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PS 11 GeneralPhysics I for the Life Sciences

PS 11 GeneralPhysics I for the Life Sciences. Rotational Motion Dr. Benjamin chan Associate Professor Physics Department february 2014. Questions and Problems for Contemplation. Chapter 8 Questions: 1, 4, 8, 10, 15, 18, 22, 24

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PS 11 GeneralPhysics I for the Life Sciences

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  1. PS 11 GeneralPhysics I for the Life Sciences Rotational Motion Dr. Benjamin chan Associate Professor Physics Department february 2014

  2. Questions and Problems for Contemplation • Chapter 8 • Questions: 1, 4, 8, 10, 15, 18, 22, 24 • Problems: 1, 2, 4, 8, 10, 15, 22, 27, 30, 38, 45, 52, 55, 61, 64 • General Problems: 72, 80, 81

  3. Describing Rotational Motion • Angular displacement q • Let O be the axis of rotation • How far the object has rotated • Only 2 directions possible: clockwise(-) and counter-clockwise(+) • Measured in radians • 1 radian (rad) is the angle subtended by an arc whose length is equal to the radius of motion

  4. Distance traveled • Arc length traversed • For one complete revolution • q can be expressed in revolutions

  5. Example: Bike Wheel • A bike wheel rotates 4.5 revolutions. How many radians has it rotated? • If the wheel has a diameter of 45 cm, what is the distance traveled by a point on the rim of the wheel?

  6. Example: Bird of Prey • A bird’s eye can distinguish objects that subtend an angle no smaller than 3  10-4rad. How many degrees is this? • How small an object can the bird just distinguish when flying at a height of 100m? • For small angles (<15), arc length and chord length are nearly the same

  7. Angular Velocity w • Average w • Instantaneous w • Dt must be very small • Velocity v of a point on a rotating wheel • Changes direction as vector turns • Increases in proportion to distance from the axis of rotation

  8. Angular Acceleration a • Average a • Instantaneous a • Make Dt as small as possible • Tangential acceleration • Radial acceleration

  9. Review of Linear and Angular Quantities • Frequency = number of complete revolutions per second = f • w = 2pf • Period = time required to complete one revolution = T = 1/f

  10. Equations of Motion • Zero angular acceleration • a = 0, w = constant • Uniform circular motion • q = wt + qo • Linear velocity is not constant • Magnitude is constant: v = wr • Direction is changing • Acceleration is not constant • atan= 0 but aR= rw2 = constant (centripetal) • Direction is changing

  11. Example: Earth’s Rotation • How fast is the earth’s equator turning? • w = 2p/T = (2prad)/84,600s = 7.27 x 10-5rad/s • v = rw = (6,380 km)(7.27 x 10-5rad/s) = 464 m/s • How will your speed change as you go to the North or South pole? • v = (r cosf)w = (464 cosf) m/s • f = 14.5°, v = 449 m/s • f = 30°, v = 402 m/s • f = 60°, v = 232 m/s • f = 90°, v = 0 m/s

  12. The Coriolis Effect • As you go from the equator towards the N pole, you are moving faster than the ground you are moving into: veer to your right (earth rotates west to east) • As you go from the N pole towards the equator, you are moving slower than the ground you are moving into: veer to your right • Clockwise flow! • To or from the S pole: veer to the left! • Counterclockwise flow!

  13. Example: Hard Drive • The platter of the hard drive of a computer rotates at 7200 rpm. What is the angular velocity of the platter? • If the reading head of the drive is 3.00 cm from the axis of rotation, how fast is the disk moving right under the head?

  14. Example (continued) • If a single bit requires 0.50 mm of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis? • The number of bits passing the head per second is or 45 megabits/s (Mbps)

  15. Constant Angular Acceleration • a = constant • w = wo + at • q = qo + wot + ½ at2 • Eliminate t between w and q • w2 = wo2 + 2aq

  16. Total Acceleration • atotal = atan + aR • atan • Constant magnitude, changing direction • aR • Variable magnitude, variable direction

  17. Example: Centrifuge • A centrifuge motor is accelerated from rest to 20,000 rpm in 30s. Determine its angular acceleration and how many revolutions it makes while it is accelerating. • Solution • Assuming constant angular acceleration

  18. Example continued • Where the final angular velocity w is • The angular displacement in 30s is then • We divide by 2p to convert to revolutions

  19. Rolling Motion • Translational + rotational motion • No Slipping • Static friction between object and rolling surface

  20. Example: Bicycle • A bicycle slows down uniformly from a velocity of 8.40 m/s to rest over a distance of 115 m. The overall diameter of the tire is 68.0 cm. Determine the initial angular velocity of the wheels.

  21. Example (continued) • Determine the number of revolutions each wheel undergoes before stopping. • The rim of the wheel turns 115m before stopping. Thus, • Determine the angular acceleration of the wheel

  22. Example continued • Determine the time it took the bicycle to stop • Note: when the bike tire completes one revolution, the bike advances a distance equal to the outer circumference of the tire (no slipping or sliding).

  23. Announcements • FINAL EXAM • Wednesday, March 19 • 7.30 - 10.30 • F-113 • Long Test 4 • Thursday, March 13 • 6.00 – 7.30 • Room TBA c/o Paulo

  24. Center of Mass You can reduce an object to a point and describe its translational motion by considering the motion of this point (called its center of mass)

  25. Determining Center of Mass Consider masses m1, m2, m3, … with coordinates (x1, y1), (x2, y2), (x3, y3), …

  26. CM for a Leg • Determine the center of mass of a leg when a) stretched out and b) bent at 90°. Assume the person is 1.70 m tall. • Solution • a) Straight leg • Essentially 1-D • Measure distance from hip joint • CM is 52.1-20.4 = 31.7 units from base of foot • For a height of 172 cm, xcm = 54.5 cm above the bottom of the foot units

  27. CM of Leg • b) Bent leg • For a height of 172 cm xcm = (172 cm)(0.149) = 25.6 cm ycm = (172 cm)(0.23) = 39.6 cm • Center of mass of bent leg is 39.6 cm above the floor and 25.6 cm from thehip joint! units units

  28. CM Trajectory Center of mass of swimmer in flight follows projectile motion (parabolic) trajectory Center of mass of wrench follows constant velocity trajectory

  29. Torque • What causes an object to rotate? • Torque = force x lever arm

  30. More Torque • Units: Nm (Newton-meter) • Reserve J for work and energy • Torque is a vector quantity • Direction determined by the right hand rule

  31. Newton’s First law • Translational Equilibrium • All forces cancel out: SF = 0 • Rotational Equilibrium • Torques must balance out: SG = 0 • When is an object in equilibrium?

  32. Newton’s Second Law • F = ma • G = Ia • I = moment of inertia • a = angular acceleration • Only two possible directions • Counter-clockwise rotation • Clockwise rotation

  33. Moment of Inertia of Particles • For a single moving object with mass m • t = rF = rma = rmra =mr2a • I = mr2 • For several objects rigidly attached to each other • St = (Smiri2)a • I = Smiri2

  34. Changing Moment of Inertia Determine the change in the moment of inertia of a particle as the radius of its orbit doubles Solution It increases by

  35. Changing Your I • Vertical axis of rotation • Arms on the side • R = 25 cm, M = 9.6 kg • Raise your arms in a crucifixion pose • R = 57.5 cm, M = 9.6 kg • 433% increase

  36. Moments of Inertia for Various Objects

  37. Rotational Kinetic Energy

  38. Example: Ball Rolling Down an Inclined Plane • Determine the speed of a solid sphere of mass M and radius R when it reaches the bottom of an inclined plane if it starts from rest at a height H and rolls without slipping. Assume no slipping occurs. Compare the result to an object of the same mass sliding down a frictionless inclined plane.

  39. Solution • Initial mechanical energy • PE = MgH • KEtrans = 0 • KErot = 0 • Final mechanical energy • PE = 0 • KEtrans = ½ Mv2 • KErot= ½ Iw2 • Conservation of Energy MgH = ½ (Mv2 + Iw2)

  40. Solution (continued) • I = (2/5)MR2 for a solid sphere rotating about an axis through its center of mass • w = v/R • Thus MgH = ½ Mv2+ ½(2/5)MR2(v/R)2 (1/2 + 1/5) v2 = gH v = [(10/7)gH]1/2 • v does not depend on the mass and radius of the sphere!!

  41. Frictionless Incline • Ball slides down the incline and does not roll • Thus, ½ Mv2 = MgH v = (2gH)1/2 • The speed is greater! • None of the original PE is converted into rotational energy.

  42. Work Done on a Rotating Body • W = FDl = F rDq • W = tDq • Power P = W/Dt P = tDq/Dt = tw

  43. Angular Momentum L • L = Iw • Newton’s second law becomes • Thus,

  44. Conservation of Angular Momentum • If the net torque acting on a rotating object is zero, then its angular momentum remains constant.

  45. The Ice Skater • How can the ice skater spin so fast?

  46. The Diver • How can the diver make somersaults? • Does she have to rotate initially? • What trajectory does she follow?

  47. The Hanging Wheel • Why is the wheel standing up? • Why does it turn around about the point of support?

  48. Rotating Disk Demo • What happens when you tilt the rotating disk? HINT:

  49. Drunk Driver Test/Tightrope Artist • Follow the line walk • Increase your moment of inertia to minimize rotations

  50. Quiz 8 1. A 4 kg mass sits at the origin, and a 10 kg mass sits at x = + 21 m. Where is the center of mass on the x-axis? (a) + 7 m (b) + 10.5 m (c) + 14 m (d) + 15 m 2. An object moving in a circular path experiences (a) free fall. (b) constant acceleration. (c) linear acceleration. (d) centripetal acceleration. 3. A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular velocity? a) The boy b) The girl c) Both have the same non-zero angular velocity. d) Both have zero angular velocity.

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