12798726.ppt

Physics 211
Lecture 15
Today’s Concepts:
a) Parallel Axis Theorem
b) Torque & Angular Acceleration
Mechanics Lecture 15, Slide 1

Parallel Axis Theorem
Smallest when D = 0

A solid ball of mass M and radius is connected to a thin rod
of mass m and length L as shown. What is the moment of
inertia of this system about an axis perpendicular to the
other end of the rod?
M
L
m
R
axis
ACT
2
2
2
3
1
5
2
ML
mL
MR
I 

=
2
2
2
3
1
5
2
mL
ML
MR
I 

=
2
2
3
1
5
2
mL
MR
I 
=
2
2
3
1
mL
ML
I 
=
A)
B)
C)
D)

A ball of mass 3M at x = 0 is connected to a ball of mass M at x = L by a
massless rod. Consider the three rotation axes A, B, and C as shown, all
parallel to the y axis.
For which rotation axis is the
moment of inertia of the object
smallest? (It may help you to
figure out where the center of
mass of the object is.)
B
3M M
C
A
L/2
L/4
0
x
y
L
CheckPoint
Demo: rod w/spheres
Hooray!

A ball of mass 3M at x = 0 is connected to a ball of mass M at x = L by a
massless rod. Consider the three rotation axes A, B, and C as shown, all
parallel to the y axis.
For which rotation axis is the
moment of inertia of the object
smallest? (It may help you to
figure out where the center of
mass of the object is.)
B
3M M
C
A
L/2
L/4
0
x
y
L
CheckPoint
Xcm = L/4, and that is at axis B
Iaxis = Icm + MtotD2
Mtot = 4 M
IB = Icm
IA = Icm + MtotD2 = cm + (4 M)(L/4) 2
IC = Icm + MtotD2 = cm + (4 M)(L/4) 2

Your Comments
Yes, D is the distance from CM to the new axis of rotation.
Good, the moment of inertia has the smallest value for rotation about the center of mass!
Any other axis will give a larger moment of inertia.
When we say “moment of inertia” we’re not finished. We always have to say — or think —
“moment of inertia about some particular axis of rotation”.
Be careful. Moment of inertia is about an axis, not with respect to a point.
The CM is on axis B.
Great!

Right Hand Rule for finding Directions
I didn't really get how to figure out
the direction of angular acceleration
by the right hand rule. Why is it
perpendicular to the direction of
rotation? So confusing!!!!!! Please
help me!!!!!!!!!!!
Demo: bike wheel

ACT
A ball rolls across the floor, and then starts up a ramp as
shown below. In what direction does the angular velocity
vector point when the ball is rolling up the ramp?
A) Into the page
B) Out of the page
C) Up
D) Down
Demo: roll ball up board

ACT
shown below. In what direction does the angular
acceleration vector point when the ball is rolling up the ramp?
A) Into the page
B) Out of the page

ACT
shown below. In what direction does the angular
acceleration vector point when the ball is rolling back down
the ramp?
A) into the page
B) out of the page

Torque
t = rF sin(q )
Demo: wrench & bolts

Torque and F=ma
I am still not comfortable with the fact that the torque equation
is somehow the same as the net force equation. Can we go over
how that is possible in lecture?
F ma
=

r r
F ma
q q
=

F mr
q 
=

2
rF mr
q 
=

2
r F mr 
 =

r
r
I
t 
=

r
r
F
q
v

In Case 1, a force F is pushing perpendicular on an object a
distance L/2 from the rotation axis. In Case 2 the same force is
pushing at an angle of 30 degrees a distance L from the axis.
In which case is the torque due to the force about the rotation
axis bigger?
A) Case 1 B) Case 2 C) Same
F
L/2 90o
Case 1
axis
L
F
30o
Case 2
axis
CheckPoint

In which case is the torque due to the force about
the rotation axis bigger?
A) Perpendicular force
means more torque.
B) F*L = torque. L is
bigger in Case 2 and the
force is the same.
C) Fsin30 is F/2 and its
radius is L so it is FL/2 which
is the same as the other one
as it is FL/2.
F
L/2 90o
Case 1
axis
L
F
30o
Case 2
axis
CheckPoint

In which case is the torque due to the force about
the rotation axis bigger?
C) Fsin30 is F/2 and its
radius is L so it is FL/2 which
is the same as the other one
as it is FL/2.
F
L/2 90o
Case 1
axis
L
F
30o
Case 2
axis
CheckPoint
Be careful; both
expressions have a
factor of 0.5 in them but
for different reasons.

Your Comments
Great!
Great!
And it is — for different reasons!
And it is — for different reasons!
Great!
Great!
I got carried away with “Great!” but I’m happy. It seems like “torque” may be well understood.

Similarity to 1D motion

ACT
Strings are wrapped around the circumference of two solid
disks and pulled with identical forces. Disk 1 has a bigger
radius, but both have the same moment of inertia.
Which disk has the bigger angular acceleration?
A) Disk 1
B) Disk 2
C) same F
F
w1
w2

CheckPoint
Two hoops can rotate freely about fixed axles through their
centers. The hoops have the same mass, but one has twice
the radius of the other. Forces F1 and F2 are applied as
shown.
How are the magnitudes of the two forces related if the
angular acceleration of the two hoops is the same?
A) F2 = F1
B) F2 = 2F1
C) F2 = 4F1
F1
F2
Case 1 Case 2

CheckPoint
A) F2 = F1
B) F2 = 2F1
C) F2 = 4F1
A) If the angular acceleration is the same,
they must have the same tangential force.
B) Twice the radius=twice the force.
C) twice the radius means 4 times the moment of
inertia, thus 4 times the torque, thus 4 times the
force
M, R
M, 2R
F1
F2
Case 1 Case 2

CheckPoint
A) F2 = F1
B) F2 = 2F1
C) F2 = 4F1
I2 = 4 I1 because of the radius.
Torque1 = R F1 = I1 alpha
Torque2 = 2R F2 = I2 alpha
F2 = I2 alpha/2R
F1 = I1 alpha/R
F2/F1 = (I2/2)/(I1) = (1/2)(I2/I1) = (1/2)(4)
F2/F1 = 2 or F2 = 2 Fa
M, R
M, 2R
F1
F2
Case 1 Case 2

Wait! We can’t have F = R; they have different units!
But why?
? ? ?
What is “it”?
Is the “bigger” force 2x or 4x as big?
Tell me more.
Why angular velocity instead of angular accelertion.
I don’t understand the significance of “torque * I “.
Tell me more.
Tell me more.

sin
RF
t q
=
0
90
q =
0
0
q =
q
90 36 54
q =  =

sin
RF
t q
=
Direction is perpendicular to
both R and F, given by the
right hand rule
0
x
t =
0
y
t =
1 2 3
z F F F
t t t t
=  

2
1
2
DISK
I MR
=
(i)
2
1
2
K Iw
=
(iii)
I
t 
=
(ii)
Use (i) & (ii)
Use (iii)

12798726.ppt

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