1. Physics 207: Lecture 17, Pg 1 Lecture 17 Goals: Chapter 12 Chapter 12  Define center of mass  Analyze rolling motion  Introduce and analyze torque  Understand the equilibrium dynamics of an extended object in response to forces  Employ “conservation of angular momentum” concept Assignment: l HW7 due tomorrow l Wednesday, Exam Review

2. Physics 207: Lecture 17, Pg 2 Rotational Motion: Statics and Dynamics l Forces are still necessary but the outcome depends on the location from the axis of rotation l This is in contrast to the translational motion and acceleration of the center of mass. Here the position of these forces doesn’t matter (doesn’t alter the physics we see) l However: For rotational statics & dynamics: we must reference the specific position of the force relative to an axis of rotation. It may be necessary to consider more than one rotation axis! l Vectors remain the key tool for visualizing Newton’s Laws

3. Physics 207: Lecture 17, Pg 3 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

4. Physics 207: Lecture 17, Pg 4 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

5. Physics 207: Lecture 17, Pg 5 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)

6. Physics 207: Lecture 17, Pg 6 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.

7. Physics 207: Lecture 17, Pg 7 Connection with motion... l So a rigid object that has rotation and translation rotates about its center of mass! l And Newton’s Laws apply to the center of mass For a point p rotating: V CM  p p p p p p p

8. Physics 207: Lecture 17, Pg 8 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem:  K = W NET l This applies to both rotational as well as linear motion. l What if there is rolling?

9. Physics 207: Lecture 17, Pg 9 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

10. Physics 207: Lecture 17, Pg 10 Same Example : Rolling, without slipping, Motion l A solid disk 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

11. Physics 207: Lecture 17, Pg 11 Rolling without slipping motion l Again consider a cylinder rolling at a constant speed. V CM CM 2V CM

12. Physics 207: Lecture 17, Pg 12 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 | If acceleration a center of mass = -  R

13. Physics 207: Lecture 17, Pg 13 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 ? Disk 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

14. Physics 207: Lecture 17, Pg 14 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

15. Physics 207: Lecture 17, Pg 15 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….

16. Physics 207: Lecture 17, Pg 16 How do we reconcile force, angular velocity and angular acceleration?

17. Physics 207: Lecture 17, Pg 17 Angular motion can be described by vectors l With rotation the distribution of mass matters. Actual result depends on the distance from the axis of rotation. l Hence, only the axis of rotation remains fixed in reference to rotation. We find that angular motions may be quantified by defining a vector along the axis of rotation. l We can employ the right hand rule to find the vector direction

18. Physics 207: Lecture 17, Pg 18 The Angular Velocity Vector The magnitude of the angular velocity vector is ω. The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule as illustrated above. As  increased the vector lengthens

19. Physics 207: Lecture 17, Pg 19 From force to spin (i.e.,  ) ?  NET = |r| |F Tang | ≡ |r| |F| sin  l If a force points at the axis of rotation the wheel won’t turn l Thus, only the tangential component of the force matters l With torque the position & angle of the force matters F Tangential a r F radial F  A force applied at a distance from the rotation axis gives a torque F radial F Tangential  r =|F Tang | sin 

20. Physics 207: Lecture 17, Pg 20 Rotational Dynamics: What makes it spin?  NET = |r| |F Tang | ≡ |r| |F| sin  l Torque is the rotational equivalent of force Torque has units of kg m 2 /s 2 = (kg m/s 2 ) m = N m F Tangentiala r F radial F  A force applied at a distance from the rotation axis   NET = r F Tang = r m a Tang = r m r  = (m r 2 )  For every little part of the wheel

21. Physics 207: Lecture 17, Pg 21 For a point mass  NET = m r 2  l This is the rotational version of F NET = ma Moment of inertia, is the rotational equivalent of mass. Moment of inertia, I ≡  i  m i r i 2, is the rotational equivalent of mass. If I is big, more torque is required to achieve a given angular acceleration. F Tangentiala r F randial F  The further a mass is away from this axis the greater the inertia (resistance) to rotation (as we saw on Wednesday)  NET = I 

22. Physics 207: Lecture 17, Pg 22 Rotational Dynamics: What makes it spin?  NET = |r| |F Tang | ≡ |r| |F| sin  l A constant torque gives constant angular acceleration if and only if the mass distribution and the axis of rotation remain constant. F Tangential a r F radial F  A force applied at a distance from the rotation axis gives a torque

23. Physics 207: Lecture 17, Pg 23 Torque, like , is a vector quantity Magnitude is given by (1)|r| |F| sin   (2)|F tangential | |r| (3)|F| |r perpendicular to line of action | l Direction is parallel to the axis of rotation with respect to the “right hand rule” And for a rigid object  = I  r F F cos(90 °  ) = F Tang. r F F radial F a r F r  90 °  r sin  line of action

24. Physics 207: Lecture 17, Pg 24 Example : Rolling Motion Notice rotation CW (i.e. negative) when a x is positive! Combining 3 rd and 4 th expressions gives f = Ma x / 2 Top expression gives Ma x + f = 3/2 M a x = Mg sin  So a x =2/3 Mg sin  M l Newton’s Laws:  Mg f N x dir

25. Physics 207: Lecture 17, Pg 25 Exercise Torque Magnitude A. Case 1 B. Case 2 C. Same l In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin  or the tangential force component times perpendicular distance L L F F axis case 1case 2

26. Physics 207: Lecture 17, Pg 26 Exercise Torque Magnitude l In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin  or the tangential force component times perpendicular distance (A) (A) case 1 (B) (B) case 2 (C) same L L F F axis case 1case 2

27. Physics 207: Lecture 17, Pg 27 Example: Rotating Rod Again 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. What is the initial angular acceleration  ? L m

28. Physics 207: Lecture 17, Pg 28 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free … What is its initial angular acceleration ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! L m mg  F = 0 occurs only at the hinge but  z = I  z = - r F sin 90° at the center of mass and I End  z = - (L/2) mg and solve for  z

29. Physics 207: Lecture 17, Pg 29 Statics Equilibrium is established when In 3D this implies SIX expressions (x, y & z)

30. Physics 207: Lecture 17, Pg 30 Example l Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. The teeter-totter is a uniform bar of 30 kg its moment of inertia about the support point is 30 kg m 2. l Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)? l For the static case:

31. Physics 207: Lecture 17, Pg 31 Example: Soln. l Draw a Free Body diagram (assume g = 10 m/s 2 ) l 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5 m from pivot point 30 kg 1 m 60 kg 30 kg 0.5 m 300 N 600 N N

32. Physics 207: Lecture 17, Pg 32 Recap Assignment: l HW7 due tomorrow l Wednesday: review session