PHY PHYSICS 231 Lecture 20: Angular momentum Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom Neutron star

PHY 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!!

PHY 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

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

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

PHY 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 !!!!!!

PHY 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 his body. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate. 0.4m 0.8m I initial =0.5M lec R 2 +2(M w R w 2 )+2(0.33M arm ) = = 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)

PHY ‘2001 a space odyssey’ revisited (lec. 16, sh. 12) A spaceship has a radius of 100m and I=5.00E+8 kgm 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

PHY 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

PHY 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

PHY Demo: defying gravity!

PHY 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 =2.5x x10 33 = x10 38  f =  i I i /I f =7.3x10 -5 * The length of the day has increased by *24 hrs=0.83 s.

PHY Two for the weekend N 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

PHY 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