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PHYSICS 231
Lecture 21: Angular momentum
Neutron star
Remco Zegers
Walk-in hour: Monday 9:15-10:15 am
Helproom
PHY 231
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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 (like the stick)
M
I=(miri2)
=(m1+m2+…+mn)R2
=MR2
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Some common cases
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Falling bars (demo)
mass: m
L
Ibar=mL2/3
Fg
Compare the angular acceleration for 2 bars of different
mass, but same length.
=I=mL2/3 also =Fd=mgL/2 so =3g/(2L)
independent of mass!
Compare the angular acceleration for 2 bars of same mass,
but different length
=3g/(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?
=I
=a/r so =0.3/0.5=0.6 rad/s
I=(miri2)=MR2=5(0.52)=1.25 kgm2
0.5m
F
25N
0.3m
friction=-25*0.5=-12.5
paddles=F*0.3+F*0.3=0.6F
0.6F-12.5=1.25*0.6, so F=22.1 N
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Rotational kinetic energy
Consider a object rotating
with constant velocity. Each point
moves with velocity vi. The total
kinetic energy is:

1
1
1




 2 m v  2 m  r  2    m r
2
i i
i
i
2 2
i i
i
2
i i
i
2
 1 2
  2 I

KEr=½I2
Conservation of energy for rotating object:
[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.
Block: [½mv2+mgh]initial= [½mv2+mgh]final
1*9.8*1 = 0.5*1*v2 so v=4.4 m/s
Ball: [½mv2+mgh+½I2]initial= [½mv2+mgh+½I2]final
I=0.4*MR2=0.064 kgm2 and =v/R=2.5v
1*9.8*1 = 0.5*1*v2+0.5*0.064*(2.5v)2
so v=3.7 m/s Part of the energy goes to the rotation: slower!
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Rotational kinetic energy
KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
Example.
A
mA=mB
B
1m
Same initial gravitational PE
Same final total KE
A has lower final linear KE, higher final rotational KE
A has lower final linear velocity
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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Conservation laws:
In a closed system:
•Conservation of energy E
•Conservation of linear momentum p
•Conservation of angular momentum L
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Neutron star
Sun: radius: 7*105 km
Supernova explosion
Neutron star: radius: 10 km
Isphere=2/5MR2 (assume no mass is lost)
sun=2/(25 days)=2.9*10-6 rad/s
Conservation of angular momentum:
Isunsun=Insns
(7E+05)2*2.9E-06=(10)2*ns
ns=1.4E+04 rad/s so period Tns=5.4E-04 s !!!!!!
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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)
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).
Iinitial=0.5MlecR2+2(MwRw2)+2(0.33Marm0.82)
2
=1.2+1.3+1.0=
3.5
kgm
0.4m
0.8m
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)
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‘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?
Initial: I=Iship+Icrew=(5.00E+8) + 150*(65*1002)=5.98E+8 kgm2
Fperson=mac=m2r=mg so =(g/r)=0.31 rad/s
Final: I=Iship+Icrew=(5.00E+8) + 1*(65*1002)=5.01E+8 kgm2
Conservation of angular momentum Iii=Iff
(5.98E+8)*0.31=(5.01E+8)*f so f=0.37 rad/s
m2r=mgcaptain so gcaptain=13.69 m/s2
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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
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Demo: defying gravity!
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Global warming
The polar ice caps contain 2.3x1019 kg of
ice. If it were all to melt, by how much
would the length of a day change?
Mearth=6x1024 kg Rearth=6.4x106 m
Before global warming: ice does not give moment of inertia
Ii=2/5*MearthR2earth=2.5x1038 kgm2
i=2/(24*3600 s)=7.3x10-5 rad/s
After ice has melted:
If=Ii+2/3*MR2ice=2.5x1038+2.4x1033=2.500024x1038
f=iIi/If=7.3x10-5*0.9999904
The length of the day has increased by
0.9999904*24 hrs=0.83 s.
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Two more examples
What is the tension in the tendon?
Does not move!
Rotational equilibrium:
30N
12.5N
0.2L
0.5L
T=0.2LTsin(155o)=0.085LT
w=0.5L*30sin(40o)=-9.64L
F=L*12.5sin(400)=-8.03L
=-17.7L+0.085LT=0
T=208 N
L
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A top
A top has I=4.00x10-4 kgm2. By pulling
a rope with constant tension F=5.57 N,
it starts to rotate along the axis AA’.
What is the angular velocity of the top
after the string has been pulled 0.8 m?
Work done by the tension force:
W=Fx=5.57*0.8=4.456 J
This work is transformed into kinetic energy of the top:
KE=0.5I2=4.456 so =149 rad/s=23.7 rev/s
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