PHY 231 1 PHYSICS 231 Lecture 20: Angular momentum Remco Zegers Walk-in hour: Thursday 11:30-13:30...
-
date post
19-Dec-2015 -
Category
Documents
-
view
218 -
download
0
Transcript of PHY 231 1 PHYSICS 231 Lecture 20: Angular momentum Remco Zegers Walk-in hour: Thursday 11:30-13:30...
PHY 2311
PHYSICS 231Lecture 20: Angular momentum
Remco ZegersWalk-in hour: Thursday 11:30-13:30 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 2313
Rotational kinetic energyKEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
Example.
1m
PHY 2314
Angular momentum
0L then 0 if
0
00
t
L
t
LL
IL
t
II
tII
Conservation of angular momentumIf the net torque equal zero, theangular momentum L does not change
Iii=Iff
PHY 2315
Conservation laws:
In a closed system:
•Conservation of energy E
•Conservation of linear momentum p
•Conservation of angular momentum L
PHY 2316
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 !!!!!!
PHY 2317
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 hisbody. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate.
0.4m0.8m
Iinitial=0.5MlecR2+2(MwRw2)+2(0.33Marm0.82)
=1.2+1.3+1.0= 3.5 kgm2
Ifinal =0.5MlecR2=1.2 kgm2
Conservation of angular mom. Iii=Iff
3.5*2=1.2*f
f=18.3 rad/s (approx 3 rev/s)
PHY 2318
‘2001 a space odyssey’ revisited (lec. 16, sh. 12)
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?
Initial: I=Iship+Icrew=(5.00E+8) + 150*(65*1002)=5.98E+8 kgm2
Fperson=mac=m2r=mg so =(g/r)=0.31 rad/sFinal: I=Iship+Icrew=(5.00E+8) + 1*(65*1002)=5.01E+8 kgm2
Conservation of angular momentum Iii=Iff
(5.98E+8)*0.31=(5.01E+8)*f so f=0.37 rad/sm2r=mgcaptain so gcaptain=13.69 m/s2
PHY 2319
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 23110
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 23112
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
Before global warming: ice does not give moment of inertiaIi=2/5*MearthR2
earth=2.5x1038 kgm2
i=2/(24*3600 s)=7.3x10-5 rad/sAfter ice has melted:If=Ii+2/3*MR2
ice=2.5x1038+2.4x1033=2.500024x1038
f=iIi/If=7.3x10-5*0.9999904The length of the day has increased by 0.9999904*24 hrs=0.83 s.
PHY 23113
Two for the weekend...
12.5N
30N
0.2L
0.5L
L
Does not move!
What is the tension in the tendon?
Rotational equilibrium:
T=0.2LTsin(155o)=0.085LT
w=0.5L*30sin(40o)=-9.64L
F=L*12.5sin(400)=-8.03L
=-17.7L+0.085LT=0T=208 N
PHY 23114
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