PHYS 1443 – Section 004
Lecture #17
Thursday, Oct. 23, 2014
Dr. Jaehoon Yu
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Torque & Vector product
Moment of Inertia
Calculation of Moment of Inertia
Parallel Axis Theorem
Torque and Angular Acceleration
Rolling Motion and Rotational Kinetic Energy
Today’s homework is homework #9, due 11pm, Thursday, Oct. 30!!
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
1
Announcements
•  Mid-term grade discussion
–  Coming Tuesday, Oct. 28
–  Will have 40min of class followed by a 45min grade
discussion
–  In my office, CPB342
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
Torque
Torque is the tendency of a force to rotate an object about an axis.
Torque, τ, is a vector quantity.
F
φ#
Consider an object pivoting about the point P by
the force F being exerted at a distance r from P.
r
The line The line that extends out of the tail of the force
of Action vector is called the line of action.
P
l2
l1
The perpendicular distance from the pivoting point
F2
Moment arm
P to the line of action is calledthe moment arm.
τ ≡ ( Magnitude of the Force)
Magnitude of torque is defined as the product of the force
× ( Lever Arm )
exerted on the object to rotate it and the moment arm.
= ( F )( r sin φ ) = Fl1
When there are more than one force being exerted on certain
points of the object, one can sum up the torque generated by each
force vectorially. The convention for sign of the torque is positive
if rotation is in counter-clockwise and negative if clockwise.
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
∑τ = τ
1
+τ 2
= Fl1 1 − F2l2
Unit? N ⋅ m 3
Ex. Torque
A one piece cylinder is shaped as in the figure with core section protruding from the larger
drum. The cylinder is free to rotate around the central axis shown in the picture. A rope
wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder,
and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A)
What is the net torque acting on the cylinder about the rotation axis?
R1
The torque due to F1
R2
τ1 = −R1F1 and due to F2 τ 2 = R2 F2
So the total torque acting on
the system by the forces is
∑τ = τ
1
+ τ 2 = − R1F1 + R2 F2
Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque
about the rotation axis and which way does the cylinder rotate from the rest?
Using the
above result
∑τ =
− R1F1 + R2 F2
= −5.0 ×1.0 + 15.0 × 0.50 = 2.5 N ⋅ m
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
The cylinder rotates in
counter-clockwise.
4
Torque and Vector Product
z
Let’s consider a disk fixed onto the origin O and
the force F exerts on the point p. What happens?
τ=rxF#
O
r
p
y
The disk will start rotating counter clockwise about the Z axis
The magnitude of the torque given to the disk by the force F is
τ = Fr sin θ
θ F
x
But torque is a vector quantity, what is the direction?
How is torque expressed mathematically?
τ ≡ r×F
What is the direction?
The direction of the torque follows the right-hand rule!!
C ≡ A× B
   

C = A × B = A B sin θ
The above operation is called the
Vector product or Cross product
What is the result of a vector product?
Another vector
Thursday, Oct. 23, 2014
What is another vector operation we’ve learned?
   
Scalar product C ≡ A ⋅ B = A B cos θ
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
Result? A scalar
5
Properties of Vector Product
Vector Product is Non-commutative
What does this mean?
   
If the order of operation changes the result changes A × B ≠ B × A

Following the right-hand rule, the direction changes  
Vector Product of two parallel vectors is 0.

A × B = −B × A

 
 
 
C = A × B = A B sin θ = A B sin 0 = 0 Thus,
 
A× A= 0
If two vectors are perpendicular to each other
 
   
 
A × B = A B sin θ = A B sin 90 = A B = AB
Vector product follows distribution law
      
A× B+C = A× B+ A×C
(
)
The derivative of a Vector product with respect to a scalar variable is
 


d A× B
d A   d B
=
× B+ A×
dt
dt
dt
(
Thursday, Oct. 23, 2014
)
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
6
More Properties of Vector Product
     
i × i = j × j = k × k= 0
 
  
i× j = −j×i = k
  
 
j × k = −k × j = i
  
 
k × i = −i × k = j
The relationship
 between

unit vectors, i, j and k
Vector product of two vectors can be expressed in the following determinant form
 
A× B=

i

j

k
Ax
Ay
Az
Bx
By
Bz
(
= Ay Bz − Az By
 Ay
=i
By

) i −(A B
x
Thursday, Oct. 23, 2014
z
Az
Bz

−j
Ax
Az
Bx
Bz

+k
Ax
Ay
Bx
By


− Az Bx ) j + Ax By − Ay Bx k
(
)
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
7
Moment of Inertia
Rotational Inertia:
For a group
of particles
The measure of resistance of an object
to changes in its rotational motion.
Equivalent to mass in linear motion.
I ≡ ∑ mi ri2
i
For a rigid
body
I ≡ ∫ r 2 dm
What are the dimension and
2
2
kg
⋅
m
⎡
⎤
ML
unit of Moment of Inertia?
⎣
⎦
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Dependent on the axis of rotation!!!
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
8
Ex. The Moment of Inertia Depends on
Where the Axis Is.
Two particles each have mass m1 and m2 and are
fixed at the ends of a thin rigid rod. The length of
the rod is L. Find the moment of inertia when this
object rotates relative to an axis that is
perpendicular to the rod at (a) one end and (b) the
center.
2
2
(a) I = ∑ mr 2 = m1r1 + m2 r2
( )
r1 = 0 r2 = L
m1 = m2 = m
() ( )
(b) I = ∑ ( mr ) = m r + m r
2
2
I = m 0 + m L = mL2
2
11
2
m1 = m2 = m
( )
2
Which case is easier to spin?
2
2 2
Case (b)
Why? Because the moment
of inertia is smaller
r1 = L 2 r2 = L 2
( )
2
2
I = m L 2 + m L 2 = 12 mL
Thursday, Oct. 23, 2014
9
Rotational Kinetic Energy
y
vi
ri
O
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
mi
θ#
x
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
2
Kinetic energy of a masslet, mi, K i = mi vi = mi ri 2ω 2
2
2
moving at a tangential speed, vi, is
Since a rigid body is a collection of masslets, the total kinetic energy of the
rigid object is
1⎛
1
2⎞ 2
2 2
K R = ∑ K i= ∑ mi ri ω = ⎜ ∑ mi ri ⎟ω
2⎝ i
2 i
i
⎠
2
Since moment of Inertia, I, is defined as I = ∑ mi ri
i
The above expression is simplified as
Thursday, Oct. 23, 2014
1
K R = Iω 2
2
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
10
Example for Moment of Inertia
In a system of four small spheres as shown in the figure, assuming the radii are negligible
and the rods connecting the particles are massless, compute the moment of inertia and
the rotational kinetic energy when the system rotates about the y-axis at angular speed ω.
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
M
x
b
I = ∑ mi ri2 = Ml 2+ Ml 2 + m ⋅ 02 + m ⋅ 02 = 2Ml 2
i
This is because the rotation is done about y axis,
and the radii of the spheres are negligible.
1
1
K R = Iω 2= 2Ml 2 ω 2 = Ml 2ω 2
2
2
Why are some 0s?
m
(
Thus, the rotational kinetic energy is
)
Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.
(
2
2
2
I = ∑ mi ri 2 = Ml+
Ml 2+ mb+2 mb 2 = 2 Ml + mb
i
Thursday, Oct. 23, 2014
)
1
1
K R = Iω 2 = (2Ml 2 + 2mb 2 )ω 2 = (Ml 2 + mb 2 )ω 2
2
2
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
11
Calculation of Moments of Inertia
Moments of inertia for large rigid objects can be computed, if we
assume the object consists of small volume elements with mass, Δmi.
2
2
The moment of inertia for the large rigid object is I = Δmlim→0 ∑ ri Δmi = ∫ r dm
i
It is sometimes easier to compute moments of inertia in terms
of volume of the elements rather than their mass
dm
Using the volume density, ρ, replace ρ =
dV
dm in the above equation with dV.
i
How can we do this?
dm = ρdV The moments of
inertia becomes
I = ∫ ρr 2 dV
Example: Find the moment of inertia of a uniform hoop of mass M and radius R
about an axis perpendicular to the plane of the hoop and passing through its center.
y
O
The moment
of inertia is
dm
R
x
Thursday, Oct. 23, 2014
I = ∫ r 2 dm = R 2 ∫ dm = MR 2
What do you notice
from this result?
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
12
Ex. Rigid Body Moment of Inertia
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis perpendicular to the rod and passing through its center of mass.
M
The line density of the rod is λ =
y
L
so the mass of a masslet is dm = λ dx = M dx
L
dx
The moment
of inertia is
x
x
L
=
What is the moment of inertia
when the rotational axis is at
one end of the rod.
Will this be the same as the above.
Why or why not?
L /2
M ⎡1 3⎤
x2 M
x
dx =
I = ∫ r dm = ∫
− L /2
L ⎢⎣ 3 ⎥⎦ − L / 2
L
L /2
2
3
3
M ⎡⎛ L ⎞
⎛ L ⎞ ⎤ M ⎛ L3 ⎞ ML2
=
⎢ ⎜ ⎟ − ⎜ − ⎟ ⎥=
⎝ 2 ⎠ ⎥⎦ 3L ⎜⎝ 4 ⎟⎠
3L ⎢⎣⎝ 2 ⎠
12
I
L
M ⎡1 3⎤
x2 M
x
dx=
= r dm = ∫0
L ⎢⎣ 3 ⎥⎦ 0
L
M
M 3
ML2
⎡( L ) 3 − 0 ⎤ =
=
L
=
⎦ 3L
3L ⎣
3
∫
2
L
( )
Since the moment of inertia is resistance to motion, it makes perfect sense
for it to be harder to move when it is rotating about the axis at one end.
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
13
Parallel Axis Theorem
Moments of inertia for a highly symmetric object is easy to compute if the
rotational axis is the same as the axis of symmetry. However if the axis of
rotation does not coincide with axis of symmetry, the calculation can still be
done in a simple manner using the parallel-axis theorem. I = I CM + MD 2
y
Moment of inertia is defined as
yCM
y
y’
(x,y)
r
D
xCM
What does this
Thursday, Oct.
2014
theorem
tell23,you?
)
x = xCM + x ' y = yCM + y'
One can substitute x and y in Eq. 1 to obtain
2
2
I = ∫ ⎡( xCM + x ') + ( yCM + y') ⎤ dm
⎣
⎦
CM
2
2
(xCM,yCM)
= xCM + yCM ∫ dm + 2xCM ∫ x 'dm + 2yCM ∫ y'dm + ∫ x '2 + y'2 dm
(
x’
x
Since x and y are
(
I = ∫ r 2 dm = ∫ x 2 + y 2 dm (1)
)
(
)
Since x’ and y’ are the distance
∫ x'dm = 0 ∫ y'dm = 0
x from CM, by definition
Therefore, the parallel-axis theorem
2
2
I = ( xCM
+ yCM
) ∫ dm + ∫ ( x '2 + y'2 )dm = MD 2 + I CM
Moment of inertia of any object about any arbitrary axis are the same as
the sum of PHYS
moment
of inertia
for a rotation about the CM and that
1443-004,
Fall 2014
14 of
Dr. Jaehoon
Yu axis.
the CM about the
rotation
Example for Parallel Axis Theorem
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis that goes through one end of the rod, using parallel-axis theorem.
The line density of the rod is λ =
y
M
L
so the mass of a masslet is dm = λdx =
CM
dx
x
L
x The moment of
inertia about
the CM
I CM
M
dx
L
L /2
M ⎡1 3⎤
x2 M
x
= ∫ r dm = ∫
dx =
− L /2
L ⎢⎣ 3 ⎥⎦ − L / 2
L
L /2
2
M ⎡⎛ L ⎞
⎛ L⎞
=
⎢⎜ ⎟ − ⎜ − ⎟
⎝
⎠
⎝ 2⎠
3L ⎢⎣ 2
3
3
⎤ M ⎛ L3 ⎞ ML2
⎥=
⎜ ⎟ = 12
⎥⎦ 3L ⎝ 4 ⎠
2
I = I CM + D 2 M =
Using the parallel axis theorem
ML2 ⎛ L ⎞
ML2 ML2 ML2
+⎜ ⎟ M =
+
=
12 ⎝ 2 ⎠
12
4
3
The result is the same as using the definition of moment of inertia.
Parallel-axis theorem is useful to compute moment of inertia of a rotation of a rigid
object with complicated shape about an arbitrary axis
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
15
Check out Table
10 – 2 for
moment of
inertia for
various shaped
objects
Thursday, Oct. 23, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
16