Download Physics 231 Topic 8: Rotational Motion Wade Fisher October 12-19 2012

Physics 231
Topic 8: Rotational Motion
Wade Fisher
October
12-19
2012
MSU Physics
231 Fall 2012
1
Key Concepts: Rotational Motion
Rotational Kinematics
Eqns of Motion for Rotation
Tangential Motion
Kinetic Energy and Rotational Inertia
Moment of Inertia
Rotational dynamics
Rolling bodies
Vector quantities
Mechanical Equilibrium
Angular Momentum
Covers chapter 8 in Rex & Wolfson
MSU Physics 231 Fall 2012
2
quiz: extra credit
The orange block collides with the blue block (same mass).
Which of the following is not true?
MSU Physics 231 Fall 2012
3
Radians & Radius
Circumference=2r
Part s=r
r
s

=s/r
 in radians!
360o=2 rad = 6.28… rad
 (rad) = /180o  (deg)
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Angular speed and acceleration
 f i



t f  ti
t

  lim
t 0 t
Average angular
velocity
Instantaneous
Angular velocity
Angular velocity : rad/s
rev/s
rpm (revolutions per
minute)
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example
What is the angular velocity of earth around the sun?
Give in rad/s, rev/s and rpm
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Angular acceleration
Definition: The change in angular velocity per time unit
 f  i
 Average angular


t f  ti
t acceleration

  lim
t 0 t
Instantaneous angular
acceleration
Unit: rad/s2
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Equations of motion
Linear motion
Angular motion
X(t)=x(0)+v(0)t+½at2
(t)= (0)+(0)t+½t2
V(t)=V(0)+at
(t)= (0)+t
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example
=0
A person is rotating a wheel. The
handle is initially at =90o. For 5s
the wheel gets an constant angular
acceleration of 1 rad/s2. After that
The angular velocity is constant.
Through what angle will the wheel
have rotated after 10s.
MSU Physics 231 Fall 2012
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question
A child sits on a merry go-around and makes 10 revolutions.
Through what angle has the child traveled?
a) 10 rad
b) 20 rad
c) 20 rad
d) 2 rad
MSU Physics 231 Fall 2012
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Clicker Quiz!
A pendulum with length L of 1 meter and a mass of 1
kg, is oscillating harmonically.
The period of oscillation is T.
If L is increased to 4 m, and the mass replaced by one
of 0.5 kg, the period becomes:
A) 0.5T
B) 1T
C) 2T
D) 4T
E) 8T
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Previously…
 f i
 Average angular


t f  ti
t velocity (rad/s)
 Instantaneous
  lim
t 0 t
Angular velocity
 f  i
 Average angular


2)
acceleration
(rad/s
t f  ti
t
 Instantaneous angular
  lim
t 0 t
acceleration
MSU Physics 231 Fall 2012
2 rad  3600
10  2/360 rad
1 rad 360/2 deg
12
Equations of motion
Linear motion
Angular motion
x(t) = xo + vot + ½at2
(t) = o + ot + ½t2
v(t) = vo + at
(t) = o + t
Angular motion = Rotational Motion!
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Angular  Linear Velocities
s
 
r
 1 s

t r  t
v

r
v  r
V
v is called the tangential velocity
or linear velocity
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An Example
What is the angular speed about the rotational axis of the
earth for a person standing on the surface?
a) 7.3x10-5 rad/s
b) 3.6x10-3 rad/s
c) 6.28x10-2 rad/s
d) 3.14x100 rad/s
e) ????
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Angular  Linear Accelerations
v  r
v    r
v 

r
t
t
a  r
velocity
Change in velocity
Change in velocity per time unit
acceleration
The linear acceleration equals the angular acceleration
times the radius of the orbit
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An Example
A
B
The angular velocity of A is 2 rad/s.
1) What is its tangential velocity?
10 m/s

5m
If B is keeping pace with A,
2) What is its angular velocity?
3) What is its linear velocity?
6m
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A Rolling Coin
D
d
0=18 rad/s
=-1.80 rad/s
d=0.02 m
a) How long does the coin roll before stopping?
b) What is the average angular velocity?
c) How far (D) does the coin roll before coming to rest?
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Clicker Quiz!
Two planets A and B travel around a star. The radius of
orbit of A is twice larger than that of planet B. If
both travel once around the star in the same period,
which of the following is not true?
a) The angular velocities of A and B are the same
b) The linear velocities of A and B are the same
c) The tangential velocities of A and B are different
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gears
r1 = 0.3
r3=0.7 m
If 1=3 rad/s, how fast
Is the bike going?
b) What if r1=0.1 m ?
r2=0.15
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question
2
r1 = 0.3
r3=0.7 m
r2=0.15
3.5 m/s
If the bike is going
at 3.5 m/s and 1=3 rad/s,
what is the radius of r1?
a)
b)
c)
d)
e)
0.15 m
0.20 m
0.25 m
0.3 m
0.35 m
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Driving a car through a bend
Is there a force that pushes
you away from the center of
the circle?
• Newton’s first law: If no net force is acting on an object,
it will continue with the same velocity (inertia of mass)
• Velocity is a vector (points to a direction)
• If no net force is acting on an object, it will not change
its direction.
• A force is acting on the car (steering+friction) but you
tend to go in the same direction as you were going!
• It is not a force that pushes you, but the lack of it!
• The side door/friction of chair will keep you from sliding
out: it exerts a force on you and you exert a force on the
door/car seat (F21=-F12)
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Centripetal acceleration
ac=v2/r directed to the center of the circular motion
Also v=r, so ac=2r
This acceleration can be caused by various forces:
• gravity (objects attracted by earth)
• tension (object making circular motion on a rope)
• friction (car driving through a curve)
• etc
This acceleration is NOT caused by a mysterious force
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Car takes
turn
GPS goes
straight
When you drive through the
turn, the GPS seems to move to
the right, as if a force is acting
on it, but it is in fact the car
‘bending’ away from underneath
the tape. There is no force on
the tape!
When you drive through a bend, anything loose on your dash
board can slip (like the GPS above). It seems like there is a
force acting on the tape (sometimes referred to as the
centrifugal force), but it is in fact the lack of the
centripetal force acting on the GPS (not enough friction)
that makes it slide, while the car turns.
There is no “centrifugal force” acting on the GPS!! Anytime
the term centrifugal force is used, it really means “lack of
centripetal force”)
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A race car accelerating on a track.
v
ac
Tangential speed is constant
Velocity is changing direction…
so there must be an acceleration!
It is pointing towards the center
ac
v
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A general strategy
• As usual, make a drawing of the problem, if not given.
• Draw all the forces that are acting on the object(s)
under investigation.
• Decompose each of these into directions toward the
center of the circular path and perpendicular to it.
• Realize that Fto center = mac = mv2/r
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A car going through a bend
A car is passing through a corner with
radius 100 m. The static coefficient
of friction of the tires on the road is
0.5. What is the maximum velocity the
car can have without flying out of the
bend?
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Car on banked highway
Assuming there is no friction (I.e. the
frictional force cannot provide the centripetal
force), a car can be kept on the road by
banking the road. The centripetal force is
then provided by the horizontal component of
the normal force
If the radius is 100 m, what is the maximum velocity the
car can travel if the highway is banked at an angle of 15o?
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answer

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Centripetal Acceleration and
Tension.
Object swinging on a rope.
T
F=ma
T
T=mac
T=mv2/r=m2r
An object with m=1 kg is swung with a rope of length 3 m
around with angular velocity =2 rad/s. What is the tension
in the rope?
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Lifting by swinging
Swinging mass (m1) with velocity v
T
r
T
What is the relation between v and r
that will keep m2 stationary?
v out of paper
m1: T=m1a a=ac=centripetal accel.
m2: -T=m2g
Combine: a=-m2g/m1
Hanging mass (m2) Also: ac=(-)v2/r v=linear or tangential
velocity of m1
Fg=m2g
v2/r= m2g/m1
If m1 slows down, r must go down
so m2 sinks.
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2001: A space odyssey
A space ship rotates with
a linear velocity of 50 m/s.
What should the distance
from the central axis to
the crew’s cabin’s be so that
the crew feels like they are
on earth? (the floor of the
cabins is the inside of the
outer edge of the spaceship)
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Conical motion

What is the speed of the mass undergoing conical
motion if the mass is 1 kg and =20o?
The length of the string swinging the mass is 1 m.
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Previously…
v
ac
r
Centripetal acceleration:
ac=v2/r=2r
Caused by force like:
•Gravity
•Tension
•Friction
•F=mac for rotating object
The centripetal acceleration is caused by a change
in the direction of the linear velocity vector, not
a change in magnitude
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quiz (extra credit)
3 children are sitting on a rotating disc in a playground.
The disc starts to spin faster and faster.
Which of the three is most likely to start sliding first?
top view
a) child A
A
B
b) child B
c) child C
d) the same for all three
C
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Torque
d
Top view
O
It is much easier to swing the
door if the force F is applied
as far away as possible (d) from
the rotation axis (O).
F
Torque: The capability of a force to rotate an object about
an axis.
Torque
=F·d
(Nm)
Torque is positive if the motion is counterclockwise
Torque is negative if the motion is clockwise
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Multiple force causing torque.
100 N
0.3 m
Top view
0.6 m
50 N
Two persons try to go
through a rotating door
at the same time, one on
the l.h.s. of the rotator and
one the r.h.s. of the rotator.
If the forces are applied as
shown in the drawing, what
will happen?
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question
B
A
dA
dB
Two people are on a see saw. Person A is lighter than
person B. If person A is sitting at a distance dA from the
center of the seesaw, is it possible to find a distance dB so
that the seesaw is in balance?
A) Yes
B) NO
If MA=0.5MB and dA=3 m, then to reach balance, dB=…
A) 1 m B) 1.5 m C) 2 m D) 2.5 m E) 3 m
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Center of gravity.
Fpull
dpull
1 2 3………………………n
Fgravity
dgravity?
=Fpulldpull+Fgravitydgravity
Vertical direction
(I.e. side view)
We can assume that
for the calculation
of torque due to gravity,
all mass is concentrated
in one point:
The center of gravity:
the average position of
the mass
dcg=(m1d1+m2d2+…+mndn)
(m1+m2+…+mn)
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Center of Gravity; more general
Note: although “center of gravity” and “center if mass” are strictly
speaking not the same, in the situations we will consider they are
equal because the gravitational forces are assumed to be uniform
The center of gravity
m x

m
i
xCG
x1,y1
i
m1g
i
i
xcg,ycg
i
m y

m
i
y CG
O
x2,y2
m2g
x3,y3
m3g
i
i
i
i
Note 2: freely rotating systems (I.e. not held fixed at some point)
must rotate around their center of gravity.
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examples: A lot more in
the book!
Where is the center of gravity?
m x

m
i
xCG
16  0  1  0.1  cos(530 )  1  0.1  cos(530 ) 0.12


 0.0067
16  1  1
18
i
i
i
i
m y

m
i
yCG
i
i
i
16  0  1  0.1  sin(530 )  1  0.1  sin(530 )

0
16  1  1
i
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question
Consider a hoop of mass 1 kg.
The center of gravity is located
at:
a) The edge of the hoop
b) The center of the hoop
c) Depends on how the hoop is
positioned relative to the
earth’s surface.
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Object in equilibrium
Top view Fp
-d d
CG
-Fp
Newton’s 2nd law: F=ma
Fp+(-Fp)=ma=0
No acceleration, no movement…
But the block starts to rotate!
=Fpd+(-Fp)(-d)=2Fpd
There is movement!
Translational equilibrium: F=ma=0 The center of gravity
does not move!
Rotational equilibrium:
=0
rotate
The object does not
Mechanical equilibrium: F=ma=0 & =0 No movement!
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Some things to keep in mind
If an object is in equilibrium, it does not matter where
one chooses the axis of rotation: the net torque must
always be zero.
If it is given that an object is at equilibrium, you can
choose the rotation axis at a convenient location to solve
the problem.
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Previously
Torque: =Fd
Center of
Gravity:
m x

m
i
xCG
i
i
i
m y

m
i
i
y CG
i
i
i
i
Translational equilibrium: F=ma=0 The center of gravity
does not move!
Rotational equilibrium:
=0
rotate
The object does not
Mechanical equilibrium: F=ma=0 & =0 No movement!
MSU Physics 231 Fall 2012
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question
20N
1m
1m
20N
20N
COG
Is this “freely rotating” object in mechanical equilibrium?
a) Yes b) No
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Weight of board: w
What is the tension in each of the
wires (in terms of w) given that the
board is in equilibrium?
0.75m
0.25m
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s=0.5 coef of friction between the wall and
the 4.0 meter bar (weight w). What is the
minimum x where you can hang a weight w
for which the bar does not slide?
37o
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Torque: =Fd
quiz (extra credit)
40N
20N
-2
40N
0
1
2
m
A wooden bar is initially balanced. Suddenly, 3 forces are
applied, as shown in the figure. Assuming that the bar can
only rotate, what will happen (what is the sum of torques)?
a) the bar will remain in balance
b) the bar will rotate counterclockwise
c) the bar will rotate clockwise
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MSU Physics 231 Fall 2012
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Torque and angular acceleration
F
r
m
Newton 2nd law: F=ma
Ft=mat
Ftr=mrat
Ftr=mr2
Used at=r
=mr2
Used =Ftr
The angular acceleration goes linear with the torque.
Mr2=moment of inertia
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Two masses
F
r
m
r
m
=mr2
=(m1r12+m2r22)

If m1=m2 and r1=r2
=2mr2
Compared to the case with only one mass, the angular
acceleration will be twice smaller when applying the
same torque, if the mass increases by a factor of two.
The moment of inertia has increased by a factor of 2.
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Two masses at different radii
F
r
m
r
m
=mr2
=(m1r12+m2r22)

If m1=m2 and r2=2r1
=5mr2
When increasing the distance between a mass and the
rotation axis, the moment of inertia increases quadraticly.
So, for the same torque, you will get a much smaller
angular acceleration.
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A homogeneous stick
m
m
m
m
m
m
m
m
m
m
F
=mr2
=(m1r12+m2r22+…+mnrn2)
=(miri2)
=I
Moment of inertia I:
I=(miri2)
Rotation point
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Two inhomogeneous sticks
r
m
m
m
m
5m
5m
m
m
m
m
F
5m
m
m
m
m
m
m
m
m
5m
=(miri2)
118mr2
18m
Easy to rotate!
F
=(miri2)
310mr2
18 m
Difficult to rotate
MSU Physics 231 Fall 2012
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More General
m2
r2
r1
=I
Moment of inertia I:
I=(miri2)
r3
m3
m1
compare with:
F=ma
The moment of inertia
in rotations is similar to
the mass in Newton’s 2nd law.
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F
A simple example
F
r
r
A and B have the same total mass. If the same
torque is applied, which one accelerates faster?
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0.2 kg
The rotation axis matters!
0.3 kg
0.5 m
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In the previous episode..
=I
(compare to F=ma)
Moment of inertia I: I=(miri2)
: angular acceleration
I depends on the choice of rotation axis!!
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Extended objects
F
I=(miri2)
=(m1+m2+…+mn)R2
=MR2
M
r
mi
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Some common cases
R
hoop/thin cylindrical
shell: I=MR2
R
disk/solid cylinder
I=1/2MR2
L
solid sphere
I=2/5MR2
thin spherical shell
I=2/3MR2
Long thin rod with
rotation axis through Long thin rod with
rotation axis through
center: I=1/12ML2
end: I=1/3ML2
MSU Physics 231 Fall 2012
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61
Falling Bars
mass: m
L
Ibar=mL2/3


FgL=Fgcos=mgcos perpendicular component
Fg=mg
Compare the angular acceleration for 2 bars of different
mass, but same length.
=I=mL2/3 also =Fd=mg(cos)L/2
So mg(cos)L/2=mL2/3 and =3g(cos)/(2L)
independent of mass!
Compare the angular acceleration for 2 bars of same mass,
but different length
=3g(cos)/(2L) so if L goes up,  goes down!
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Example
A monocycle (bicycle with one wheel) has a wheel that
has a diameter of 1 meter. The mass of the wheel is 5
kg (assume all mass is sitting at the outside of the wheel).
The friction force from the road is 25 N. If the cycle
is accelerating with 0.3 m/s2, what is the force applied on
each of the paddles if the paddles are 30 cm from the
center of the wheel?
0.5m
F
0.3m
25N
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v
ri

i
Rotational kinetic energy
mi
 
Consider a object rotating
with constant velocity. Each point
moves with velocity vi. The total
kinetic energy is:

1 m v 2  1 m  2 r 2  1  2   m r 2   1  2 I
i i i
i i
2 i i
2
2
2
i
 i

KEr=½I2
Conservation of energy for rotating object must include:
rotational and translational kinetic energy and potential energy
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
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Example.
1m
Consider a ball and a block going down the same 1m-high slope.
The ball rolls and both objects do not feel friction. If both
have mass 1kg, what are their velocities at the bottom (I.e.
which one arrives first?). The diameter of the ball is 0.4 m.
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Objects going of a slope
Assume an object is rolling of a slope.
Its mass is m and the height of the
slope is h. The radius of the object is r.
h
The moment of inertia I of the objects is: kmr2
With k:
object
I
k
cube
0 (no rotation)
0
Cylindrical shell
mr2
1
Solid cylinder
1/2mr2
0.5
Thin spherical shell
2/3mr2
0.666
Solid sphere
2/5mr2
0.4
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Which one goes fastest on the slope
Use conservation of mechanical energy.
At top (at rest): [½mv2+mgh+½I2] = mgh
At bottom: [½mv2+mgh+½I2]=
[½mv2+½(kmr2)v2/r2]
(used: I=kmr2 and =v/r and h=0)
Simplify: [½mv2+½kmv2]=(1+k)½mv2
Combine:
Etop=Ebottom
mgh=(1+k)½mv2
v=[2gh/(1+k)]
No mass
dependence!!
object
k
1+k
v
v
fastest
cube
0
1
[2gh/(1)]
1.41
1st place
Cylindrical shell
1
2
[2gh/(2)]
1
5
Solid cylinder
0.5
1.5
[2gh/(1.5)]
1.15
3
Thin spherical shell 0.666
1.66
[2gh/(1.66)] 1.09
4
Solid sphere
1.4
[2gh/(1.4)]
1.20
2
0.4
The larger the moment of inertia, the slower it rolls!
The available Ekin is spread over rotational and translational (linear) Ekin
Or: besides moving the object, you spend energy rotating it
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Angular momentum
    0  I  I 0
  I  I 

t
 t 
L  I
L  L0 L


t
t
if    0 then L  0
Conservation of angular momentum
If the net torque equals zero, the
angular momentum L does not change
Iii=Iff
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Note…
L=I
For a point-like mass orbiting a central axis:
I=mr2 and =v/r
So L=mr2 x v/r and thus L=mrv
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Conservation laws:
In a closed system:
•Conservation of energy E
•Conservation of linear momentum p
•Conservation of angular momentum L
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The spinning lecturer…
A lecturer (60 kg) is rotating on a platform with =2 rad/s
(1 rev/s). He is holding two 1 kg masses 0.8 m away from his
body. He then puts the masses close to his body (R=0.0 m).
Estimate how fast he will rotate (marm=2.5 kg).
0.8m
Iinitial=0.5MlecR2+2(MwRw2)+2(0.33Marm0.82)
=1.2+1.3+1.0= 3.5 kgm2
0.4m
Ifinal =0.5MlecR2=1.2 kgm2
Conservation of angular mom. Iii=Iff
3.5*2=1.2*f
f=18.3 rad/s (approx 3 rev/s)
For details see next slide
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71
Details of spinning lecturer
Iinitial=Ibody+2Imass+2Iarms
Ibody= 0.5MlecRlec2 (solid cylindrical shell) M=60, R=0.2
Imass=MmassRmass2 (point-like mass at radius R) M=1 R=0.8
Iarms=1/3MarmRarm2 (arm is like a long rod with rotation axis through
one of its ends) M=2.5 kg R=0.8
Iinitial=0.5MlecRlec2+2(MmassRmass2)+2(0.33MarmRarm2)
=1.2+1.3+1.0= 3.5 kgm2
Ifinal =0.5MlecR2=1.2 kgm2 (assumed that masses and arms do not
contribute to moment of inertia anymore)
Conservation of angular mom. Iii=Iff
3.5*2=1.2*f i= was given to be 2 rad/s
f=18.3 rad/s (approx 3 rev/s)
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72
‘2001 a space odyssey’ revisited
A spaceship has a radius of 100m and
I=5.00E+8 kgm2. 150 people (65 kg pp)
live on the rim and the ship rotates
such that they feel a ‘gravitational’
force of g. If the crew moves to the
center of the ship and only the captain
would stay behind, what ‘gravity’ would
he feel?
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The direction of the rotation axis
L
L
The rotation in the horizontal plane is
reduced to zero: There must have been a large
net torque to accomplish this! (this is why you
can ride a bike safely; a wheel wants to keep turning
in the same direction.)
The conservation of angular momentum not only holds
for the magnitude of the angular momentum, but also
for its direction.
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Rotating a bike wheel!
L
L
A person on a platform that can freely rotate is holding a
spinning wheel and then turns the wheel around.
What will happen?
Initial: angular momentum: Iwheelwheel
Closed system, so L must be conserved.
Final: -Iwheel wheel+Ipersonperson
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person=
2Iwheel wheel
Iperson
75
Neutron star
Sun: radius: 7*105 km
Period of rotation: 25 days
Supernova explosion
Neutron star: radius: 10 km
What is the period of rotation T???
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Kepler’s second law
Area(D-C-SUN)=Area(B-A-SUN)
A line drawn from the sun to the elliptical orbit of a planet
sweeps out equal areas in equal time intervals.
This is the same as the conservation of angular momentum:
IABAB=ICDCD Iplanet at x=(Mplanet at x)(Rplanet at x)2
planet at x =(vplanet at x)/(Rplanet at x)
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For Next Week
Chapter 9:
Gravitation
Homework Set 6
Due Wednesday Oct 17!
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Centripetal acceleration
v f  vi
v
a

t f  ti
t
The change in velocity is
not the change in speed
but in direction.
Sin(/2)=(s/2)/r
Sin(/2)=(v/2)/v
(s/2)/r=(v/2)/v
v=s*(v/r)
t t
ac=v2/r
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vs/t
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