PHY 2311
PHYSICS 231Lecture 19: Angular momentum
Remco ZegersWalk-in hour: Tue 4:00-5:00 am
Helproom
Neutron star
PHY 2312
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!!
PHY 2315
Falling bars (demo)
L
mass: m
Fg
Compare the angular acceleration for 2 bars of differentmass, 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!
Ibar=mL2/3
PHY 2316
ExampleA monocycle (bicycle with one wheel) has a wheel thathas a diameter of 1 meter. The mass of the wheel is 5kg (assume all mass is sitting at the outside of the wheel).The friction force from the road is 25 N. If the cycleis 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?
25N
0.5m
0.3mF
PHY 2317
Rotational kinetic energy
Consider a object rotatingwith constant velocity. Each pointmoves with velocity vi. The totalkinetic energy is:
Irmrmvmi i
iiiiii
ii222222
21
21
21
21
KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
PHY 2318
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 bothhave mass 1kg, what are their velocities at the bottom (I.e.which one arrives first?). The diameter of the ball is 0.4 m.
Rotational kinetic energyKEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]finalExample.
1m
Same initial gravitational PESame final total KEA has lower final linear KE, higher final rotational KEA has lower final linear velocity
A BmA=mB
PHY 23110
Angular momentum
0L then 0 if
0
00
t
L
t
LL
IL
t
II
tII
Conservation of angular momentumIf the net torque equals zero, theangular momentum L does not change
Iii=Iff
PHY 23111
Conservation laws:
In a closed system:
•Conservation of energy E
•Conservation of linear momentum p
•Conservation of angular momentum L
PHY 23112
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:Isunsun=Insns
(7E+05)2*2.9E-06=(10)2*ns
ns=1.4E+04 rad/s so period Tns=5.4E-04 s !!!!!!
Kepler’s second law
A line drawn from the sun to the elliptical orbit of a planetsweeps out equal areas in equal time intervals.
Area(D-C-SUN)=Area(B-A-SUN)
This is the same as the conservation of angular momentum:
IABAB=ICDCD Iplanet at x=(Mplanet at x)(Rplanet at x)2
planet at x =(vplanet at x)/(Rplanet at x)
PHY 23114
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 hisbody. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate (marm=2.5 kg).
PHY 23115
‘2001 a space odyssey’ revisited
A spaceship has a radius of 100m andI=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?
PHY 23116
The direction of the rotation axis
The conservation of angular momentum not only holdsfor the magnitude of the angular momentum, but alsofor its direction.
The rotation in the horizontal plane isreduced to zero: There must have been a largenet torque to accomplish this! (this is why youcan ride a bike safely; a wheel wants to keep turningin the same direction.)
L
L
PHY 23117
Rotating a bike wheel!L
LA 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: Iwheelwheel
Closed system, so L must be conserved.
Final: -Iwheel wheel+Ipersonperson
person=2Iwheel wheel
Iperson
PHY 23119
Global warming
The polar ice caps contain 2.3x1019 kg ofice. If it were all to melt, by how muchwould the length of a day change?Mearth=6x1024 kg Rearth=6.4x106 m
PHY 23120
A top
A top has I=4.00x10-4 kgm2. By pullinga rope with constant tension F=5.57 N,it starts to rotate along the axis AA’.What is the angular velocity of the topafter the string has been pulled 0.8 m?
Work done by the tension force:W=Fx=5.57*0.8=4.456 JThis work is transformed into kinetic energy of the top:KE=0.5I2=4.456 so =149 rad/s=23.7 rev/s