Methods for finding I for non-standard shapes and axes
For a single point mass:
I = mr ⊥2
I=
For collection of point masses:
∑ m ir⊥2,i
For rigid body made of continuous masses:
I ≡
2
∫ r ⊥ dm
For a standard shaped body about some axis, with I known about cm:
IP = Icm + Mh 2
parallel axis theorem – look up Icm
For complex shapes made of standard bodies:
Use parallel axis theorem for each standard
component body.
Add moments of
f inertia together
g
about
common axis.
Example: Disk with holes cut out
I
IP = II − III − IIII
II
III
P
For complex shapes, integrate analytically when there is
some symmetry. If not, integrate numerically
Optional Example: Find I for a long, thin rod as
before, but rotation axis is now a distance “h” from
one end
1
Concepts for torque, moment arm, levers,…
A pure twist that causes angular acceleration.
Effect depends on applied force AND point of application
relative to a chosen rotation axis
Torque:
Greek
letter
“TAU”
TAU
not “T”
τ = force x moment arm = Fr sin( θ ) = r × F
Moment
Arm:
units: forceXdistance
Newtonxmeters,
lb.ft.
NOT Joules
The perpendicular distance from axis to line of action of force
Also called “lever arm”
rotation axis
(toward viewer)
F
l
90o
r
θ
force F
applied
here
moment arm
l =r.sin(θ)
90o
line of
action of F
…versus…
mg
F
l
mg
Torque and Moment arm (lever arm)
• Torque τ measures the effect a force applied at a point has in
causing an object to rotate about some axis
ƒ
ƒ
ƒ
ƒ
Torque is a vector, magnitude is: τ = r F sin φ = F d
F is the force
φ is the angle the force makes with the displacement
d is the moment arm (or lever arm) of the force
• The moment arm, d, is the
perpendicular distance
from the axis of rotation to
a line drawn along the
direction of the force
d = r sin Φ
• Direction: counterclockwise Æ torque is positive
clockwise Æ torque is negative
2
Torque in 3D – Force applied at a point Æ a vector
Axis of rotation along z, through P
φ
90o
r
Moment arm ⊥ = r sin(φ)
Torque Frsin(φ) (CCW about z)
or (equivalent view)
Transverse component of the
force
⊥= F sin(φ)
Moment arm is simply r
F
F⊥
φ
y
F
Frad
φ
Torque is still Frsin(φ)
F applied anywhere along its
line of action would produce the
same torque
Vector picture: torque vector is
perpendicular to plane of r and F,
( l
(along
z-axis
i in
i sketch,
k
h CCW sense)
)
90o
r
P
line of action
of F
r⊥
x
z
G
τ = force x moment arm = r F sin( φ ) → r × F
Net Torque
• Add up torque vectors using superposition; i.e.
G
G G
G
τnet = τ1 + τ2 + ...+ τn
E
Example:
l
same axis
G
ƒ The force F1 will tend to
cause a counterclockwise
rotation about
G O
ƒ The force F2 will tend to
cause a clockwise rotation
about O
ƒ Superposition about
common rotation axis
through point “O” gives
net torque
ƒ Στ = τ1 + τ2 = F1d1 – F2d2
3
Examples: Evaluating torques
• Massless, horizontal bar: φ = 90o
• Torque τ = - mgl
CCW Æ +
CW Æ -
l
• Moment arm = l sin(90o) = l
φ
clockwise
F = mg
moment arm
• Bar as above but NOT horizontal
• Moment arm = l sin(φ)
• Torque τ = + m g l sin(φ)
l
φ
CCW
φ
F = mg
You can add torques if they are referenced to the same rotation axis
N
l1
m1
l2
W1= m1g
m2
W2=
m2g
Free Body Diagram
• Horizontal see saw, unequal arms
• Calculate torques around pivot point
τnet = τ1 + τ2 + τN
• Moment arm = 0 for force N about pivot
∴ τnet = + m 1 g l 1 − m 2 g l 2
See Saw
2.4 Suppose in the last example that the masses and lengths are:
m1 = 3 kg, m2 = 2 kg, l1 = 2 m, l2 = 3 m.
Which of the following statements are correct?
1.
2
2.
3.
4.
5.
6.
the net torque is negative
the net torque is positive
the net torque is zero
the moment of inertia is positive
the moment of inertia is negative
the moment of inertia is zero
A.
B.
C.
D.
E.
1
2
3
3
2
and
and
and
and
and
m1
l1
W1= m1g
N
m2
l2
W2=
m2g
5
4
6
4
6
∴ τnet = + m 1 g l 1 − m 2 g l 2
4