A mass m is attached to a long , massless rod. The mass is close to
one end of the rod. Is it easier to balance the rod on end with the
mass near the top or near the bottom? Hint: Small α means sluggish behavior and α = τ/I. A) Easier with mass near top. B) Easier with mass near bottom. C) No difference. τ ≡ rF⊥ = Iα
I point = mr 2
L35 F 11/14/14 a*er lecture 1 A mass m is attached to a long , massless rod. The mass is close to
one end of the rod. Is it easier to balance the rod on end with the
mass near the top or near the bottom? Hint: Small α means sluggish behavior and α = τ/I. F⊥ = mgsin θ
L
θ
θθ
mg
L35 F 11/14/14 a*er lecture τ = Iα
τ F⊥ L mgsin θ
α= =
=
2
I mL
mL
gsin θ 1
=
∝
L
L
2 A mass m is attached to a long , massless rod. The mass is close to
one end of the rod. Is it easier to balance the rod on end with the
mass near the top or near the bottom? Hint: Small α means sluggish behavior and α = τ/I. A) Easier with mass near top. B) Easier with mass near bottom. C) No difference. 1
α∝
⇒ Large L means a small α , so the bar is
L
easier to balance.
L35 F 11/14/14 a*er lecture 3 Assignments
Announcements:
•  No HW due in recitation next week. CAPA 12 is now live.
•  You should have read Ch. 10 and 11.
Coming Up:
•  Midterm exam 3: NEXT week -- Thurs Nov 20, same time, same
place. Primary focus will be on Chs. 8-11.
•  Old exam 3 has been placed on D2L.
Today:
•  Continuing with rotational motion, found in Chs. 10 & 11: torque,
moment of inertia, rotational kinetic energy, conservation of energy
including rotational motion. Then moving on to angular momentum
and its conservation: re-express all quantities as vectors.
L35 F 11/14/14 a*er lecture 4 Rotational Kinetic Energy and Conservation of Energy
ω
M
v
ω
v
(same units: J)
(v/r)2
L35 F 11/14/14 a*er lecture 5 Rotational Kinetic Energy and Conservation of Energy
1
1
Mv 2 + Iω 2
2
2
2
I sphere = MR 2
5
I hoop = MR 2
KEtot =
I disk =
Sphere:
2MgH
10
vsphere = R
=
gH
2
I + MR
7
> vdisk > vhoop
L35 F 11/14/14 a*er lecture 1
MR 2
2
Hoop:
vhoop = gH
Disk:
vdisk =
4
gH
3
6 Rotational Kinetic Energy and Conservation of Energy
1
1
Mv 2 + Iω 2
2
2
10
vsphere =
gH
7
KEtot =
Which has the greater speed at the bottom of the ramp, the sphere
that rolls down the ramp or a block of the same mass that slides
down the ramp? (Assume sliding friction is negligible.)
A) Block L35 F 11/14/14 a*er lecture B) Sphere C) Both the same.
7 Rotational Kinetic Energy and Conservation of Energy
1
1
Mv 2 + Iω 2
2
2
10
vsphere =
gH
7
KEtot =
What’s the speed of the block at the bottom of the frictionless ramp?
11
A)
7
gH
B) 2gH
1
2
MgH = Mvblock
→
2
L35 F 11/14/14 a*er lecture 19
C)
gH
7
D) 3gH
vblock = 2gH > vsphere
8 Rotational Kinetic Energy and Conservation of Energy
H
1
⎛
2
2⎞
I
=
MR
,
I
=
MR
⎜⎝ hoop
⎟⎠
disk
2
Which object will go furthest up the incline?
A) Puck B) Disk
C) Hoop Because the hoop has the largest moment of inertia and
therefore the
highest total kinetic energy.
L35 F 11/14/14 a*er lecture D) Same height. KEi = PE f
KEtrans + KErot = MgH
2
1
1 ⎛ v⎞
2
Mv + I ⎜ ⎟ = MgH
2
2 ⎝ r⎠
9 Two Miscellaneous Subjects – before going on to angular
momentum
1. Center of Mass
y
mi

ri

rCM =

∑ mi ri
i
∑m
i
x
Example:
y
1
i
m1 = 2 kg
m2 = 1 kg
(2,4)
x
(5,1) 2
yCM
x
L35 F 11/14/14 a*er lecture xCM
⎧
mi xi
∑
⎪x = i
⎪ CM
mi
∑
⎪
i
⎨
∑ mi yi
⎪
⎪ yCM = i
⎪
∑ mi
i
⎩
m1 x1 + m2 x2 2(2) + 1(5)
9
=
=
m = m = 3m
m1 + m2
2 +1
3
m1 y1 + m2 y2 2(4) + 1(1)
9
=
=
m = m = 3m
m1 + m2
2 +1
3
(3,3)
10 Two Miscellaneous Subjects – before going on to angular
momentum
1. Center of Mass
y

ri
mi

rCM =

∑ mi ri
i
∑m
i
x
i
⎧
mi xi
∑
⎪x = i
⎪ CM
mi
∑
⎪
i
⎨
∑ mi yi
⎪
⎪ yCM = i
⎪
∑ mi
i
⎩
Remember:
1. You can choose to place your origin anywhere, but your center of mass will be
relative to your origin.
2. It’s often useful to place your center of mass at one of your masses.
3. If there is a continuous distribution of masses, then the center of mass formula
is revised into an integral. L35 F 11/14/14 a*er lecture 11 ∑m x
=
∑m
i i
Two Miscellaneous Subjects
xCM
1. Center of Mass
i
i
i
A hyrogen (m = 1u) and a chlorine (m = 35u) atom are separated by
about 100 pico-meter (i.e., 10-10 m) in a HCl molecule. How far from
the chlorine atom is the center of mass of the molecule?
y
Cl
H x
10-10 m
L35 F 11/14/14 a*er lecture 1 −10
1 −10
A)
10 m
B)
10 m
36
35
35 −10
C)
10 m
D) 10 −10 m
36
mH xH
1
xcm =
=
10 −10 m
mH + mCl 1+ 35
1 −10
= 10 m
36
12 Two Miscellaneous Subjects – before going on to angular
momentum
2. Parallel Axis Theorem
If you know the moment of inertia of an object of
mass M about a particular axis through the object’s
center-of-mass, then its moment of inertia about any
axis parallel to the original axis is:
cm
L
cm
I cm
L35 F 11/14/14 a*er lecture 1
= ML2
12
I = I cm + Md
2
d = L/2
2
1
1
⎛ L⎞
I = I cm + M ⎜ ⎟ = ML2 + ML2
⎝ 2⎠
12
4
1
= ML2
3
13 Two Miscellaneous Subjects – before going on to angular
momentum
2. Parallel Axis Theorem
I cm
2
= Mr 2
5
I = I cm + Md
2
If the moment of inertia through the
center of mass of a sphere of radius
r and mass M is Icm as given at left,
what’s the moment of inertia about
a parallel axis tangent to the edge of
the sphere as shown at right?
12
A) Mr
D)
Mr 2
5
2
7
2
2
2
I = I cm + Mr = Mr + Mr = Mr 2
5
5
L35 F 11/14/14 a*er lecture 14 2
2
B) Mr 2
5
7
C) Mr 2
5
Transition to Discussion of Angular Momentum
Previously: considered only rotating systems with a fixed axis
angular position (rad)
angular velocity (rad/s)
angular acceleration (rad/s2)
moment of inertia (kg m2)
Torque (N m)
L35 F 11/14/14 a*er lecture θ
ω
α
I
τ
All scalar quantities but with a sign.
15 Transition to Discussion of Angular Momentum
Now: consider rotating systems with a moveable axis
angular position (rad)
angular velocity (rad/s)
angular acceleration (rad/s2)
moment of inertia (kg m2)
Torque (N m)
L35 F 11/14/14 a*er lecture θ

ω

α
I

τ
Some quantities become vectors.
And we introduce another quantity,
angular momentum, which is a
vector quantity:

L
(kg m2/s)
16 Transition to Discussion of Angular Momentum

Angular velocity as a vector quantity:
ω
 dθ
ω =
dt
Direction of angular velocity is given by the
“Right Hand Rule”:
magnitude of angular velocity
L35 F 11/14/14 a*er lecture 17 Transition to Discussion of Angular Momentum

 

 dω Δω ω 2 − ω 1
≈
=
Angular acceleration as a vector quantity:
α =
dt
Δt
Δt
L35 F 11/14/14 a*er lecture 18 Transition to Discussion of Angular Momentum

 

 dω Δω ω 2 − ω 1
≈
=
Angular acceleration as a vector quantity:
α =
dt
Δt
Δt

A disk is spinning as shown with angular velocity ω . It begins to slow down. While it is slowing, what is the direction of its 
vector angular acceleration α ? 

ω
2
ω1

Δω



ω 2 = ω 1 + Δω


α points in the direction Δω
L35 F 11/14/14 a*er lecture 19