Lectre17 - LSU Physics

Physics 2101
S ti 3
Section
March 8rd : Ch. 10
Announcements:
• Next Quiz is March 12 on
SHW#6
• I will be at the March APS
meeting the week of 151519th.
• Try to solve the SHW#5
as possible
ibl as you can.
Class Website:
http://www.phys.lsu.edu/classes/spring2010/phys2101--3/
http://www.phys.lsu.edu/classes/spring2010/phys2101
http://www.phys.lsu.edu/~jzhang/teaching.html
Moment of inertia of a Pencil
It depends on where the rotation axis is considered…
I = 121 ML2
I = ML
1
3
2
I = 150 × 10−6 kg ⋅ m 2
I = 38 × 10−6 kg ⋅ m 2
I = 12 MR2
I = 0.3 × 10 −6 kg ⋅ m 2
Consider a 20g pencil 15cm long and 1cm wide …
… somewhat like MASS, you can feel the
diff
difference
iin th
the rotational
t ti
l iinertia
ti
Example #1
A bicycle wheel has a radius of 0.33m
0 33m and a rim of mass 11.2
2 kg
kg.
The wheel has 50 spokes, each with a mass 10g.
What is the moment of inertial about axis of rotation?
What is moment of inertia about COM?
I tot,com = I rim,center + 50I spoke
→ What is Ispoke (p
(parallel-axis)?
)
Ispoke = Irod .com + Mh 2 = 121 ML2 + M (12 L ) = 13 ML2
2
= 13 (0.01kg)(0.33m) ≅ 3.6 ×10−4 kg ⋅ m 2
2
→ Putting together
I tot,com
= I rim,center
50I spokek
t t
i
t + 50
= M wheel R 2 + 50I spoke
= (1.2kg)(0.33m) + 50(3.6 ×10−4 kg ⋅ m 2 ) = 0.149kg ⋅ m 2
2
Example #2
How much work did Superman exert on earth in
order to stop it?
What is the kinetic energy of the earth’s rotation
about its axis?
Energy of rotational motion is found from:
KErot = 12 Iω 2
What is earth’s moment of inertia, I?
Iearth = Isphere = 25 Mr 2
= 25 (6 ×10 24 kg)(6.4 ×10 6 m) ≅ 1×10 38 kg ⋅ m 2
2
What is earth’s angular velocity, ω?
From T = 2 π
ω
⇒
di
ω = 2π radians
= 2π
day
(3600 × 24)
= 7.3 ×10−5 rad / s
Now plug-’n-chug:
1⎛
⎜
2⎝
KErot = 1×10 kg ⋅ m
38
29
7.3
×10
rad
/
s
=
2.6
×10
J
(
)
2⎞
⎟
⎠
−5
2
Rotational dynamics: Torque
What causes
something to rotate
Ability of force F to rotate body
What does it depend on??
Depends on distance from pivot point
Depends on angle between force and radius
Moment of force about an axis = torque - “to twist”
τ = r ×F
τ depends on: r, F and θ
τ = r F sinϑ
Units of torque: N·m (not Joule)
Sec. 3.7 Multiplying vectors:
2) Cross Product (Vector × Vector = Vector)
iˆ
ˆj
kˆ
C = A × B = Ax
Ay
Az
Bx
By
Bz
C = A B sin θ
If A⊥B then A×B=max
If A||B then A×B=0
Two vectors A and B define a p
plane. Vector C is p
perpendicular
p
to
plane according to rightright-hand rule.
Rotational dynamics - Torque
vector
τ =r ×F
=
magnitude
+
τ = r F sin ϑ
direction
vector τ is always
perpendicular to both
vectors r and F
If Ft is tangential component of force
If r⊥ is moment arm
τ = r (F sin ϑ )= rFt
= (r sin
i ϑ )F = r⊥ F
Units of torque: N·m (not Joule)
The figure shows an overhead view of a meter stick
that can pivot about he dot at the position marked 20
(for 20 cm). All five horizontal forces on the stick
have the same magnitude. Rank those forces
according to the magnitude of the torque that they
produce,
d
greatest
t t fi
first.
t
| τ | = r F sinθ
|τ1| = |τ3| > |τ4| > |τ2| = |τ5| = 0
Checkpoint
Problem 1010-47
A body is pivoted at O and two forces act on it as shown. Find an expression for the net
torque on the body about the pivot
What is vector equation for net torque?
τ net = ∑ τ i =∑ ri × Fi = r1 × F1 + r2 × F2
→ What is magnitude of torque?
τ = τ = rF sin θ
Thus expression is:
τ net = [r1F1 sin θ1 − r2 F2 sin θ2 ]kk̂
(out − of − page)
Note: When using
g the right-hand
g
rule …
Use angle between F and r which is less than 180°
τ = rF sin θ
θ'
r
θ
Whichh angle?
Whi
l ?
The one less than 180° → θ
F
Rotational dynamics - Newton’s 2nd Law
Force about a point
causes torque
τ =r ×F
Torque causes rotation
τ net = ∑ τ i = Iα
{F
net
= ∑ Fi = ma
ma}
• Torque is positive when the force tends to produce a counterclockwise rotation about an
axis is negative when the force tends to produce a clockwise rotation about an axis
(→ VECTOR! positive if pointing in +z, negative if pointing in -z )
• Net torque and angular acceleration are parallel
Example #1 A uniform disk, with mass M and radius R, is mounted
on a fixed horizontal axle. A block of mass M hangs from a massless cord
that is wrapped around the rim of the disk.
a) Find the acceleration of the disk, and the tension in the cord.
b) b) Initially at rest, what is the speed of the block after dropping a
distance d?
M
Example #1
A wheel (mass M,
radius R, and I =1/2MR2)
i attached
is
h d to block
bl k with
ih
Using force
+
torque
equal mass M (i.e.
Mdisk=Mblock). Initially
yˆ : T − Mg = −Ma
zˆ : τ net = r × F
at rest, what is the speed
⇒ T = M (g − a)
= (RT)(−zˆ)
of the block after
dropping a distance d?
τ = Iα
Equal
masses
M
a
d
M
(RT )(−zˆ) = (Iα)(−zˆ)
at = α R &
Using conservation of Energy
ΔKEtot = −ΔU
ΔU
(
1
2
[
Mv2 + 12 Iω 2 )final − (0 + 0)init = − (0) final − (Mgd)init
Mv +
2
1
2
(
1
2
a(12 M + M ) = Mg
⎛v⎞
MR )⎜ ⎟ = Mgd
⎝ R⎠
a=
2
Mgg 2
= g
3
M 3
2
2
v 2 + (12 )(v) = 2gd
2
v (1+
2
2
v = ω R & I = 12 MR2
Substituting
b i i
1
2
]
2 ⎛a⎞
1
R
M
g
−
a
=
MR
(
)
[
]
(
)⎜⎝ R ⎟⎠
2
I = MR ⇒
1
2
1
2
)=
v=
3
2
v = 2gd
2
4
3
gd
Now use 1-D kinematics
v 2 = v02 + 2ad
v 2 = 0 + 2d (23 g)
v=
4
3
gd
Newton’s 2nd law for rotation
τ net = ∑ τ i = Iα
{F
net
= ∑ Fi = ma}
W k and
Work
dR
Rotational
t ti
l Ki
Kinetic
ti Energy
E
NET Work done
ON system
W = ΔKE rot
= 12 I (ω 2f − ω i2 )
θ2
Rotational work,
fixed axis rotation
(if torque is const)
Power,
Power
fixed axis rotation
W = ∫ τ net dθ
θ1
= τ (θ f − θ i )
dW d
P=
= (τθ
θ ) = τω
dt dt
⎧W = ΔKEtrans
⎫
⎨
⎬
2
2
1
⎩ = 2 m(v f − vi )⎭
x
⎧
⎨W = ∫ Fdx
⎩
x
2
1
⎫
1− D motion⎬
⎭
⎧
⎫
dW d
= (F • x )= F •v
• v⎬
⎨P =
⎩
⎭
dt dt
Block and Pulley Accelerating
m, r
T
M
Probelme 10-54:
10 54: A disk wheel of radius r (m) and
mass m (kg) is mounted on a frictionless horizontal
axis. The rotational inertia of the wheel is I kg/m2. A
massless cord wrapped around the wheel is attached
to a M kg block that slides on a horizontal frictionless
surface. If a horizontal force of magnitude P is
applied,
pp
, what is the magnitude
g
of the angular
g
acceleration of the wheel? Assume that the string
does not slip on the wheel.
Force about a p
point
causes torque
τ = Fl
Torque causes rotation
τ net = ∑ τ i = Iα
Trans - rotational
relationship
aT = rα
P − T = MaT
2
IIaT
IIaT
mr
Tr = τ = Iα =
or T = 2
For a disc I=
r
r
2
IaT
P
P − 2 = MaT
aT =
m
r
M+
P
2
aT =
I
M+ 2
Check M =0, m=0
r

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