PS 11 GeneralPhysics I for the Life Sciences

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1 PS 11 GeneralPhysics I for the Life Sciences ROTATIONAL MOTION D R. B E N J A M I N C H A N A S S O C I A T E P R O F E S S O R P H Y S I C S D E P A R T M E N T F E B R U A R Y 0 1 4

2 Questions and Problems for Contemplation Chapter 8 Questions: 1, 4, 8, 10, 15, 18,, 4 Problems: 1,, 4, 8, 10, 15,, 7, 30, 38, 45, 5, 55, 61, 64 General Problems: 7, 80, 81

3 Describing Rotational Motion Angular displacement q Let O be the axis of rotation How far the object has rotated Only 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 θ l r

4 Distance traveled Arc length traversed l rq For one complete revolution l r q can be expressed in revolutions 1 rev rad

5 Example: Bike Wheel A bike wheel rotates 4.5 revolutions. How many radians has it rotated? θ 4.5 rad revs 8 rev rad If the wheel has a diameter of 45 cm, what is the distance traveled by a point on the rim of the wheel? d 45 cm l q 8 rad 630 cm

Example: Bird of Prey A bird s eye can distinguish objects that subtend an angle no smaller than 3 10-4 rad. How many degrees is this? 4 360 θ 310 rad 0.

6 Example: Bird of Prey A bird s eye can distinguish objects that subtend an angle no smaller than rad. How many degrees is this? θ 310 rad rad 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 4 l rq (100m)(310 rad) 3 cm

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

8 Angular Acceleration a Average a Instantaneous a a Make Dt as small as possible Tangential acceleration D w Dt a ra tan Radial acceleration a R v r ( rw) r rw

9 Review of Linear and Angular Quantities Frequency = number of complete revolutions per second = f w = f 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 + q o Linear velocity is not constant Magnitude is constant: v = wr Direction is changing Acceleration is not constant a tan = 0 but a R = rw = constant (centripetal) Direction is changing

11 Example: Earth s Rotation How fast is the earth s equator turning? w = /T = ( rad)/84,600s = 7.7 x 10-5 rad/s v = rw = (6,380 km)(7.7 x 10-5 rad/s) = 464 m/s How will your speed change as you go to the North or South pole? v = (r cos f)w = (464 cos f) m/s f = 14.5, v = 449 m/s f = 30, v = 40 m/s f = 60, v = 3 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 700 rpm. What is the angular velocity of the platter? 700 rev 1 min 10rev w f 754 min 60 sec s rad s 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? v rw (310 m)(754rad / s).6m / s

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.6m / s bits/ m/ bit or 45 megabits/s (Mbps) s

15 Constant Angular Acceleration a = constant w = w o + at q = q o + w o t + ½ at Eliminate t between w and q w = w o + aq

16 Total Acceleration a total = a tan + a R a tan Constant magnitude, changing direction a R Variable magnitude, variable direction

17 Example: Centrifuge A centrifuge motor is accelerated from rest to 0,000 rpm in 30s. Determine its angular acceleration and how many revolutions it makes while it is accelerating. Solution Assuming constant angular acceleration a w wo t 100rad / 30s s 0 70rad / s

18 Example continued Where the final angular velocity w is rad 0000rev / min w f 100rad / rev 60s / m The angular displacement in 30s is then s q 0 (1/ )(70rad / s )(30s) rad We divide by to convert to revolutions rad 3 q rev rad / rev

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

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. v o 8.40m / s w 4.7 rad o r 0.340m / s

21 Example (continued) Determine the number of revolutions each wheel undergoes before stopping. The rim of the wheel turns 115m before stopping. Thus, 115m 115m 53.8 rev r (0.340m) Determine the angular acceleration of the wheel a w wo q 0 (4.7rad / s) (rad / rev)(53.8rev ) 0.90rad / s

22 Example continued Determine the time it took the bicycle to stop w wo 0 4.7rad / s t 7. 4s a 0.90rad / s 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 F-113 Long Test 4 Thursday, March 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 m 1, m, m 3, with coordinates (x 1, y 1 ), (x, y ), (x 3, y 3 ), i i i i i cm m m x m m m m x m x x m x i i i i i cm m m y m m m m y m y m y y

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 xcm (1.5)(9.6) (9.6)(33.9) (3.4)(50.3) units CM is = 31.7 units from base of foot For a height of 17 cm, x cm = 54.5 cm above the bottom of the foot

CM of Leg b) Bent leg xcm (1.5)(9.6) (9.6)(3.6) (3.4)(3.6) 1.5 9.6 3.4 14.9 units ycm (3.4)(1.8) (9.6)(18.) (1.5)(8.5) 1.5 9.6 3.4 3.

27 CM of Leg b) Bent leg xcm (1.5)(9.6) (9.6)(3.6) (3.4)(3.6) units ycm (3.4)(1.8) (9.6)(18.) (1.5)(8.5) units For a height of 17 cm x cm = (17 cm)(0.149) = 5.6 cm y cm = (17 cm)(0.3) = 39.6 cm Center of mass of bent leg is 39.6 cm above the floor and 5.6 cm from the hip joint!

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 r F rf sinq

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 = rf = rma = rmra =mr a I = mr For several objects rigidly attached to each other S = (Sm i r i )a I = Sm i r i

34 Changing Moment of Inertia Determine the change in the moment of inertia of a particle as the radius of its orbit doubles Solution I I i f mr m(r) mr 4mR 1 4 It increases by I f Ii % 100% I 1 i 300%

35 Changing Your I Vertical axis of rotation Arms on the side R = 5 cm, M = 9.6 kg I side MR ( 9.6kg)(0.5m) 0.60kgm Raise your arms in a crucifixion pose R = 57.5 cm, M = 9.6 kg I cross MR 433% increase ( 9.6kg)(0.575m) 3.0kgm

36 Moments of Inertia for Various Objects

37 Rotational Kinetic Energy ) ( ) ( ) ( w w i i i i i i r m m r m v KE 1 w I KE rotational 1 1 I CM Mv CM KE Total w

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 KE trans = 0 KE rot = 0 Final mechanical energy PE = 0 KE trans = ½ Mv KE rot = ½ Iw Conservation of Energy MgH = ½ (Mv + Iw )

40 Solution (continued) I = (/5)MR for a solid sphere rotating about an axis through its center of mass w = v/r Thus MgH = ½ Mv + ½(/5)MR (v/r) (1/ + 1/5) v = gh v = [(10/7)gH] 1/ 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 Mv = MgH v = (gh) 1/ 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 = Dq Power P = W/Dt P = Dq/Dt = w

43 Angular Momentum L L = Iw Newton s second law becomes Thus, I a I D w Dt I w I Dt w o DL Dt

44 Conservation of Angular Momentum If the net torque acting on a rotating object is zero, then its angular momentum remains constant. 0 0 L i L DL Dt L f I w i i L I f f i w f

45 The Ice Skater How can the ice skater spin so fast? w f Ii w i f I

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: dl dt

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 = + 1 m. Where is the center of mass on the x-axis? (a) + 7 m (b) m (c) + 14 m (d) + 15 m. 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.

51 Quiz 8 4. A wheel starts at rest, and has an angular acceleration of 4 rad/s. Through what angle does it turn in 3 s? a) 36 rad b) 18 rad c) 1 rad d) 9 rad 5. A wheel of diameter 6 cm turns at 1500 rpm. How far will a point on the outer rim move in s? a) 314 cm b) 4084 cm c) cm d) cm 6. What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70 cm when the bike is moving 8 m/s? a) 91 m/s b) 183 m/s c) 06 m/s d) 66 m/s 7. A bicycle is moving 4 m/s. What is the angular speed of a wheel if its radius is 30 cm? a) 0.36 rad/s b) 1. rad/s c) 4.8 rad/s d) 13.3 rad/s

52 Quiz 8 8. An ice skater is in a spin with his arms outstretched. If he pulls in his arms, what happens to his kinetic energy? a) It increases. b) It decreases. c) It remains constant but non-zero. d) It remains zero. 9. What is the quantity used to measure an object's resistance to changes in rotation? a) mass b) moment of inertia c) linear momentum d) angular momentum 10. A wheel of moment of inertia of 5.00 kg-m starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s? a) 57.6 J b) 64.0 J c) 78.8 J d) 1 J

Chapter 9-10 Test Review

Chapter 9-10 Test Review Chapter Summary 9.2. The Second Condition for Equilibrium Explain torque and the factors on which it depends. Describe the role of torque in rotational mechanics. 10.1. Angular