Rotational motion pt1

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Rotational motion pt1

  1. 1. Center of gravity & Center of mass• The center of gravity, (aka center of mass) of an object is where all the mass of an object can be considered to be concentrated – It is the point at which the force due to gravity acts through.• The center of gravity can also be thought of as the balancing point on an object
  2. 2. How do we find C of G?• In regularly shaped objects where the mass of the object is evenly distributed, it’s quite easy to find!
  3. 3. How do we find C of G?• In more complex objects it’s a little more difficult
  4. 4. How do we find C of G?We can find the point at which Hang the object from several points; draw linean object will balance straight down and find intersect
  5. 5. How do we find C of G?1. 3. 2.
  6. 6. How do we find C of G?• The C of G location for humans is REALLY complicated: – It varies based on gender and age – It varies based on our position!
  7. 7. Center of Gravity• C of G of women is typically 55% of their height• C of G of men is typically 57% of their height
  8. 8. Center of Gravity• Remains fixed as long as object does not change shape – Varies constantly in humans as we move!• Humans spend most of their time adjusting their positions to the type of equilibrium best suited to the task and environment• Demo: line of gravity must remain within its base. – stand up and stand on one foot – how does your position change? – Try doing this again, but lean against the wall so your posture does not change. – Lean as far forward as you can while standing
  9. 9. Center of Gravity Fig 14.9 & 14.10
  10. 10. Center of Gravity• Objects are much more stable (less likely to topple) if – the center of gravity is located over the base – The center of gravity is lower to the ground • Think of cars: top-heavy vehicles are much more likely to topple than cars that are lower to the ground – They have a wide base
  11. 11. Center of gravity & topplingAn object is less likely to toppleif C of G is lower and centered
  12. 12. Torque• Torque is a twisting force that is required to turn an object about its axis of rotation• Think of a door on its hinges…what influences its ability to rotate on its axis? • Distance from axis of rotation • Magnitude of the force • Direction of the force • More effective if force applied is perpendicular to the lever arm
  13. 13. Torque!Τ=FrsinθUnits: Nm = N*m
  14. 14. What is Torque?Torque is defined as the Force that is applied TANGENT to the circle and at somelever arm distance causing rotation around a specific point. Lever Arm Distance, r POR – Point of RotationCircular Path of the handle
  15. 15. Lever Arm Distance, r POR – Point of Rotation Torque TWO THINGS NEED TO BE UNDERSTOOD: 1) The displacement from a point of rotation is necessary. Can you unscrew a bolt without a wrench? Maybe but it isnt easy. That extra distance AWAY from the point of rotation gives you the extra leverage you need.Circular Path of the handle THUS we call this distance the LEVER (EFFORT) ARM (r) . 2) The Force MUST be perpendicular to the displacement. Therefore, if the force is at an angle, sinq is needed to meet the perpendicular requirement.
  16. 16. Torque• When there is no net torque on an object, it remains in equilibrium (no turning)• When there is a net torque on an object, it will rotate about its axis of rotation • We usually designate: – counterclockwise motion to be positive – Clockwise motion to be negative
  17. 17. ExampleA 150 kg man stands 2 m from the end of adiving board. How much torque does heapply at the base where the board isattached assuming the board remainshorizontal?   Fr sin q Torque takes the units of Force and Displacement   mgr,q  90    (150)(9.8)(2)  2940 Nm
  18. 18. Example: torque applied at an angle• An upward force of 10 Newtons is applied at a 62 degree angle to the end of a wrench that has a 30 cm long lever arm. Calculate the torque on the wrench.
  19. 19. Example: Rotational EquilibriumTwo masses are placed on a see saw. At what distance must the3.0 kg mass be placed in order for the seesaw to be in equilibrium(net torque = 0). Recall that when an object is in ROTATIONAL EQUILIBRIUM, its net torque equals zero! - We must designate one turning direction as positive, and the other as negative.
  20. 20. Example: Rotational EquilibriumTwo masses are placed on a see saw. At what distance must the3.0 kg mass be placed in order for the seesaw to be in equilibrium(net torque = 0). Recall that when an object is in ROTATIONAL EQUILIBRIUM, its net torque equals zero! - We must designate one turning direction as positive, and the other as negative.
  21. 21. Example: Finding net torque on an objectA basketball is being pushed by two players during tip-off. One player exerts a downward force of11 N at a distance of 7.0 cm from the axis of rotation. The second player applies an upward forceof 15 N at a perpendicular distance of 14 cm from the axis of rotation. The forces are applied onopposite sides of the ball. Find the net torque acting on the ball.
  22. 22. In which picture is the torque the most?

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