Lecture 8
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Parallel Axis Theorem
Torque
Newton’s Second Law for
Rotation
Work, Power and Rotational
Kinetic Energy
More on Rotational Variables as
Vectors
More Torque
Parallel Axis Theorem
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Moment of inertia of body about axis
through c. of m. is Icm. Calc. moment
of inertia about parallel axis.
P
c. of m.
mass M
h
Parallel axis theorem cont.
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Parallel axis theorem cont.
IP = ∫ x 2 + y 2 dm − 2a ∫ x dm −
View from above, place origin at c.
of m., z axis along initial axis of
rotation.
M
M
2b ∫ y dm + ∫ a 2 + b 2 dm
M
y
dm
P
Now 2nd and 3rd integrals zero from
definition of c. of m. and choice of
position of origin. First integral just
definition of Icm. We also see:
c. of m.
r
h
b
IP = ∫ r 2 dm
M
= ∫ (x − a)2 + (y − b)2 dm
M
y-b
∫a
x-a
a
M
x
M
2
+ b 2 dm = ∫ h2 dm = Mh 2
M
Hence parallel axis theorem:
IP = Icm + Mh2
Consequence, minimum moment
of inertia for axes through c. of m.
Parallel Axis Theorem, an
Example
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Parallel Axis Theorem, an
Example cont.
Consider again rectangular prism
c b
2 2 a
Ie = ρ ∫
∫∫x
c b
2
+ y 2 dxdydz
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z
◆
− − 0
2 2
c
b
2 a
= cρ ∫ ∫ x 2 + y 2 dxdy
a
b
b
− 0
2
a
b
y3 2
2
c
x
y
dx
= ρ∫
+
b
12
0
−
2
a
abcρ
 x 3 b 2 x
4a 2 + b 2 )
= bcρ +
(
 =
12
 3 12  0
=
M
(4a 2 + b 2 )
12
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Moment of inertia about c. of m.
M
Icm = (a 2 + b 2 )
12
Using parallel axis theorem
2
a

Ie = Icm + M 
 2
M 2
M
(a + b 2 ) + a 2
=
12
4
M
(4a 2 + b 2 )
=
12
Exercise, repeat check for axis along
corner of prism and for other shapes.
Torque
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Newton’s Second Law
A force applied to a body may tend to
rotate the body about an axis.
Quantify using concept of torque
(from latin “to twist”)
Force in plane normal to axis
rotation
axis
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Consider rotation of a (simple)
rigid body
y
massless Ft
r
Ft
r
O
φ
rT
τ = Ftr
= FrT
= Fr sin φ
θ
F
φ
units Nm
Note τ = r × F
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Fr
F
Fr
mass m
x
Relate tangential acceleration to
torque
Ft = ma t
τ = Ftr
= ma tr
Newton’s second law cont.
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Consider rotational acceleration
dω 1 dv t a t
α=
=
=
dt r dt
r
a t = αr
Substitute for at
τ = m(αr )r = mr 2 α
Recall definition of moment of inertia
I = mr 2
Hence Newton’s Second Law for
rotation
τ = Iα
If many forces applied
∑ τ = Iα
Work and Rotation
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Consider same rigid body rotating
through dθ under influence of force F
y
F
massless Ft
mass m
Fr
ds
r
dθ
x
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Calculate work done
dW = F ⋅ ds = Ft r dθ = τ dθ
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For finite angular displacement
θf
W = ∫ τ dθ
θi
xf
cf. W = ∫ F dx
xi
Power and Work K.E.
Relation
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Consider power
d
dθ
P= W =τ
= τω
dt
dt
Now relate work to K.E. of rotation
θf
θf
θf
θi
θi
θi
W = ∫ τ dθ = ∫ Iα dθ = ∫ I
ωf
= ∫I
ωi
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Derive vector angular velocity
dω
dθ
dt
2
= K f − Ki
2
|∆r|=rsinθ∆φ
v
ω
θ
ωf
= 12 Iωf − 12 Iωi
∆φ
rsinθ
dθ
dω = ∫ Iω dω
dt
ωi
= ∆K
z
φ
r
y
x
lim ∆r
lim r sin θ∆φ
=
∆t → 0 ∆t ∆t → 0 ∆t
dφ
= r sin θ
= rω sin θ = r × v
dt
As v tangential to circle it is normal
to plane containing r and ω, hence
v= ×
(using right hand rule)
v =
◆
More on Rotational
Variables as Vectors
More Torque
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More torque cont.
For force in plane normal to axis can
write torque as τ = r × F
Generalise for all forces.
What is torque about given axis?
Example, torque about z axis.
z
F
rT
Ft
r
x
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y
Write F=(fx,fy,fz) and r=(rx,ry,rz) then
Ft=(fx,fy,0) and rT=(rx,ry,0).
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From previous discussion we know
torque about z is
τ z = rT × Ft
In terms of vector components this is
τ z = (rx , ry ,0) × (fx , fy ,0)
i
= rx
fx
j
ry
fy
= rx fy − ry fx
k
0
0
More torque cont.
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Consider torque as vector and take
component of vector along z axis to
get torque about z
τ = r ×F
i
= rx
fx
j
ry
fy
k
rz
fz
= (ry fz − rz fy ,rz fx − rx fz ,rx fy − ry fx )
◆
 ry fz − rz fy   0 


τz = τ ⋅ kˆ =  rz fx − rx fz  ⋅  0 
 
 rx fy − ry fx   1 

  
= rx f y − ry fx
The correct result.