1. Chapter 8 Rotational Equilibrium and Rotational Dynamics

2. Rotational kinetic energy We consider a system of particles participating in rotational motion Kinetic energy of this system is Then

3. Moment of inertia From the previous slide Defining moment of inertia (rotational inertia) as We obtain for rotational kinetic energy

4. Moment of inertia: rigid body There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter, its distribution in the rigid object and also depends upon the location of the axis of rotation For a rigid body the sum is calculated over all the elements of the volume This sum can be calculated for different shapes and density distributions For a constant density and the rotation axis going through the center of mass the rotational inertia for 9 common body shapes is given in Table 8-1

5. Moment of inertia: rigid body

6. The rotational inertia of a rigid body depends on the position and orientation of the axis of rotation relative to the body

7. Moment of inertia: rigid body Example: moment of inertia of a uniform ring The hoop is divided into a number of small segments, m 1 …, which are equidistant from the axis

8. Chapter 8 Problem 29 Four objects are held in position at the corners of a rectangle by light rods. Find the moment of inertia of the system about (a) the x-axis, (b) the y-axis, and (c) an axis through O and perpendicular to the page.

9. Torque The door is free to rotate about an axis through O Three factors that determine the effectiveness of the force in opening the door: 1) The magnitude of the force 2) The position of the application of the force 3) The angle at which the force is applied

10. Torque We apply a force at point P to a rigid body that is free to rotate about an axis passing through O Only the tangential component F t = F sin φ of the force will be able to cause rotation

11. Torque The ability to rotate will also depend on how far from the rotation axis the force is applied Torque (turning action of a force): SI unit: N*m (don’t confuse with J)

12. Torque Torque: Moment arm (lever arm): r ┴ = r sinφ Torque can be redefined as: force times moment arm τ = F r ┴

13. Torque Torque is the tendency of a force to rotate an object about some axis Torque is a vector The direction is perpendicular to the plane determined by the position vector and the force If the turning tendency of the force is CCW (CW), the torque will be positive (negative)

14. Torque When two or more torques are acting on an object, they are added as vectors The net torque is the sum of all the torques produced by all the forces If the net torque is zero, the object’s rate of rotation doesn’t change Forces cause accelerations, whereas torques cause angular accelerations

15. Newton’s Second Law for rotation Consider a particle rotating under the influence of a force For tangential components Similar derivation for rigid body

16. Newton’s Second Law for rotation

17. Chapter 8 Problem 35 A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?

18. Center of mass The force of gravity acting on an object must be considered In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at a single point We wish to locate the point of application of the single force whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles This point is called the center of mass of the object

19. Center of mass In a certain reference frame we consider a system of particles, each of which can be described by a mass and a position vector For this system we can define a center of mass:

20. Center of mass of two particles A system consists of two particles on the x axis Then the center of mass is

21. Center of mass of a rigid body For a system of individual particles we have For a rigid body (continuous assembly of matter) the sum is calculated over all the elements of the volume

22. Chapter 8 Problem 8 A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The bonds are 0.100 nm in length, and the angle between the two bonds is 106°. Use the coordinate axes shown, and determine the location of the center of gravity of the molecule. Take the mass of an oxygen atom to be 16 times the mass of a hydrogen atom.

23. Center of gravity Gravitational force on a body effectively acts on a single point, called the center of gravity If g is the same for all elements of a body (which is not always so: see for example Chapter 7) then the center of gravity of the body coincides with its center of mass

24. Angular momentum Angular momentum: SI unit: kg*m 2 /s Recall:

25. Chapter 8 Problem 45 A light rigid rod 1.00 m in length rotates about an axis perpendicular to its length and through its center. Two particles of masses 4.00 kg and 3.00 kg are connected to the ends of the rod. What is the angular momentum of the system if the speed of each particle is 5.00 m/s? (Neglect the rod’s mass.)

26. Conservation of angular momentum Newton’s Second Law for rotational motion If the net torque acting on a system is zero, then If no net external torque acts on a system of particles, the total angular momentum of the system is conserved (constant)

27. Conservation of angular momentum

29. Equilibrium Equilibrium: Static equilibrium: Stable equilibrium: the body returns to the state of static equilibrium after having been displaced from that state Unstable equilibrium: the state of equilibrium is lost after a small force displaces the body

30. Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation

31. Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation

32. Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation

33. Center of mass: stable equilibrium We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation

34. The requirements of equilibrium For an object to be in equilibrium, we should have two requirements met Balance of forces: the vector sum of all the external forces that act on the body is zero Balance of torques: the vector sum of all the external torques that act on the body, measured about any possible point, is zero

35. Equilibrium: 2D case If an object can move only in 2D ( xy plane) then the equilibrium requirements are simplified: Balance of forces: only the x- and y-components are considered Balance of torques: only the z-component is considered (the only one perpendicular to the xy plane)

36. Examples of static equilibrium

40. Chapter 8 Problem 28 One end of a uniform 4.00-m-Iong rod of weight F g is supported by a cable. The other end rests against the wall, where it is held by friction. The coefficient of static friction between the wall and the rod is μ s = 0.500. Determine the minimum distance x from point A at which an additional object, also with the same weight F g can be hung without causing the rod to slip at point A.

41. Indeterminate structures Indeterminate systems cannot be solved by a simple application of the equilibrium conditions In reality, physical objects are not absolutely rigid bodies Concept of elasticity is employed

42. Total energy of a system Conservation of mechanical energy (no non- conservative forces present) In the case where there are non-conservative forces such as friction, we use the generalized work-energy theorem instead of conservation of energy:

43. Questions?

44. Answers to the even-numbered problems Chapter 8 Problem 20 (b) T = 343 N, H = 171 N, V = 683 N ; (c) 5.14 m

45. Answers to the even-numbered problems Chapter 8 Problem 40 10.9 rads

46. Answers to the even-numbered problems Chapter 8 Problem 54 12.3 m/s 2