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Torque and Center of Mass

Torque and Center of Mass. Julius Sumner Miller. Center of Mass:. The center of mass (or mass center) is the mean location of all the mass in a system.

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Torque and Center of Mass

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  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.

  3. 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.

  4. 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:

  5. 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.

  6. (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…

  7. 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.

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

  9. 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:

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

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

  12. 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.

  13. 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.

  14. applied force Equation of Torque: 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

  15. 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.

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

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

  18. 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

  19. 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.

  20. Static Equilibrium:Torque and Center of Mass

  21. 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.

  22. 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.

  23. 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.

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

  25. 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.

  26. 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.

  27. 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.

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

  29. 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.

  30. Solve F1 from Newton’s laws:

  31. 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.

  32. 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.

  33. Static Equilibrium:Day #2

  34. 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.

  35. Balance the net ccw and cw torque:

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

  37. 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.

  38. Balance the net ccw and cw torque:

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

  40. 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.

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

  42. 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 = 1 400 N, R = 30 cm, and h = 10 cm.

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

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