1. Torque and Center of Mass
Julius Sumner Miller

2. Center of Mass:
The center of mass (or mass center) is the mean location of all the mass in a system.
The motion of an object can be characterized by this point in space. All the mass of the object can be thought of being concentrated at this location. The motion of this point matches the motion of a point particle.

5. Finding the Center of Mass:
Uniform geometric figures have the center of mass located at the geometric center of the object.
Note that the center of mass does not have to be contained inside the volume of the object.

6. Collections of Point Masses:
The center of mass for a collection of point masses is the weighted average of the position of the objects in space.
Each object will have a position in space. The center of mass is found as:

7. Example #1: A 10. 0 kg mass sits at the origin, and a 30
Example #1: A 10.0 kg mass sits at the origin, and a 30.0 kg mass rests at the 12.0 m mark on the x – axis. (a) Find the center of mass for this system.

8. (b) Find the center of mass for this system relative to the mass at the right.
Although numerically different, it is the same point in space relative to the masses…

9. Example #2: A 10. 0 cm long wire has a mass of 4. 00 grams
Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.

10. Center of Mass of Both
Sticks together.
Mass 1
Mass 2
2 cm

11. Treat as two objects: 6 cm object: 4 cm object:
Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.
Treat as two objects:
6 cm object:
4 cm object:

12. Center of Mass
X =
Y =
Mass 1
2.40 g
Mass 2
1.60 g
2 cm

14. Center of Mass
X = .800 cm
Y = 1.80 cm
Mass 1
2.40 g
Mass 2
1.60 g
2 cm

15. Example #3: Determine the center of mass of the following masses, as measured from the left end. Assume the blocks are of the same density. This is homework.

18. Torque
Torque is the rotational equivalent of force. A torque is the result of a force applied to an object that tries to make the object rotate about some pivot point.

19. Note that torque is maximum when the angle q is 90º.
Equation of Torque:
applied force
distance from pivot to applied force
angle between direction of force and pivot distance.
pivot point
Note that torque is maximum when the angle q is 90º.
The units of torque are Nm or newton · meter

20. The torque is also the product of the distance from the pivot times the component of the force perpendicular to the distance from the pivot.

21. The torque is also the product of the force times the lever arm distance, d.

22. Example #4: Calculate the torque for the force shown below.

23. Example #5: Calculate the total torque about point O on the figure below. Take counterclockwise torques to be positive, and clockwise torques to be negative.
60degrees

24. Example #6: The forces applied to the cylinder below are F1 = 6
Example #6: The forces applied to the cylinder below are F1 = 6.0 N, F2 = 4.0 N, F3 = 2.0 N, and F4 = 5.0 N. Also, R1 = 5.0 cm and R2 = 12 cm. Determine the net torque on the cylinder.

25. Static Equilibrium: Torque and Center of Mass

26. Static Equilibrium:
Static equilibrium was touched on in the unit of forces. The condition for static equilibrium is that the object is at rest. Since the object is not moving, it is not accelerating. Thus the net force is zero. Shown at right is a typical example from that unit: Find the force of tension in each rope.
A new condition can now be added into this type of problem: Since the object is at rest, it must not be rotating, as that would also require an acceleration.

27. Answer: The pivot point can be put anyplace you want!
If there is no rotation, there must not be a rotational acceleration. Thus the net torque must be zero. This is an application of Newton’s 2nd law to rotational motion.
If the net torque is zero, then all the counterclockwise (ccw) torques must balance all the clockwise (cw) torques.
If there is no rotation, where is the pivot point for calculating torque?
Answer: The pivot point can be put anyplace you want!
Hint: Put the pivot point at one of the unknowns. This eliminates the unknown from the torque equation.

28. Example #1: A meter stick has a mass of 150 grams and has its center of mass located at the 50.0 cm mark. If the meter stick is supported at each of its ends, then what forces are needed to support it?
Show that the two forces are equal through torque. Put the pivot point at the left end.
Force F1 does not contribute to torque. {force applied to pivot point!}
Force F2 makes a ccw torque.
Force mmsg makes a cw torque.

29. The other unknown must also equal half the weight, so:
Balance the net ccw and cw torque:
The other unknown must also equal half the weight, so:

30. Example #2: Suppose the meter stick above were supported at the 0 cm mark (on the left) and at the 75 cm mark (on the right). What are the forces of support now?
Find the two unknown forces through torque. Put the pivot point at the left end.
Force F1 does not contribute to torque. {force applied to pivot point!}
Force F2 makes a ccw torque.
Force mmsg makes a cw torque.

31. Balance the net ccw and cw torque:
If force F2 holds 2/3 the weight, then F1 must hold the remaining 1/3 of the weight.

32. Example #3: A meterstick is found to balance at the 49
Example #3: A meterstick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
The meterstick behaves as if all of its mass was concentrated at its center of mass.
Calculate the torque about the pivot point. The support force of the fulcrum will not contribute to the torque in this case.

33. Force maddedg makes a ccw torque.
Force mmsg makes a cw torque.
Balance the net ccw and cw torque:

34. Example #4: A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 200 N and is 3.00 m long. What is the tension in each rope when the 700-N worker stands 1.00 m from one end?
Put the pivot point on the left end. The force F1 does not contribute torque. Solve for F2.

35. Solve F1 from Newton’s laws:

36. Put the pivot point at the left end and balance the torques.
Example #5: A cantilever is a beam that extends beyond its supports, as shown below. Assume the beam has a mass of 1,200 kg and that its center of mass is located at its geometric center. (a) Determine the support forces.
Put the pivot point at the left end and balance the torques.

37. Balance the net ccw and cw torque:
Solve FA from Newton’s laws:
= 11,760 N
The fact FA is negative means that the force really points downwards.
When the wrong direction is chosen for a force, it just comes out negative at the end.

38. Static Equilibrium: Day #2

39. Example #6: Calculate (a) the tension force FT in the wire that supports the 27.0 kg beam shown below.
Put the pivot point at the left end. The wall support does not contribute to torque.
Note that q and 40° are supplements, so it does not matter which is used in the sine function.
Beam length is L.

40. Balance the net ccw and cw torque:

41. Balance forces in each component direction.
(b) Determine the x and y components to the force exerted by the wall.
Balance forces in each component direction.

42. Example #7: A shop sign weighing 245 N is supported by a uniform 155 N beam as shown below. Find the tension in the guy wire and the horizontal and vertical forces exerted by the hinge on the beam.
Put the pivot point at the left end. The wall support does not contribute to torque.

43. Balance the net ccw and cw torque:

44. The fact Fy is negative means that the force really points downwards.

46. Example #8: A person bending forward to lift a load “with his back” (see figure below) rather than “with his knees” can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in the figure below of a person bending forward to lift a 200-N object. The spine and upper body are represented as a uniform horizontal rod of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. Let the distance from the hinge point to the weight be distance, L. The angle between the spine and this muscle is 12.0°. Find the tension in the back muscle and the compressional force in the spine.

47. Put the pivot point at the left end
Put the pivot point at the left end. The hip support does not contribute to torque.

49. Example #9: A person in a wheelchair wishes to roll up over a sidewalk curb by exerting a horizontal force to the top of each of the wheelchair’s main wheels (Fig. P8.81a). The main wheels have radius r and come in contact with a curb of height h (Fig. P8.81b). (a) Assume that each main wheel supports half of the total load, and show that the magnitude of the minimum force necessary to raise the wheelchair from the street is given by
where mg is the combined weight of the wheelchair and person. (b) Estimate the value of F, taking mg = N, R = 30 cm, and h = 10 cm.

50. Minimum F to lift comes when the normal force becomes zero.
Balance torques about contact point A:

51. 2F Find the Clockwise Torque Φ X 2R-h A Φ h
Ƭcw = 2F X sinΦ There are two hands at work here
Ƭcw = 2F X 2R-h X
Ƭcw = 2F (2R-h)

52. Find the Counterclockwise Torque
R-h A d h
Ƭccw = F X sinΦ
Pythagorean Theorem
Ƭccw = mg d

53. Do you see why ramps are needed around town?
Ƭcw = 2F(2R-h)
F = 313 N = 64 pounds.
Do you see why ramps are needed around town?

54. Example #10: A circular disk 0
Example #10: A circular disk m in diameter, pivoted about a horizontal axis through its center, has a cord wrapped around its rim. The cord passes over a frictionless pulley P and is attached to an object that weighs 240 N. A uniform rod 2.00 m long is fastened to the disk, with one end at the center of the disk. The apparatus is in equilibrium, with the rod horizontal. (a) What is the weight of the rod?

55. (b) What is the new equilibrium direction of the rod when a second object weighing 20.0 N is suspended from the other end of the rod, as shown by the broken line in the image below? That is, what angle does the rod then make with the horizontal?

57. Inertia and Rotary Motion Moment of Inertia
Handout
HW #8

58. Final Schedule: Mosig’s Class 2013
Monday 12/9 Notes
Tuesday 12/10 Finish notes
Wednesday 12/11 Study / Tower practice day
Thursday 12/12 Study / Tower practice day
Friday 12/13 Final Exam {covers current unit}
Monday 12/16 Towers p. 0, 1, & 6
Tuesday 12/17 Towers p. 2 & 3
Wednesday 12/18 Towers p. 4 & 5
Thursday 12/19 Elf Dance / Non – Academic day
Friday 12/20 No School

59. Definition: Inertia is the ability of an object to resist a change in its motion.
Inertia was introduced earlier in the Force unit. For straight line motion, the inertia of an object is measured through mass: The more massive an object, the more it is able to resist changes to its (straight line) motion.
Force and acceleration were related through inertia:
There is an equivalent to inertia in rotary motion. Here, inertia would try to resist a change to the angular motion. This form of inertia will depend on mass, just as before, but it will also depend on the distribution of the mass.
Demonstration:

60. When the mass is distributed close to the center of rotation, the object is relatively easy to turn. When the mass is held much further away, it is more difficult to rotate the object.
For example, if you had a ring (hoop) and a disk with the same radius and same mass, the ring would show more resistance to rotation than would the disk. The disk has mass uniformly distributed across its body, so some of the mass is near the center of rotation. The ring has more mass concentrated at the outside edge of the body than does the disk, so it will show more inertia (even though the mass and radius are the same).

61. The dependence of the inertia on mass and distribution can be built up through the kinetic energy of an object moving through a circular path.
The kinetic energy of a mass m moving at a speed v around the circle is:
Let the circle have a radius r, and let the angular speed of the mass be w. Write the kinetic energy in terms of the angular speed:
Define the moment of inertia I of this point mass to be:
Then: This is now the definition.

62. Now build up to a more complicated object:
The total kinetic energy becomes:
Mass m1 travels in a circle of radius r1, etc. All masses have the same angular velocity, w. Write the kinetic energy in terms of this:

63. In general, the moment of inertia is found as:
The farther the mass is from the center of rotation, the higher the moment of inertia.
Example #1: (a) If m = 2.00 kg and d = m for the image below, determine the moment of inertia of this group of objects.

64. (b) If this apparatus is rotating at 4
(b) If this apparatus is rotating at 4.00 rad/s, what is its kinetic energy?

65. Eeek!!! Calculus! Run for your lives!
For more complicated objects, the moment of inertia is tabulated below. The calculations of these are complex and beyond the scope of the class.
Turns into:
Eeek!!! Calculus! Run for your lives!

66. Example of calculating the rotational inertia of a solid ball
Example of calculating the rotational inertia of a solid ball. You are not responsible for knowing these calculations…

67. Newton’s 2nd Law and Rotational Motion:
There is an equivalent to Newton’s 2nd law for rotational motion.
This can be combined with the kinematics equations from earlier in the unit to solve uniform motion problems;

68. Example #2: A force of N is applied to the edge of a disk that can spin about its center. The disk has a mass of 240 kg and a diameter of 3.20 m.
If the force is applied for 24.0 s, how fast will the disk be turning if it starts from rest?

70. Example #3: Variation of the Atwood’s machine. A 6
Example #3: Variation of the Atwood’s machine. A 6.00 kg mass is tied via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm. What is the acceleration of the hanging mass downwards?
The tangential acceleration of the disk is equal to the linear acceleration of the falling mass.
The net torque on the disk gives one equation:

71. Next use Newton’s 2nd law on the falling mass:
Combine the equations:

73. Inertia and Rotary Motion Day #2
Handout
HW #8

74. Rolling and Energy Conservation.
When an object rolls along the ground, the tangential speed of the outside edge of the object is the same as the speed of the center of mass of the object relative to the ground.

75. The rolling object has two parts to its motion
The rolling object has two parts to its motion. First is the motion of the center of mass, and second is the rotation around the center of mass. The total kinetic energy is the sum of the kinetic energy associated with each part.
The kinetic energy associated with the center of mass moving in a straight line is given by the term:
The portion associated with rotating around the center of mass is:
Icm is the moment of inertia of the object about its center of mass. Refer to the given table for values.

76. Example #4: A solid ball of radius 10. 0 cm and mass 10
Example #4: A solid ball of radius 10.0 cm and mass 10.0 kg rolls at a given speed vo of 5.00 m/s. (a) What is the total kinetic energy of this rolling ball?
For a rolling ball: &

77. (b) What percentage of the total kinetic energy is rolling?

78. Example #5: Four different objects are placed at the top on an incline, as shown below. A point particle can slide down without friction. The other three objects will roll down the incline. In what order will the objects reach the bottom, from fastest to slowest?
(a) What is the speed of the sliding point particle when it reaches the bottom?
Energy conservation!

79. (b) Solve for the speed of the sphere (solid ball) at the bottom.
Energy conservation!
Note that there is a fixed starting energy, and this is split between linear motion and rotation. The rolling objects are slower!

81. (c) Solve for the speed of the hoop at the bottom.
Energy conservation!
The hoop has the slowest speed, and thus takes the longest to reach the bottom. The disk will be between the ball and the hoop.

83. Example #6: Variation of the Atwood’s machine. A 6
Example #6: Variation of the Atwood’s machine. A 6.00 kg mass is tied via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm.
(a) How fast will the mass be traveling after it falls a distance of h = 4.00 m downwards?
Energy conservation!
Speed of falling mass equals tangential speed of disk:

85. (b) Solve for the acceleration of the hanging mass:
Shortcut: Remember the kinematics equation…
Same result as before!

86. Momentum and Rotary Motion
Handout
HW #9