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Circular Motion

Circular Motion. 1. Rotation and Revolution. Rotation. the motion or spin on an internal axis. Revolution . the motion or spin on an external axis. Rotational Speed. Rotational Speed. Number of rotations per unit of time Rpm or Rps All objects that rotated on same axis have the

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Circular Motion

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  1. Circular Motion 1

  2. Rotation and Revolution

  3. Rotation the motion or spin on an internal axis

  4. Revolution the motion or spin on an external axis

  5. Rotational Speed

  6. Rotational Speed Number of rotations per unit of time Rpm or Rps All objects that rotated on same axis have the same rotational speed. Also called Frequency (cycles/s or Hertz) Period (seconds) is the inverse of Frequency

  7.  Gymnast on a High Bar A gymnast on a high bar swings through two rotations or cycles in a time of 1.90s. Find the average rotational speed (in rps) or frequency (in Hz) of the gymnast.

  8. Given: t = 1.90 s & 2 rotation (cycle) • Find the average rotational speed (in rps) • Rps= rotations • second • = 2 rotation • 1.90 seconds • = 1.05 rps = 1.05 cycles/second

  9.   A Helicopter Blade Find the rotational speed or frequency at #1 if it takes 0.154 s for one rotation (cycle)?

  10. Given: t = 0.154 s & 1 rotation (cycle) • Find the average rotational speed (in rps) • Rps= rotations • second • = 1 rotation • 0.154 seconds • = 6.49 rps = 6.49 cycles/second

  11.   A Helicopter Blade Find the rotational speed or frequency at #2 if takes 0.154 s for one rotation (cycle)?

  12. Given: t = 0.154 s & 1 rotation (cycle) • Find the average rotational speed (in rps) • Rps= rotations • second • = 1 rotation • 0.154 seconds • = 6.49 rps = 6.49 cycles/second

  13. Do Frequency/ Rotational Speed Problems

  14. Tangential speed The speed in m/s of something moving along a circular path. It always tangent to the circle.

  15. Linear (Tangential) Speed The distance moved per unit of time. Linear speed is greater on the outer edge of a rotating object than it is closer to the axis. Linear Speed

  16. Linear (Tangential) Speed Distance traveled in one period is the circumference 2πr Time for one “cycle” is the “period” (T) Tangential Speed = Circumference / Period Tangential Speed = 2πr T

  17. Linear (Tangential) Speed But remember that period is the inverse of frequency So instead of dividing by period you multiply by frequency Tangential Speed = Circumference x Frequency Tangential Speed = 2πr x cycle cycles

  18.   A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 1 on the blade, find the tangential speed

  19. Given: r = 3.00 m & • Angular speed = 6.50 rps • Tangential Speed = 2πr x cycle • cycles • = 2π 3.00m x 6.50 cycle • cycles • = 122 m/s

  20.   A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 2 on the blade, find the tangential speed

  21. Given: r = 6.70 m & • Angular speed = 6.50 rps • Tangential Speed = 2π 6.70mx 6.50 cycle • cycles • = 273 m/s • Notice that the tangential speed at 3 meter is 122 m/s while at 6.70 meters is 273 m/s

  22. Do Tangential Speed • Linear Velocity • Problems

  23. Centripetal Acceleration (centripetal acceleration)

  24.   A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 1 on the blade, find the tangential acceleration

  25. Given: r = 3.00 m & • Tangential Speed = 122 m/s • Tangential • Acceleration = (122 m/s)2 / 3.00m • = 4,960 m/s2 • = 4.96 x 103m/s2

  26.   A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 2 on the blade, find the tangential acceleration

  27. Given: r = 6.70 m & • Tangential Speed = 273 m/s • Tangential • Acceleration = (273 m/s)2 / 6.70m • = 11,200 m/s2 • = 1.12 x 104m/s2

  28. Do Centripetal • Acceleration • Problems

  29. Centripetal Force

  30. Centripetal Force • Fc = mac • Fc = mvT2 • r

  31. Centrifugal force: Center-fleeing, away form center Center-fleeing

  32. Vertical drum rotates, you’re pressed against wall Friction force against wall matches gravity Seem to stick to wall, feel very heavy Rotating Drum Ride Real Forces: Perceived Forces: Friction; up Centripetal; inwards Gravity (weight); down Centrifugal; outwards Gravity (weight); down Perceived weight; down and out The forces real and perceived

  33. Simulated Gravity

  34. Weight the force due to gravity on an object Weight = Mass  Acceleration of Gravity W = m g Weightlessness - a conditions wherein gravitational pull appears to be lacking Examples: Astronauts Falling in an Elevator Skydiving Underwater Weight and Weightlessness

  35. Rotating Space Stations Simulate Gravity Just like spinning drum in amusement park, create gravity in space via rotation Where is the “floor”? Where would you still feel weightless? Note the windows on the face of the wheel From 2001: A Space Odyssey rotates like bicycle tire

  36. Do Centripetal • Force Problems

  37. What makes something rotate? TORQUE How do I apply a force to make the rod rotate about the axle? Not just anywhere! AXLE

  38. Torque = force times lever arm Torque = F L

  39. Torque example What is the torque on a bolt applied with a wrench that has a lever arm of 30 cm with a force of 30 N? F Torque = F x L = 30 N x 0.30 m = 9 N m L For the same force, you get more torque with a bigger wrench  the job is easier!

  40. Net Force = 0 , Net Torque ≠ 0 10 N 10 N > The net force = 0, since the forces are applied in opposite directions so it will not accelerate. .

  41. Net Force = 0 , Net Torque ≠ 0 10 N 10 N > However, together these forces will make the rod rotate in the clockwise direction.

  42. Net torque = 0, net force ≠ 0 The rod will accelerate upward under these two forces, but will not rotate.

  43. Balancing torques 20 N 10 N 0.5 m 1 m Left torque = 10 N x 1 m = 10 n m Right torque = 20 N x 0.5 m = 10 N m

  44. Balancing torques 20 N 10 N 0.5 m 1 m Left torque = 10 N x 1 m = 10 n m Right torque = 20 N x 0.5 m = 10 N m How much force is exerted up by the Fulcrum?

  45. Torque = force times lever arm Torque = F L

  46. Equilibrium • To ensure that an object does not accelerate or rotate two conditions must be met: • net force = 0 • net torque = 0

  47. Example 1 Given M = 120 kg. Neglect the mass of the beam. Find the Torque exerted by the mass Torque = F L = 120 kg (9.8 m/s2) (7 m) = 8232 N m

  48. Example Given: Wbox=300 N Find: FTR= FCC = ? N TorqueC=TorqueCC FCL =FCCL 300 N (6 m) =FCC (8 m) 225 N =FCC A C D

  49. Example Given: Wbox=300 N Find: FTL= FCC= ? N TorqueC=TorqueCC FCL =FCCL FC (8 m) =300N (2 m) FC = 75 N A C D

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