PHY 231 1 PHYSICS 231 Lecture 21: Angular momentum Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom Neutron star.

PHY 2311

PHYSICS 231Lecture 21: Angular momentum

Remco ZegersWalk-in hour: Monday 9:15-10:15 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

Extended objects (like the stick)

I=(miri2)

=(m1+m2+…+mn)R2

=MR2

M

PHY 2314

Some common cases

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

=I=a/r so =0.3/0.5=0.6 rad/sI=(miri

2)=MR2=5(0.52)=1.25 kgm2

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

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.Block: [½mv2+mgh]initial= [½mv2+mgh]final

1*9.8*1 = 0.5*1*v2 so v=4.4 m/sBall: [½mv2+mgh+½I2]initial= [½mv2+mgh+½I2]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!

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).

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 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?

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 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 23118

Demo: defying gravity!

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

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 23120

Two more examples

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 23121

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