Transcript Physics 106P: Lecture 1 Notes

Physics 101: Chapter 9

Today’s lecture will cover Textbook Sections 9.1 - 9.6
UB, Phy101: Chapter 9, Pg 1
See text: chapter 8
Rotation Summary
(with comparison to 1-D kinematics)
Angular
Linear
  constant
a  constant
ω  ω0  αt
v  v 0  at
1
  0   0 t  t 2
2
x  x 0  v 0t 
 2   02  2Δθ
v2  v20  2ax
1 2
at
2
And for a point at a distance R from the rotation axis:
x = Rv = R
a = R
See Table 8.1
UB, Phy101: Chapter 9, Pg 2
See text: chapter 9
New concept: Torque
Rotational analog of force
Torque = (magnitude of force) x (lever arm)
t=Fl
UB, Phy101: Chapter 9, Pg 3
Comment on axes and sign
(i.e. what is positive and negative)
Whenever we talk about rotation, it is implied that there
is a rotation “axis”.
This is usually called the “z” axis (we usually omit the z
subscript for simplicity).
Counter-clockwise (increasing ) is usually
called positive.
Clockwise (decreasing ) is usually
called negative.

z
UB, Phy101: Chapter 9, Pg 4
Chapter 9, Preflight
The picture below shows three different ways of using a wrench to
loosen a stuck nut. Assume the applied force F is the same in each case.
In which of the cases is the torque on the nut the biggest?
1. Case 1
2. Case 2
3. Case 3
CORRECT
UB, Phy101: Chapter 9, Pg 5
Chapter 9, Preflight
The picture below shows three different ways of using a wrench to
loosen a stuck nut. Assume the applied force F is the same in each case.
In which of the cases is the torque on the nut the smallest?
1. Case 1
2. Case 2
3. Case 3
CORRECT
UB, Phy101: Chapter 9, Pg 6
UB, Phy101: Chapter 9, Pg 7
UB, Phy101: Chapter 9, Pg 8
Static Equilibrium
A system is in static equilibrium if and only if:
acm = 0  Fext = 0
  = 0  text = 0 (about any axis)

pivot
torque about pivot due to
gravity:
tg = mgd
d
Center of mass
W=mg
(gravity acts at center of mass)
This object is NOT in static equilibrium
UB, Phy101: Chapter 9, Pg 9
pivot
Not in equilibrium
Equilibrium
pivot
d
Center of mass
W=mg
Center of mass
Torque about pivot  0
Torque about pivot = 0
UB, Phy101: Chapter 9, Pg 10
Homework Hints

Painter is standing to the right of the support B.
FA
FB
Mg
mg
What
is the maximum distance the painter can move to the
right without tipping the board off?
UB, Phy101: Chapter 9, Pg 11
Homework Hints

If its just balancing on “B”, then FA = 0
the only forces on the beam are:
x
FB
Mg
Using FTOT = 0:
FB = Mg + mg
mg
This does not tell us x
UB, Phy101: Chapter 9, Pg 12
Homework Hints

Find net torque around pivot B: (or any other place)
FB
d1
Mg
d2
mg
t (FB ) = 0 since lever arm is 0
t (Mg ) = Mgd1
t (mg ) = -mgd2
Total torque = 0 = Mgd1 -mgd2
So d2 = Md1 /m and you can use d1 to find x
UB, Phy101: Chapter 9, Pg 13
Homework Hints

Painter standing at the support B.
Find total torque
about this axis
D
FA
FB
d
Mg
mg
t(FA) = - FAD
t(Mg) = Mgd
t(FB) = 0 (since distance is 0)
t(mg) = 0 (since distance is 0)
Total torque = 0 = Mgd -FAD
So FA = Mgd /D
UB, Phy101: Chapter 9, Pg 14

MORE EXAMPLES (bar and weights suspended by the
string):
Find net torque around this (or any other) place
T
x
t (m1g) = 0 since lever arm is 0
m 1g
Mg
m2g
UB, Phy101: Chapter 9, Pg 15
T
L/2
t (m1g) = 0 since lever arm is 0
m 1g
Mg
m2g
t (Mg ) = -Mg L/2
UB, Phy101: Chapter 9, Pg 16
T
x
t (m1g) = 0 since lever arm is 0
m 1g
Mg
m2g
t (Mg ) = -Mg L/2
t (T ) = T x
UB, Phy101: Chapter 9, Pg 17
T
L
t (m1g) = 0 since lever arm is 0
m 1g
Mg
m2g
t (Mg ) = -Mg L/2
t (T ) = T x
t (m2g ) = -m2g L
All torques sum to 0: Tx = MgL/2 + m2gL
So x = (MgL/2 + m2gL) / T
UB, Phy101: Chapter 9, Pg 18
UB, Phy101: Chapter 9, Pg 19
UB, Phy101: Chapter 9, Pg 20
UB, Phy101: Chapter 9, Pg 21
Moment of Inertia & Rotational KE
Textbook Sections 9.4 - 9.5:
UB, Phy101: Chapter 9, Pg 22
Torque and Stability
Center of mass outside of base:
Center of mass over base:
--> unstable
--> stable
UB, Phy101: Chapter 9, Pg 23
UB, Phy101: Chapter 9, Pg 24
UB, Phy101: Chapter 9, Pg 25
Moments of Inertia of Common Objects
Hollow cylinder or hoop about central axis
I = MR2
Solid cylinder or disk about central axis
I = MR2/2
Solid sphere about center
I = 2MR2/5
Uniform rod about center
I = ML2/12
Uniform rod about end
I = ML2/3
UB, Phy101: Chapter 9, Pg 26
UB, Phy101: Chapter 9, Pg 27
UB, Phy101: Chapter 9, Pg 28
Chapter 9, Preflight
The picture below shows two different dumbbell shaped objects.
Object A has two balls of mass m separated by a distance 2L, and
object B has two balls of mass 2m separated by a distance L.
Which of the objects has the largest moment of inertia for rotations
around the x-axis?
m
1. A
CORRECT
2. B
2m
3. Same
2L
L
x
2m
m
A
I = mL2 + mL2
= 2mL2
B
I = 2m(L/2)2 + 2m(L/2)2
= mL2
UB, Phy101: Chapter 9, Pg 29
Rotational Kinetic Energy
Translational kinetic energy:
KEtrnas = 1/2 MV2cm
Rotational kinetic energy:
KErot = 1/2 I2
Rotation plus translation:
KEtotal = KEtrans + KErot = 1/2 MV2cm + 1/2 I2
UB, Phy101: Chapter 9, Pg 30
Angular Momentum

Textbook Section 9.6
UB, Phy101: Chapter 9, Pg 31
See text: chapters 8-9
Define Angular Momentum
Momentum
Angular Momentum
p = mV
L = I
conserved if Fext = 0
conserved if text =0
Vector
Vector!
units: kg-m/s
units: kg-m2/s
See Table 8.1
UB, Phy101: Chapter 9, Pg 32
Chapter 9, Pre-flights
You are sitting on a freely rotating bar-stool with your arms stretched out
and a heavy glass mug in each hand. Your friend gives you a twist and you
start rotating around a vertical axis though the center of the stool. You can
assume that the bearing the stool turns on is frictionless, and that there is
no net external torque present once you have started spinning.
You now pull your arms and hands (and mugs) close to your body.
UB, Phy101: Chapter 9, Pg 33
Chapter 9, Preflight
What happens to your angular momentum as you pull in your arms?
1. it increases
2. it decreases
3. it stays the same
CORRECT
L1
L
2
This is like the spinning skater example in the
book. Since the net external torque is zero (the
movement of the arms and hands involve internal
torques), the angular momentum does not change.
UB, Phy101: Chapter 9, Pg 34
Chapter 9, Preflight
What happens to your angular velocity as you pull in your arms?
1
1. it increases
CORRECT
2. it decreases
3. it stays the same
2
I2
I1
L
L
as with the skater example given in the book....as you pull
your arms in toward the rotational axis, the moment of
inertia decreases, and the angular velocity increases.
My friends and I spent a good half hour doing this once,
and I can say...based on a great deal of nausea, that the
angular velocity does increase.
UB, Phy101: Chapter 9, Pg 35
Chapter 9, Preflight
What happens to your kinetic energy as you pull in your arms?
1
1. it increases
CORRECT
2. it decreases
3. it stays the same
2
I2
I1
L
K
1
1 2 2
I 2 
I 
2
2I

1 2
L
2I
L
(using L = I )
Your angular velocity increases and moment of inertia
decreases, but angular velocity is squared, so KE will
increase with increasing angular velocity
UB, Phy101: Chapter 9, Pg 36
Spinning disks

Two different spinning disks have the same angular momentum,
but disk 2 has a larger moment of inertia than disk 1.
Which one has the biggest kinetic energy ?
(a) disk 1
(b) disk 2
UB, Phy101: Chapter 9, Pg 37
K
1
1 2 2
I 2 
I 
2
2I

1 2
L
2I
(using L = I )
If they have the same L, the one with the smallest I
will have the biggest kinetic energy.
L  I1 1
L  I2  2
1
2
I1
disk 1
<
I2
disk 2
UB, Phy101: Chapter 9, Pg 38
Preflights: Turning the bike wheel
A student sits on a barstool holding a bike wheel. The wheel is
initially spinning CCW in the horizontal plane (as viewed from
above). She now turns the bike wheel over. What happens?
1. She starts to spin CCW.
2. She starts to spin CW.
3. Nothing
CORRECT
UB, Phy101: Chapter 9, Pg 39
Turning the bike wheel...

Since there is no net external torque acting on the studentstool system, angular momentum is conserved.
 Remenber,
L has a direction as well as a magnitude!
Initially: LINI = LW,I
Finally: LFIN = LW,F + LS
LS
LW,I
LW,I = LW,F + LS
LW,F
UB, Phy101: Chapter 9, Pg 40
See text: chapters 8-9
Rotation Summary
(with comparison to 1-d linear motion)
Angular
  constant
  0  t
1
  0   0 t  t 2
2
t  I
KErot 
1
I 2  L2 / 2 I
2


L  
See Table 8.1
Linear
a  constant
v  v 0  at
x  x 0  v 0t 
1 2
at
2
F  ma
1
mv2  p 2 /( 2m)
2


p  mv
KEtrans 
UB, Phy101: Chapter 9, Pg 41