Lesson 8. Luis Anchordoqui. Physics 168. Thursday, October 11, 18

Transcription

1 Lesson 8 Physics 168 1

2 Rolling 2

3 Intuitive Question Why is it that when a body is rolling on a plane without slipping the point of contact with the plane does not move? A simple answer to this question is quite simply: because the body does not slip Why? Because slipping implies 2 bodies in contact moving relative to each other Here there is no slipping Therefore the point of contact and the plane don t move relative to each other Therefore the point of contact does not move However for those still unconvinced We ll work out a mathematical argument which should help 3

4 Nonslip conditions Imagine a cylinder moving forward at speed ω v on a plane without slipping R v As indicated in diagram cylinder must also be rotating about its axis because it is rolling at an angular speed! It is clear that is completely determined by and vice-versa! v because if we change! (say make the cylinder move faster) must also increase (the cylinder must roll faster) Our aim is to find the relation between v 4 v and!

5 Nonslip conditions To do this consider cylinder moving forward a distance x x R θ θ As a result it will rotate through angle Now consider following argument On purely mathematical grounds length of bold arc of the circle is R However this is equal to x because the distance the circle has move forward is equal to the arc length 5

6 Nonslip conditions We have found thus x = R Consider differentiating each side of this equation dx dt dx dt = R d dt d dt BUT is the velocity and is the angular velocity As such v = R! v = R! These two equations are the non-slip conditions is the same relationship that we had obtained for circular motion However it s important to realize that the v in each equation is different In circular motion refers to a point on the rim of the circle Here v v refers to the velocity of the centre of mass Let s convince ourselves they have to be the same as predicted by these equations 6

7 Point of contact Consider cylinder rolling on a plane again What are the velocities of points at very top and very bottom of circle? ω ω R+ v R v ωr v Top point moving forward at v (because the cylinder is moving forward at v) but also has an extra forward speed ωr because of the rotation Bottom point moving forward at v (for the same reason) but is now moving backwards at ωr because of the rotation (because the circle is rotating clockwise it s moving back at its bottom point) 7

8 Point of contact (cont d) However we have found that Feeding this into our diagram v =!R R ω v 2ωR 0 As indeed expected we have found that the velocity of the point of contact is zero It does not move!!! Differentiating on both sides the non-slip condition with respect to time tangential acceleration a T = R (e.g. for string not to slip on a pulley wheel) 8

9 Atwood s machine m 1 m 2 An Atwood's machine consists of two masses, and which are connected by a mass less inelastic cord that passes over a pulley R If pulley has radius and moment of inertia about its axle m 1 determine acceleration of masses and m 2 I 9

10 Assume m 2 >m 1 and so pulley will accelerate clockwise Align coordinate system with acceleration clockwise positive X Fy1 = m 1 a ) F T1 m 1 g = m 1 a X Fy2 = m 2 a ) m 2 g F T2 = m 2 a X = I ) FT2 R + F T1 R = I = I a R F T1 = m 1 g + m 1 a F T2 = m 2 g m 2 a Substituting the force relation in the torque equation a = (m 2 m 1 ) (m 1 + m 2 + I/R 2 ) g 10

11 Question 11 cm 7.2 kg A bowling ball that has radius and mass is rolling without on a horizontal ball return It continues to roll without slipping up a hill to a height slipping at 2 m/s coming to rest and then rolling back down hill Model ball as a uniform sphere and find h h before momentarily 11

12 Answer E mech =0 U f + K f = U i + K i ) Mgh = 1 2 Mv2 CM i I CM! 2 i I CM = 2 5 MR2 ) Mgh = 1 2 Mv2 CM i v 2 5 MR2 CMi R 2 Mgh = 7 10 Mv2 CM i ) h = 29 cm 12

13 Rolling with slipping When an object slips (skids) as it rolls nonslip condition v cm = R! Suppose a bowler releases a ball with no initial rotation as ball skids along bowling lane v cm >R! does not hold (! 0 = 0) Kinetic frictional force will both reduce its linear speed! and increase its angular speed until nonslip condition is reached after which balls rolls without slipping v cm = R! v cm 13

14 Conservation Theorems: Angular Momentum 14

15 Vector Nature of Rotation Torque is expressed mathematically as a vector product of ~r and F ~ ~ = ~r ~ F ~F ~r z ~ If and are both perpendicular to axis is parallel to axis z 15

16 Vector product ~A B ~ C ~ Vector product of two vectors and is a vector and has a magnitude ~ C that is perpendicular to both C = A B sin equals area of parallelogram shown A and B ~C = ~ A ~ B = AB sin 16

17 Direction of Vector product (cont d) ~A B ~ ~A B ~ is given by right-hand rule when fingers are rotated from direction of toward through angle Defines a right-handed cartesian system 17

18 Vector product (cont d) If we take vector product by going around figure in direction of arrows (clockwise) sign is positive ~ı ~ = k Going around against arrows sign is negative î ˆk = ˆ Throughout this course we adopt right handed coordinate systems 18

19 Angular momentum Angular momentum ~L of particle relative to origin is defined to be vector product of ~r and ~p O ~L = ~r ~p 19

20 Angular momentum (cont d) m attached to circular disk xy z! Figure shows particle of mass of negligible mass moving in a circle in disk is spinning about plane with its center at origin -axis with angular speed ~L = ~r ~p = ~r m~v = rmvˆk = mr 2! ˆk = mr 2 ~! Angular momentum is in same direction as angular velocity vector Because mr 2 is moment of inertia for a single particle we have ~L = I~! 20

21 Angular momentum (cont d) Angular momentum of this particle about a general point on is not parallel to angular velocity vector z axis Angular momentum computed about a point on ~L 0 ~ L 0 for same particle attached to same disk but with z axis that is not at center of circle 21

22 Angular momentum (cont d) For any system of particles that rotates about a symmetry axis We now attach a second particle of equal mass to spinning disk at a point diametrically opposite to first particle Total angular momentum is again parallel to angular ~L 0 = ~ L ~ L 0 2 velocity vector ~! In this case axis of rotation passes through center of mass of two-particle system and mass distribution is symmetric about this axis Such an axis is called a symmetry axis For any system of particles that rotates about a symmetry axis total angular momentum (which is sum of angular momenta of individual particles ) is parallel to angular velocity ~L = I~! 22

23 Conservation of angular momentum Angular momentum of a particle ~L = ~r ~p (with respect to origin from which position vector is measured ) Torque (or moment of force) with respect to same origin is ~ = ~r F ~ Position vector from origin to point where force is applied ~r ~ = ~r ~p ~L = d dt (~r ~p )=( ~r ~p )+(~r ~p ) ~r But of course ~r ~p = ~r mv = m( ~r ~r )=0 ~L = ~r ~p = ~ If ~ =0) ~ L =0) L is a vector constant in time If net external torque acting on a system about some point is zero total angular momentum of system about that point remains constant 23

24 (a) Use conservation of angular momentum to estimate angular velocity of a neutron star which has collapsed to a diameter of ( m) radius was equal to that of Sun, of mass 20 km 1.5 M and which rotated like our Sun once a month (b) By what factor rotational kinetic energy change after collapse?, from a star whose Bright dot in middle is believed to be hot young neutron star result of a supernova explosion from about 300 years ago 24

25 a Conservation of angular momentum (I!) initial =(I!) final! final =! initial Iinitial I final =! initial 2 5 MR2 initial 2 5 MR2 final! final =1rev/month m = rev/month = 1900 rev/s m b Rotational kinetic energy K = 1 2 I!2 K f K i = 1 2 I f! 2 f 1 2 I i! 2 i = Rf! f R i! i 2 =

26 Angular Momentum of a System of Particles Newton s second law for angular motion Net external torque about a fixed point acting on a system equals rate of change of angular momentum of system about same point ~ net,ext = d~ L sys dt Angular impulse ~L sys = Z tf It is often useful to split total angular momentum of a system about an arbitrary point O into orbital angular momentum and spin angular momentum t i ~ net, ext dt ~L sys = ~ L orbit + ~ L spin 26

27 Angular Momentum of a System of Particles Newton s second law for angular motion (cont d) Earth has spin angular momentum due to its spinning motion about its rotational axis and it has orbital angular momentum about center of Sun due to its orbital motion around Sun ~L orbit = ~r cm M ~v cm = M r 2 cm! yearly = kg m 2 /s 27

28 Pulling Through a Hole m A particle of mass moves with speed in a circle with radius on a frictionless table top Particle is attached to a string that passes through a hole in table String is slowly pulled downward until particle is a distance r 0 a Find final velocity in terms of, v 0 from hole b Find tension when particle is moving in a circle of radius in terms of r, m and angular momentum ~L c Calculate work done on particle by tension force ~T by integrating T ~ d Express your answer in terms of r and L 0 v 0 after which particle moves in a circle of radius and r r r r 0 r 28

29 Pulling Through a Hole (cont d) L f = L 0 ) v f = r 0v 0 r f Because particle is being pulled in slowly acceleration is virtually same as if particle were moving in a circle T m v2 r ~L = ~r ~p = rmv cos rmv 1! cos 1 T = m v2 r = m r L mr 2 = L 2 mr 3 29

30 Pulling Through a Hole (cont d) dr = dr dw = ~ T d ~` = Td`cos dr = d` cos ) dw = T dr = Tdr W = Z rf r 0 Tdr = L2 m Z rf r 0 r 3 dr = L2 2m 1 r 2 f 1 r