1. Chapter 9 Rotations of Rigid Bodies Up to this point when studying the motion of objects we have made the (implicit) assumption that these are “point objects” i.e. all the mass is concentrated at one point. In chapter 9 we will drop this assumption and study the rotation of rigid bodies. These are objects that do not change volume or shape. All that is needed to describe the rotations of rigid bodies is Newton’s laws of motion. From these we will develop the equations that describe rotational motion The following new concepts will be introduced: Moment of Inertia ( I ) Torque (  ) Angular Momentum ( L ) (9-1)

2. Cartesian coordinates: (x, y) Polar coordinates: (r,  ) How to convert Cartesian to polar coordinates and vice versa (x, y) (r,  ) O P Ax y y x  r   (9-2)

3. Consider the rotation of a flat rigid body (the hand of a clock in the picture) in the xy-plane about the origin O which is fixed. Any point P of the rigid point (the tip of the hand in the picture) remains at a constant distance from O. All we need to describe the motion of P (and thus the motion of the rigid body) is the angle  between the position vector r of point P and the x-axis. We express the angle  as function of time  (t) Note: Point O is the only point of the rotating body that does not move O (9-3)

4. In a three dimensional rigid body that rotates about a fixed axis all points on the rotation axis do not move Any point P on the rigid body rotates on a plane that is perpendicular to the rotation axis, on a circular orbit of fixed radius r (9-4) Rotation axis r

5. Consider a point P on a pot that is rotating on a potter’s wheel (fig.a). The position of P is described by the angle  Average angular velocity Instantaneous angular velocity Units: rad/s (9-5) t t +  t 

6. The angular velocity vector The vector  is defined as follows: 1. Magnitude 2. Direction and sense of  Direction: The axis of  coincides with the rotation axis Sense: Curl the fingers of the right hand in the direction of rotation. The thumb points along  (9-6)

7. Special case: Rotation with constant angular velocity  This equation is the analog in rotational motion of the equation that describes motion along the x-axis with constant velocity: x(t) = x o + vt Note: In this chapter and the next we will take advantage of the similarity in form between equations that describe motion along a line and rotational motion. (9-7)

8.  (t) =  o +  t eqs.1 Rotational motion with constant angular velocity Period T is the time required to complete one revolution (  = 2  ) Frequency f is the number of revolutions per second From the definition  Relationship between f and  : In eqs. 1 we set  o = 0 and  = 2   2  =  T =  /f  (9-8)

9. In general the angular velocity is not constant but it changes with time. For this reason we introduce the notion of angular acceleration to describe the rate at which  changes with t Average angular acceleration Units: rad/s 2 Instantaneous angular acceleration (9-9) t t +  t

10. Motion with constant angular acceleration  (9-10)

11. Non-uniform acceleration In general angular acceleration  is not constant In this most general case we decompose the vector a into two components: 1. a c (centripetal acceleration) that points towards the center C 2. a t (tangential acceleration) that points along the tangent r C Rotation axis (9-11)

12. Rotational kinetic energy of a rigid body that rotates about an axis with angular velocity . Divide the object into N elements with masses  m 1,  m 2, …,  m N mimi vivi RiRi O x y Each mass element  m i moves on a circle of radius R i with speed v i =  R i. The kinetic energy of  m i is given by: RiRi (9-12)

13. mimi vivi RiRi O x y (9-13) RiRi

14. Moment of Inertia I Recipe for the determination of I for a given object 1. Divide the object into N elements with masses  m 1,  m 2, …,  m N 2. The contribution of each element  I i =  m i R i 2 3. Sum all the terms I =  I 1 +  I 2 + …. +  I N 4. Take the limit as N   RiRi (9-14)

15. The moment of inertia of an object depends on: a. The mass and shape of the object, and b. the position of the rotation axis (9-15)

16. Moment of inertia of a rod of length L and mass M Linear mass density = M/L Divide the rod into elements of length dx and mass dm = dx A x x’ (9-16)

17. Parallel axis theorem Moment of inertia I about any axis of an object of mass m Moment of Inertia I CM about a parallel axis that passes through the center of mass =+ md 2 I = I CM + md 2 d is the distance between the rotation axis and the parallel axis that passes through the center of mass (9-17)

18. Torque  of a force acting on a point Associated with any force F we can define a new vector , known as “the torque of F”, which plays an important role in rotational dynamics Magnitude of   = Fr  r  = “arm of the moment” r  = rsin    = Frsin  rr  Direction of  If the force tends to rotate the object on which it acts the clockwise (CW) direction (as in the picture) the torque points into the plane of rotation and has a negative sign. If on the other hand the rotation is counterclockwise (CCW) the torque points out of the plane and is positive (9-18) F r O P

19. The equivalent of the “second law” in rotational dynamics Consider the object shown in the figure which can rotate about a vertical axis. We divide the object into element of masses m 1, m 2, … m N On each of these elements we apply a force F 1, F 2,... F N For simplicity we assume that these forces are perpendicular to the object. Consider one element of mass m i The force F i results in a torque  i = F i r i, F i = m i a i = m i r i    i = m i r i 2  Total torque  =  1 +  2 + …+  N =  (m 1 r 1 2 +…+ m N r N 2 ) =  I Thus: Compare this with: F = ma (9-19)

20. Example (9-6) page 250. A bucket of mass m = 12 kg is connected via a rope to a cylindrical flywheel of mass M = 88 kg and radius R = 0.5 m. The bucket is dropped and the flywheel is allowed to spin. Determine the angular velocity of the flywheel after the bucket has fallen for 5 s. (9-20)

21. a R (9-21)

22. R a The acceleration of the bucket is also the acceleration of the rope and therefore the tangential acceleration at the rim of the flywheel (9-22)

23. Torque of the gravitational force The torque of the gravitational force on a rigid body F g is equal to the torque of F g acting at the center of mass of the object Thus  = mgR  R  = Rsin    = mgRsin  RR (9-23)

24. Analogies between linear and rotational motion Linear MotionRotational Motion x (distance)  (rotation angle) v (velocity)  (angular velocity) a (acceleration)  (angular acceleration) m (mass)I (moment of inertia) F (force)  (torque) p = (linear momentum)L (angular momentum) p = mvL = I  (9-24)

25. Angular momentum L By analogy to the definition of the linear momentum we define the vector of angular momentum L as follows: units: kg.m 2 /s Since I > 0  L is parallel to  L (9-25)

26. L Conservation of angular momentum (9-26) If the torque  of all external forces on a rigid body is zero then the angular momentum L does not change (it is conserved

27. Rolling When a wheel rolls on flat ground it executes two types of motion a. All points on the rim rotate about the center of mass with angular velocity  b. The center of mass and all other points on the wheel move with velocity v cm Note 1: The total velocity of any point on the wheel is the vector some of the velocities due to these two motions Note 2:  and v cm are connected (9-27)

28. Rolling v cm L1L1 (9-28)

29. Kinetic energy of a rolling object v cm P (9-29)

30. Find the acceleration of the center of mass of the rolling object (moment of inertia about the center of mass I cm, mass m, radius R) as it rolls down an inclined plane of angle  (9-30)

31. (9-31)

32. Rolling object 1 = cylinder Rolling object 2 = hoop (9-32)