PHYSICS 231
Lecture 20: Angular momentum
Neutron star
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
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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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Rotational kinetic energy
KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
Example.
1m
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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 equal 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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The spinning lecturer…
A lecturer (60 kg) is rotating on a platform with =2 rad/s
(1 rev/s). He is holding two masses of 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.
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 (lec. 16, sh. 12)
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 for the weekend...
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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