1. Physics 207: Lecture 16, Pg 1 Lecture 16Goals: Chapter 12 Chapter 12  Extend the particle model to rigid-bodies  Understand the equilibrium of an extended object.  Analyze rolling motion  Understand rotation about a fixed axis.  Employ “conservation of angular momentum” concept Assignment: l HW8 due March 17 th

2. Physics 207: Lecture 16, Pg 2 Exercise Rotational Definitions A. The wheel is spinning counter-clockwise and slowing down. B. The wheel is spinning counter-clockwise and speeding up. C. The wheel is spinning clockwise and slowing down. D. The wheel is spinning clockwise and speeding up A friend at a party (perhaps a little tipsy) sees a disk spinning and says “Ooh, look! There’s a wheel with a negative  and positive  !” l Which of the following is a true statement about the wheel? 

3. Physics 207: Lecture 16, Pg 3 Home Exercise (review) : Wheel And Rope l A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a T = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) aTaT r

4. Physics 207: Lecture 16, Pg 4 Home exercise: Wheel And Rope l A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a T = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) Revolutions = R = (    and a T =  r      t + ½  t    R = (    + ½  a T /r  t    R = (0.5 x 10 x 100) / 6.28 aTaT r

5. Physics 207: Lecture 16, Pg 5 System of Particles (Distributed Mass): l Until now, we have considered the behavior of very simple systems (one or two masses). l But real objects have distributed mass ! l For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r. l Compare the v tangential and kinetic energies at these two points. 1 2 

6. Physics 207: Lecture 16, Pg 6 System of Particles (Distributed Mass): l The rotation axis matters too! l Twice the radius, four times the kinetic energy l KEY POINT: It matters where you put your mass! 1 K= ½ m v 2 = ½ m (  r) 2 2 K= ½ m (2v) 2 = ½ m (  2r) 2  v2v2 v1v1

7. Physics 207: Lecture 16, Pg 7 Exercise Rotational Kinetic Energy A. 1/9 B. 1/3 C. 1 D. 3 E. 9 l We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.3 m long rope. It spins around at 2 revolutions per second. l What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? Ball 1 Ball 2

8. Physics 207: Lecture 16, Pg 8 Exercise Rotational Kinetic Energy K 2 /K 1 = ½ m  r 2 2 / ½ m  r 1 2 =  2 /  2 = 4 l What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? (A) 1/9 (B) 1/3 (C) 1 (D) 3 (E) 9 Ball 1 Ball 2

9. Physics 207: Lecture 16, Pg 9 A special point for rotation System of Particles: Center of Mass (CM) l A supported object will rotate about its center of mass. l Center of mass: Where the system is balanced !  Building a mobile is an exercise in finding centers of mass. m1m1 m2m2 + m1m1 m2m2 + mobile

10. Physics 207: Lecture 16, Pg 10 System of Particles: Center of Mass l How do we describe the “position” of a system made up of many parts ? l Define the Center of Mass (average position):  For a collection of N individual point like particles whose masses and positions we know: (In this case, N = 2) y x r2r2 r1r1 m1m1 m2m2 R CM

11. Physics 207: Lecture 16, Pg 11 Sample calculation: l Consider the following mass distribution: (24,0) (0,0) (12,12) m 2m m R CM = (12,6) X CM = (m x 0 + 2m x 12 + m x 24 )/4m meters Y CM = (m x 0 + 2m x 12 + m x 0 )/4m meters X CM = 12 meters Y CM = 6 meters m at ( 0, 0) 2m at (12,12) m at (24, 0)

12. Physics 207: Lecture 16, Pg 12 System of Particles: Center of Mass l For a continuous solid, one can convert sums to an integral. y x dm r where dm is an infinitesimal mass element but there is no new physics

13. Physics 207: Lecture 16, Pg 13 Connection with motion... l So, for a rigid rotating object whose CM is moving, it rotates about its center of mass! For a point p rotating: V CM  p p p p p p p

14. Physics 207: Lecture 16, Pg 14 Rotational Kinetic Energy l Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). l The kinetic energy of this system will be the sum of the kinetic energy of each piece: K = ½  m 1 v 1  + ½  m 2 v 2  + ½  m 3 v 3  + ½  m 4 v 4  rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

15. Physics 207: Lecture 16, Pg 15 Rotation & Kinetic Energy Notice that v 1 =  r 1, v 2 =  r 2, v 3 =  r 3, v 4 =  r 4 l So we can rewrite the summation: We recognize & define a new quantity, moment of inertia or I, and write: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

16. Physics 207: Lecture 16, Pg 16 Calculating Moment of Inertia where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: mm mm L

17. Physics 207: Lecture 16, Pg 17 Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I 1 = 4 m R 2 = 4 m (2 1/2 L / 2) 2 L I = 2mL 2 I 2 = mL 2 mm mm I 1 = 2mL 2

18. Physics 207: Lecture 16, Pg 18 Home Exercise Moment of Inertia A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is I a, I b, and I c respectively.  Which of the following is correct: ( A) ( A) I a > I b > I c (B) (B) I a > I c > I b (C) (C) I b > I a > I c a b c

19. Physics 207: Lecture 16, Pg 19 Home Exercise Moment of Inertia I a = 2 m (2L) 2 I b = 3 m L 2 I c = m (2L) 2 l Which of the following is correct: ( A) ( A) I a > I b > I c (B) I a > I c > I b (C) (C) I b > I a > I c a b c L L

20. Physics 207: Lecture 16, Pg 20 Moments of Inertia Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center. I = ½ M R 2 Some examples of I for solid objects: R L r dr r dm l For a continuous solid object we have to add up the mr 2 contribution for every infinitesimal mass element dm.  An integral is required to find I : Use the table…

21. Physics 207: Lecture 16, Pg 21 Moments of Inertia Some examples of I for solid objects: Thin hoop (or cylinder) of mass M and radius R, about an axis through it center, perpendicular to the plane of the hoop is just MR 2 R Thin hoop of mass M and radius R, about an axis through a diameter. R

22. Physics 207: Lecture 16, Pg 22 Rotation & Kinetic Energy... l The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

23. Physics 207: Lecture 16, Pg 23 Moment of Inertia and Rotational Energy Notice that the moment of inertia I depends on the distribution of mass in the system.  The further the mass is from the rotation axis, the bigger the moment of inertia. l For a given object, the moment of inertia depends on where we choose the rotation axis (unlike the center of mass). In rotational dynamics, the moment of inertia I appears in the same way that mass m does in linear dynamics ! lSo where

24. Physics 207: Lecture 16, Pg 24 Exercise: Work & Energy l Strings are wrapped around the circumference of two solid disks and pulled with a force, F, for the same linear distance, d. Disk 1 has a bigger radius, but both are of identical material (i.e., their density is  = M / V ) and have the same thickness. l Both disks rotate freely around axes though their centers, and start at rest.  Which disk has the biggest angular velocity, , at the end ? Recall W = F d =  K ( =½ I  2 ) ( A) ( A) Disk 1 (B) (B) Disk 2 (C) (C) Same FF 11 22 start finish d

25. Physics 207: Lecture 16, Pg 25 Exercise Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.  Which disk has the biggest angular velocity after the end ? W = F d = ½ I 1  1 2 = ½ I 2  2 2 and  1 = (I 2 / I 1 ) ½  2 and I 2 < I 1 ( A) ( A) Disk 1 (B) (B) Disk 2 (C) (C) Same FF 11 22 start finish d

26. Physics 207: Lecture 16, Pg 26 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem:  K = W NET l This applies to both rotational as well as linear motion. l So for an object that rotates about a fixed axis l For an object which is rotating and translating

27. Physics 207: Lecture 16, Pg 27 Demo Example : A race rolling down an incline l Two cylinders with identical radii and total masses roll down an inclined plane. l The 1 st has more of the mass concentrated at the center while the 2 nd has more mass concentrated at the rim. l Which gets down first? Two cylinders with radius R and mass m M  h M who is 1 st ? M M M M M A)Mass 1 B)Mass 2 C)They both arrive at same time

28. Physics 207: Lecture 16, Pg 28 Same Example : Rolling, without slipping, Motion l A cylinder is about to roll down an inclined plane. l What is its speed at the bottom of the plane ? M  h M v ? M M M M M

29. Physics 207: Lecture 16, Pg 29 Rolling without slipping motion l Again consider a cylinder rolling at a constant speed. V CM CM 2V CM

30. Physics 207: Lecture 16, Pg 30 Motion l Again consider a cylinder rolling at a constant speed. V CM CM 2V CM CM V CM Sliding only CM Rotation only V Tang =  R Both with |V Tang | = |V CM |

31. Physics 207: Lecture 16, Pg 31 Example : Rolling Motion l A solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? l Use Work-Energy theorem M  h M v ? Ball has radius R M M M M M Mgh = ½ Mv 2 + ½ I CM  2 and v =  R Mgh = ½ Mv 2 + ½ (½ M R 2 )(v/R) 2 = ¾ Mv 2 v = 2(gh/3) ½

32. Physics 207: Lecture 16, Pg 32 Example: The Frictionless Loop-the-Loop … last time l To complete the loop the loop, how high do we have to release a ball with radius r (r <<R) ? l Condition for completing the loop the loop: Circular motion at the top of the loop (a c = v 2 / R) l Use fact that E = U + K = constant ! h ? R ball has mass m & r <<R Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is U b =mgh U=mg2R

33. Physics 207: Lecture 16, Pg 33 Example: The Loop-the-Loop … last time l If rolling then ball has both rotational and CM motion! l E= U + K CM + K Rot = constant = mgh (at top) E= mg2R + ½ mv 2 + ½ 2/5 mr 2  2 = mgh & v=  r l E= mg2R + ½ mgR + 1/5 m v 2 = mgh  h = 5/2R+1/5R h ? R ball has mass m & r <<R U b =mgh U=mg2R Just a little bit more….

34. Physics 207: Lecture 16, Pg 34 Exercise: Work Energy Example, Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. l What is its angular speed when it reaches the lowest point ? L m

35. Physics 207: Lecture 16, Pg 35 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. l What is its angular speed when it reaches the lowest point ? The hinge changes everything! m mg W = m g h = ½ I Hinge  2 W = mgL/2 = ½ (m L 2 /3)  2 and solve for  mg L/2 L

36. Physics 207: Lecture 16, Pg 36 Lecture 16 Assignment: l HW7 due March 25 th l For the next Tuesday: Catch up