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 23111

Demo: defying gravity!

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