Angular Momentum - Hong Kong University Of Science And Technology

Angular momentum Instructor: Dr. Hoi Lam TAM (譚海嵐) Physics Enhancement Programme for Gifted Students The Hong Kong Academy for Gifted Education and Department of Physics, HKBU Department of Physics Hong Kong Baptist University 1

Contents Operation of vectors Angular momentum Angular momentum of rigid body Conservation of angular momentum Examples Department of Physics Hong Kong Baptist University 2

Department of Physics Hong Kong Baptist University 3

Department of Physics Hong Kong Baptist University 4

Angular Momentum l r p m(r v ) l mrv sin . Alternatively, l rp rmv or l r p r mv Newton’s Second Law dl . dt The vector sum of all the torques acting on a particle is equal to the time rate of change of the angular momentum of that particle. Department of Physics Hong Kong Baptist University 5

Proof l m(r v ) Differentiating with respect to time, dv dr dl m r v dt dt dt dl m r a v v dt v v 0 because the angle between v and v is zero. dl m r a r ma dt Using Newton’s law, F ma. Hence dl r F r F . dt Since r F , we arrive at dl dt . Department of Physics Hong Kong Baptist University 6

The Angular Momentum of a System of Particles Total angular momentum for n particles: L l1 l2 li Newtons’ law for angular motion: dli d dL . l i dt dt i dt i includes torques acting on all the n particles. Both internal torques and external torques are considered. Using Newton’s law of action and reaction, the internal forces cancel in pairs. Hence dL . ext dt Department of Physics Hong Kong Baptist University 7

Department of Physics Hong Kong Baptist University 8

The Angular Momentum of a Rigid Body For the ith particle, angular momentum: li ri pi sin90 o ri mi vi . The component of angular momentum parallel to the rotation axis (the z component): liz li sin (ri sin )( mi vi ) ri mi vi . The total angular momentum for the rotating body Lz liz mi vi ri i i 2 mi ( ri )ri mi ri . i i This reduces to L I . Department of Physics Hong Kong Baptist University 9

Conservation of Angular Momentum dL ext dt If no external torque acts on the system, dL 0 dt L constant. Li L f . Law of conservation of angular momentum. Department of Physics Hong Kong Baptist University 10

Department of Physics Hong Kong Baptist University 11

Department of Physics Hong Kong Baptist University 12

Examples A student sits on a stool that can rotate freely about a vertical axis. The student, initially at rest, is holding a bicycle wheel whose rim is loaded with lead and whose rotational inertia I about its central axis is 1.2 kgm2. The wheel is rotating at an angular speed wh of 3.9 rev/s; as seen from overhead, the rotation is counterclockwise. The axis of the wheel is vertical, and the angular momentum Lwh of the wheel points vertically upward. The student now inverts the wheel; as a result, the student and stool rotate about the stool axis. The rotational inertia Ib of the student stool wheel system about the stool axis is 6.8 kgm2. With what angular speed b and in what direction does the composite body rotate after the inversion of the wheel? Department of Physics Hong Kong Baptist University 13

Using the conservation of angular momentum, Lwh L ( Lwh ) L 2Lwh I b b 2 I wh b 2 I wh (2)(1.2)(3.9) Ib 6.8 1.38 rev s 1 Department of Physics Hong Kong Baptist University 14

Examples A cockroach with mass m rides on a disk of mass 6m and radius R. The disk rotates like a merry-go-round around its central axis at angular speed I 1.5 rad s 1. The cockroach is initially at radius r 0.8R, but then it crawls out to the rim of the disk. Treat the cockroach as a particle. What then is the angular speed? Department of Physics Hong Kong Baptist University 15

Using the conservation of angular momentum, we have I f f I i i Rotational inertia: The disk: 1 I d MR 2 3mR 2 2 The cockroach: I ci m(0.8R) 2 0.64mR 2 and I cf mR 2 I i I d I ci 3.64mR2 I f I d I cf 4mR 2 Therefore, I i i (3.64mR 2 )(1.5) 1 f 1 . 37 rad s If 4mR 2 Department of Physics Hong Kong Baptist University 16

Precession of a Gyroscope Torque due to the gravitational force Mgr sin 90o Mgr Angular momentum L I For a rapidly spinning gyroscope, the magnitude of L is not affected by the precession, dL d L dt dt dL Ld Using Newton’s second law for rotation, dL dt Mgr L d L dt where is the precession rate d dt Mgr Mgr L I Department of Physics Hong Kong Baptist University 17

Department of Physics Hong Kong Baptist University 18

Department of Physics Hong Kong Baptist University 19

Example Rolling of a Hexagonal Prism (1998 IPhO) Consider a long, solid, rigid, regular hexagonal prism like a common type of pencil. The mass of the prism is M and it is uniformly distributed. The length of each side of the cross-sectional hexagon is a. The moment of inertia I of the hexagonal prism about its central axis is I 5Ma2/12. a) The prism is initially at rest with its axis horizontal on an inclined plane which makes a small angle with the horizontal. Assume that the surfaces of the prism are slightly concave so that the prism only touches the plane at its edges. The effect of this concavity on the moment of inertia can be ignored. The prism is now displaced from rest and starts an uneven rolling down the plane. Assume that friction prevents any sliding and that the prism does not lose contact with the plane. The angular velocity just before a given edge hits the plane is i while f is the angular velocity immediately after the impact. Show that f s i and find the value of s. b) The kinetic energy of the prism just before and after impact is Ki and Kf. Show that Kf rKi and find r. c) For the next impact to occur, Ki must exceed a minimum value Ki,min which may be written in the form Ki,min Mga. Find in terms of and r. d) If the condition of part (c) is satisfied, the kinetic energy Ki will approach a fixed value Ki,0 as the prism rolls down the incline. Show that Ki,0 can be written as Ki,0 Mga and find . e) Calculate the minimum slope angle 0 for which the uneven rolling, once started, will continue indefinitely. Pf 30o Pi E Department of Physics Hong Kong Baptist University 20

a) Angular momentum about edge E before the impact Li I CM i M (a i )(a sin 30o ) 11 Ma 2 i 12 Angular momentum about edge E after the impact 17 5 L f I E f Ma 2 Ma 2 f Ma 2 f 12 12 Using the conservation of angular momentum, Li Lf 11 17 Ma 2 i Ma 2 f 12 12 11 f i 17 11 Thus, s 17 Pf 30o Pi E Department of Physics Hong Kong Baptist University 21

b) The kinetic energy of the prism just before and after impact is Ki and Kf. Show that Kf rKi and find r. b) 1 5 17 K i Ma 2 Ma 2 i2 Ma 2 i2 2 12 24 1 5 17 K f Ma 2 Ma 2 2f Ma 2 2f 2 12 24 2 2f 11 121 K f 2 Ki Ki Ki i 289 17 121 r 289 Pf 30o Pi E Department of Physics Hong Kong Baptist University 22

c) For the next impact to occur, Ki must exceed a minimum value Ki,min which may be written in the form Ki,min Mga. Find in terms of and r. c) After the impact, the center of mass of the prism raises to its highest position by turning through an angle 90o ( 60o) 30o . Hence rKi ,min Mga Mga cos(30o ) 1 1 cos(30o ) r Department of Physics Hong Kong Baptist University 23

d) If the condition of part (c) is satisfied, the kinetic energy Ki will approach a fixed value Ki,0 as the prism rolls down the incline. Show that Ki,0 can be written as Ki,0 Mga and find . d) At the next impact, the center of mass lowers by a height of asin . Change in the kinetic energy K Mga sin Kinetic energy immediately before the next impact f ( K i ) Mga sin rKi When the kinetic energy approaches Ki,0, K i , 0 Mga sin rKi , 0 Ki ,0 Thus Mga sin 1 r sin 1 r Department of Physics Hong Kong Baptist University 24

e) For the rolling to continue indefinitely, K i , 0 K i ,min sin 1 1 cos(30o ) 1 r r r 3 1 sin 1 cos sin 1 r 2 2 1 3 r sin cos 1 2 1 r 2 A 1 / 2 2 3 / 4 cos u sin sin u cos 1 where A r/(1 r) 121/168 and 3 0.5788 2 2 A 1 / 2 3 / 4 (205 / 84) 3 1 o sin( u ) u 35.36 2 A 1 / 2 3 / 4 sin u 3/2 4 0.6683 2 (205 / 84) 3 u 41.94o. 0 6.58o. Department of Physics Hong Kong Baptist University 25

Ki,0 Mgasin rKi,0 r Mga Ki 1 sin,0 r 1 sin d) At the next impact, the center of mass lowers by a height of asin . Change in the kinetic energy Kinetic energy immediately before the next impact When the kinetic energy approaches K i,0, d) If the condition of part (c) is satisfied, the kinetic energy K i will approach a fixed value K