1. Chapter 8 Torque and Rotation  8.2 Torque and Stability  6.5 Center of Mass  8.3 Rotational Inertia Dorsey, Adapted from CPO Science DE Physics

2. Chapter 8 Objectives  Calculate the torque created by a force.  Solve problems by balancing two torques in rotational equilibrium.  Using center of mass to find the torque  Calculate the moment of inertia for rotation mass.  Know and use the Moment of Inertia of common objects  Describe the relationship between torque, angular acceleration, and rotational inertia.

3. Chapter 8 Vocabulary Terms  torque  center of mass  angular acceleration  rotational inertia  rotation  translation  center of rotation  rotational equilibrium  lever arm  center of gravity  moment of inertia  line of action

4. 8.2 Torque Key Question: How does force create rotation?

5. 8.2 Torque  A torque is an action that causes objects to rotate.  Torque is not the same thing as force.  Torque matters more for motion than force

6. Torque  Motion in which an entire object moves is called translation.  Motion in which an object spins is called rotation.

7. Torque  The point or line about which an object turns is its center of rotation.  An object can rotate and translate.

8. Torque  The line of action: Goes through the point of application in the direction of the applied force  Torque is created when the line of action….???

9. Let’s see.  Put a book on the table.  Push it in the middle  Then push it on the edge  How does it move both times?

10. Torque  Torque is created when the line of action, does not pass through the center of rotation.

11. Another Example  When you came in the classroom you used torque to get in.  What did you have to do to get in the room?

12. Opening a Door  Where is the handle of the door located with respect to the hinges?  Go to the door as a class:  try opening it really close to the hinges  then as far as you can go.

13. Opening a Door  Which was easier  Opening from the edge or near the hinges?

14. Torque  Maximum torque happens when pulling perpendicular to the shaft  The lever arm is the perpendicular distance between the center of rotation and the line of action

16. Torque  = r x F Lever arm length (m) Force (N) Torque (N. m)

17. The right hand rule: Curl your right hand following the torque arrow with your thumb in the middle. Thumb pointing out of page is + Thumb into page is -

18. Calculate a torque  A force of 50 newtons is applied to a wrench that is 30 centimeters long.  Calculate the torque if the force is applied perpendicular to the wrench so the lever arm is 30 cm.

19. Calculate a torque  A force of 50 newtons is applied down, counter clockwise, to a wrench in a way with a 30 cm long lever arm.  τ = (-50 N)(0.3 m) = -15 N. m  The torque is negative following the right hand rule.

20. When the force and lever arm are NOT perpendicular

21. 9.1 Calculate a torque  Perpendicular Force 50 N turns this bolt. What force is needed at 30 degrees:  Knowns: r =.20 m Angle2 = 30 deg  F1 = 50. N F2 = ?  Angle1= 0 deg  Eq: τ = rF.

22. Calculate a torque  Torque required to break τ = (50 N)(0.2 m) = 10 N. m  To get the same torque with a F at 30 degrees:  10 N. m = F × (0.2 m)cos30 o  10 N. m = 0.173 F F = 58 N : More force is required  Perpendicular Force 50 N turns this bolt. What force is needed at 30 degrees:  Knowns: Eq: τ = rF.  r =.20 m Angle = 30 deg  F = 50. N F(perpen) = ?

23. Rotational Equilibrium  When an object is in rotational equilibrium, the net torque applied to it is zero.  Rotational equilibrium is often used to determine unknown forces.

24. Calculate using equilibrium  A boy and his cat sit on a seesaw.  The cat has a mass of 4 kg and sits 2 m from the center of rotation.  If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?

25. Calculate using equilibrium The cat, mass of 4 kg sits 2 m from the center of rotation. Where should a boy of mass 50 kg sit to balance the system  Solve: τcat = (2 m)(4 kg)(9.8 N/kg) = + 78.4 N-m  τboy = (d)(50 kg)(9.8 N/kg) = - 490 d  For rotational equilibrium, the net torque = zero.  78.4 - 490 d = 0  d = 0.16 m The boy must sit 16 cm from the center.

26. Force on bridge supports  http://www.physics.ucla.edu/demoweb/n ewdyn/html/static_forces_bridge.htm http://www.physics.ucla.edu/demoweb/n ewdyn/html/static_forces_bridge.htm  Play with this for no longer than 3 min.  Notice how the forces on the pillars change as the truck moves across the bridge.

27. Force on bridge supports  Find the fraction of where the center of mass is to the total length of the bridge  That is the fraction of the weight that is on the further pillar  The closer pillar takes the rest of the weight

28. Force on bridge supports  You can also sum the torques of what is on the bridge and the torque the CM of the bridge applies.

29. Rotational Equilibrium  When an object is in rotational equilibrium, the net torque applied to it is zero.  Rotational equilibrium is often used to determine unknown forces.  What are the forces (F A, F B ) holding the bridge up at either end?

30. Rotational Equilibrium

31. 6.5 Center of Mass Key Question: How do objects balance?

32. Center of Mass  There are three different axes about which an object will naturally spin.  The point at which the three axes intersect is called the center of mass.

33. Finding the center of mass  Center of mass for all objects can be found by spinning the object and finding the intersection of the three spin axes.  There is not always material at an object’s center of mass.

35. Finding the center of gravity  Center of gravity is different from center of mass.  For very tall objects, such as skyscrapers, the acceleration due to gravity may be slightly different at points throughout the object.

36. Balance and center of mass  For an object to remain upright, its center of gravity must be above its area of support.  The area of support includes all the area within the supports.  An object will topple over if its center of mass is not above its area of support.

38. Center of mass and people:  Start at 1:00

39. Rotational Inertia Key Question: Does mass resist rotation the same way it resists linear acceleration?

40. Rotational Inertia  Inertia is resistance to a change in its motion  Rotational inertia describes an object’s resistance to a change in its rotational motion.  An object’s rotational inertia depends on the total mass and the way mass is distributed.  Look at the pictures on the right. Mass further means more inertia

41. Watch this!

42. Mass on the end means slower!!!

43. Watch this!

44. Linear and Angular Acceleration a =  r Radius of motion (m) Linear acceleration (m/sec 2 ) Angular acceleration (kg)

45. Rotational Inertia  Rotational motion’s equation has force replaced by the torque about the center of rotation.  The linear acceleration is replaced by the angular acceleration.

46. Rotational Inertia  A rotating mass on a rod can be described with variables from linear or rotational motion.

47. Rotational Inertia  The product of mass × radius squared (mr 2 ) is the rotational inertia for a point mass where r is measured from the axis of rotation.

48. Moment of Inertia  Is analogous to the mass of rotation  Greater the moment of inertia: the harder it is to change its motion

49. Moment of Inertia  For just one particle it is mr 2  Summing all the mass*r 2 for all the particles is the moment of inertia (I)  Each solid object has mass distributed at different distances from the center of rotation.  Rotational inertia depends on the square of the radius, so mass distribution makes a big difference for rotational inertia.

50. Moment of Inertia The moment of inertia of some simple shapes rotated around their center axes You need to write these six down and know them!!!

51. Watch This: Why is one faster than the other.

52. Rotation and Newton's 2nd Law  If you apply a torque to a wheel, it will spin in the direction of the torque.  The greater the torque, the greater the angular acceleration.

53. Solving Inertia Example  Two point mass of 1 kg are on opposite ends of a 2 m long massless rod.  If spun around the center, what is the rotational inertia, I.  If spun at one end, what is I

54. Solving Inertia Example  Two point mass of 1 kg are on opposite ends of a 2 m long massless rod.  If spun around the center, what is the rotational inertia, I.  I=m 1 r 2 + m 2 r 2 = 1*1 2 +1*1 2 = 2 kg*m 2

55. Solving Inertia Example  Two point mass of 1 kg are on opposite ends of a 2 m long massless rod.  If spun at one end, what is I.  I=m 1 r 2 = 1*2 2 = 4 kg*m 2

56. Application: Bicycle Physics