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

2. PHY 231 2 In the previous episode..  =I  (compare to  F=ma ) Moment of inertia I: I=(  m i r i 2 )  : angular acceleration I depends on the choice of rotation axis!!

3. PHY 231 3 Extended objects (like the stick) I=(  m i r i 2 ) =(m 1 +m 2 +…+m n )R 2 =MR 2 M

4. PHY 231 4 Some common cases

5. PHY 231 5 Falling bars (demo) L mass: m FgFg Compare the angular acceleration for 2 bars of different mass, but same length.  =I  =mL 2  /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! I bar =mL 2 /3

6. PHY 231 6 Example A monocycle (bicycle with one wheel) has a wheel that has a diameter of 1 meter. The mass of the wheel is 5 kg (assume all mass is sitting at the outside of the wheel). The friction force from the road is 25 N. If the cycle is accelerating with 0.3 m/s 2, 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 0.3m F  =I   =a/r so  =0.3/0.5=0.6 rad/s I=(  m i r i 2 )=MR 2 =5(0.5 2 )=1.25 kgm 2  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

7. PHY 231 7 Rotational kinetic energy Consider a object rotating with constant velocity. Each point moves with velocity v i. The total kinetic energy is: KE r =½I  2 Conservation of energy for rotating object: [PE+KE t +KE r ] initial = [PE+KE t +KE r ] final

8. PHY 231 8 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 both have 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: [½mv 2 +mgh] initial = [½mv 2 +mgh] final 1*9.8*1 = 0.5*1*v 2 so v=4.4 m/s Ball: [½mv 2 +mgh+½I  2 ] initial = [½mv 2 +mgh+½I  2 ] final I=0.4*MR 2 =0.064 kgm 2 and  =v/R=2.5v 1*9.8*1 = 0.5*1*v 2 +0.5*0.064*(2.5v) 2 so v=3.7 m/s Part of the energy goes to the rotation: slower!

9. Rotational kinetic energy KE r =½I  2 Conservation of energy for rotating object: [PE+KE t +KE r ] initial = [PE+KE t +KE r ] final Example. 1m Same initial gravitational PE Same final total KE A has lower final linear KE, higher final rotational KE A has lower final linear velocity A B m A =m B

10. PHY 231 10 Angular momentum Conservation of angular momentum If the net torque equals zero, the angular momentum L does not change I i  i =I f  f

11. PHY 231 11 Conservation laws: In a closed system: Conservation of energy E Conservation of linear momentum p Conservation of angular momentum L

12. PHY 231 12 Neutron star Sun: radius: 7*10 5 km Supernova explosion Neutron star: radius: 10 km I sphere =2/5MR 2 (assume no mass is lost)  sun =2  /(25 days)=2.9*10 -6 rad/s Conservation of angular momentum: I sun  sun =I ns  ns (7E+05) 2 *2.9E-06=(10) 2 *  ns  ns =1.4E+04 rad/s so period T ns =5.4E-04 s !!!!!!

13. Kepler’s second law A line drawn from the sun to the elliptical orbit of a planet sweeps 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: I AB  AB =I CD  CD I planet at x =(M planet at x )(R planet at x ) 2  planet at x =(v planet at x )/(R planet at x )

14. PHY 231 14 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 his body. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate (m arm =2.5 kg). 0.4m 0.8m I initial =0.5M lec R 2 +2(M w R w 2 )+2(0.33M arm 0.8 2 ) =1.2+1.3+1.0= 3.5 kgm 2 I final =0.5M lec R 2 =1.2 kgm 2 Conservation of angular mom. I i  i =I f  f 3.5*2  =1.2*  f  f =18.3 rad/s (approx 3 rev/s)

15. PHY 231 15 ‘2001 a space odyssey’ revisited A spaceship has a radius of 100m and I=5.00E+8 kgm 2. 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=I ship +I crew =(5.00E+8) + 150*(65*100 2 )=5.98E+8 kgm 2 F person =ma c =m  2 r=mg so  =  (g/r)=0.31 rad/s Final: I=I ship +I crew =(5.00E+8) + 1*(65*100 2 )=5.01E+8 kgm 2 Conservation of angular momentum I i  i =I f  f (5.98E+8)*0.31=(5.01E+8)*  f so  f =0.37 rad/s m  2 r=mg captain so g captain =13.69 m/s 2

16. PHY 231 16 The direction of the rotation axis The conservation of angular momentum not only holds for the magnitude of the angular momentum, but also for its direction. 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.) L L

17. PHY 231 17 Rotating a bike wheel! LL 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: I wheel  wheel Closed system, so L must be conserved. Final: - I wheel  wheel +I person  person  person = 2I wheel  wheel I person

18. PHY 231 18 Demo: defying gravity!

19. PHY 231 19 Global warming The polar ice caps contain 2.3x10 19 kg of ice. If it were all to melt, by how much would the length of a day change? M earth =6x10 24 kg R earth =6.4x10 6 m Before global warming: ice does not give moment of inertia I i =2/5*M earth R 2 earth =2.5x10 38 kgm 2  i =2  /(24*3600 s)=7.3x10 -5 rad/s After ice has melted: I f =I i +2/3*MR 2 ice =2.5x10 38 +2.4x10 33 =2.500024x10 38  f =  i I i /I f =7.3x10 -5 *0.9999904 The length of the day has increased by 0.9999904*24 hrs=0.83 s.

20. PHY 231 20 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(155 o )=0.085LT  w =0.5L*30sin(40 o )=-9.64L  F =L*12.5sin(40 0 )=-8.03L  =-17.7L+0.085LT=0 T=208 N

21. PHY 231 21 A top A top has I=4.00x10 -4 kgm 2. 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