kinetic energy was gained or lost in the
collision? [ ] J
m( v A + v B ) = m( v′A + v′B )
(12i + 30 j ) + (- 5i + 15.0 j) = (- 6.0i + 22 j) + v′B
∆E = 12 m(v′A2 + v′B2 ) − 12 m(v A2 + vB2 )
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PHY 113 -- Lecture 11
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From HW 10 –
7. HRW6 10.P.047. An alpha particle collides with an oxygen
nucleus, initially at rest. The alpha particle is scattered at an
angle of 64.0° above its initial direction of motion, and the
oxygen nucleus recoils at an angle of 47.0° on the opposite
side of that initial direction. The final speed of the nucleus is
1.10 ×105 m/s. In atomic mass units, the mass of an alpha
particle is 4.0 u. The mass of an oxygen nucleus is 16 u.
(a) Find the final speed of the alpha particle.
[ ] m/s
(b) Find the initial speed of the alpha particle. [ ] m/s
vi
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vf
vO
PHY 113 -- Lecture 11
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Angular motion
s
angular “displacement” Î θ(t)
dθ
angular “velocity” Î ω(t) =
dt
dω
angular “acceleration” Î
α(t) =
dt
“natural” unit == 1 radian
Relation to linear variables:
sθ = r (θf-θi)
vθ = r ω
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PHY 113 -- Lecture 11
aθ = r α
6
v1=r1ω
r1
ω
r2
v2=r2ω
Special case of constant angular acceleration: α = α0:
ω(t) = ωi + α0 t
θ(t) = θi + ωi t + ½ α0 t2
( ω(t))2 = ωi2 + 2 α0 (θ(t) - θi )
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PHY 113 -- Lecture 11
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Example: Compact disc motion
ω1
ω2
In a compact disk, each spot on the disk passes the laser-lens system
at a constant linear speed of vθ = 1.3 m/s.
ω1=vθ/r1=56.5 rad/s
ω2=vθ/r2=22.4 rad/s
What is the average angular deceleration of the CD over the time
interval ∆t=4473 s?
α = (ω2-ω1)/∆t =-0.0076 rad/s2
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PHY 113 -- Lecture 11
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Object rotating with constant angular velocity (α = 0)
ω
R
v=Rω
v=0
Kinetic energy associated with rotation:
K = ∑ 1 2mi vi2 = ∑ 1 2mi ri2 ω 2 ≡
i
i
where : I ≡ ∑ mi ri2
1
2
I
ω
;
2
“moment of inertia”
i
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PHY 113 -- Lecture 11
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PHY 113 -- Lecture 11
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Peer instruction question:
Suppose each of the following objects each has the same total
mass M and outer radius R and each is rotating counterclockwise at an constant angular velocity of ω=3 rad/s. Which
object has the greater kinetic energy?
(a) (Solid disk)
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(b) (circular ring)
PHY 113 -- Lecture 11
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Various moments of inertia:
R
R
R
solid cylinder:
I=1/2 MR2
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solid sphere:
solid rod:
I=2/5 MR2
I=1/3 MR2
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Calculation of moment of inertia:
Example -- moment of inertia of solid rod through an axis
perpendicular rod and passing through center:
R
M⎞
M ⎞R 2
1
⎛
⎛
2
I = ∑ mi ri = ∫ ⎜ ⎟dr r = ⎜ ⎟ ∫ r dr = MR 2
3
i
⎝ 2 R ⎠− R
−R ⎝ 2R ⎠
2
R
Extra credit: Write out the evaluation of I for another shape.
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PHY 113 -- Lecture 11
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How to make objects rotate.
θ
r
r sin θ
θ
n
si
F
Define torque:
τ=rxF
τ = rF sin θ
θ
F
F = ma
r × F ≡ τ = r × ma = Iα
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PHY 113 -- Lecture 11
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From HW 11 --
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PHY 113 -- Lecture 11
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Newton’s second law applied to center-of-mass motion
dv i
dv CM
⇒ Ftotal = M
∑ Fi = ∑ mi
dt
dt
i
i
Newton’s second law applied to rotational motion
Fi = mi
dv i
dv i
m
⇒ ri × Fi = ri × i
dt
dt
I ≡ ∑ mi d i2
τ i = ri × Fi
i
v i = ω × ri
ri
d (ω × ri )
⇒ τ i = mi ri ×
dt
dimi
Fi
dω
⇒ τ total = I
= Iα (for rotating about principal axis)
dt
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PHY 113 -- Lecture 11
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Another example:
A horizontal 800 N merry-go-round is a solid disc of radius
1.50 m and is started from rest by a constant horizontal force
of 50 N applied tangentially to the cylinder. Find the kinetic
energy of solid cylinder after 3 s.
F
R
K = ½ I ω2
τ= Ια
In this case I = ½ m R2
and
ω = ωi + αt = αt
τ = FR
2
(
)
F2 2
N
50
K=g
t = 9.8m/s 2
(3s ) 2 = 275.625 J
mg
800 N
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PHY 113 -- Lecture 11
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Re-examination of “Atwood’s” machine
R
T1
T1-m1g = m1a
I
T2
T2-m2g = -m2a
τ =T2R – T1R = I α = I a/R
T1
T2
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⎞
⎛
m2 − m1
⎟
a = g⎜⎜
2⎟
⎝ m2 + m1 + I / R ⎠
⎞
Ig ⎛
m2 − m1
⎟
τ = ⎜⎜
2⎟
R ⎝ m2 + m1 + I / R ⎠
PHY 113 -- Lecture 11
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Conservation of energy:
Another example:
h
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K f + Uf = K i + U i
v1
m1
m2
h/2 v2
PHY 113 -- Lecture 11
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Peer instruction question
Three objects of uniform density – a solid sphere (a), a solid
cylinder (b), and a hollow cylinder (c) -- are placed at the top of
an incline. If they all are released from rest at the same elevation
and roll without slipping, which object reaches the bottom first?
(a) solid sphere (b)solid cylinder (c)hollow cylinder
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PHY 113 -- Lecture 11
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