Rotational Motion. Chapter 10 of Essential University Physics, Richard Wolfson, 3 rd Edition

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1 Rotational Motion Chapter 10 of Essential University Physics, Richard Wolfson, 3 rd Edition 1

2 We ll look for a way to describe the combined (rotational) motion 2

3 Angle Measurements θθ ss rr rrrrrrrrrrrrrr 22 ππ rrrrrrrrrrrrrr = = 11 rrrrrrrrrrrrrrrrrrrr When you re not sure, always use radians to describe angles!! 3

4 Angular Velocity Average: ωω θθ tt Instantaneous: θθ ωω lim tt 0 tt = ddθθ dddd Units: ωω = rrrrrrrrrrrrrr ssssssssssss 4

5 Angular Velocity and Linear Velocity θθ ss rr ωω ddθθ dddd = 1 rr ddss dddd = vv rr vv = ωω rr 5

6 Example 10.1 rr = 28 mm ωω = 21 rrrrrr vv = ωω rr ωω = 2.2 rrrrrr/ss vv = 62 mm/ss 6

7 Angular Acceleration Average: Instantaneous: αα ωω tt ωω αα lim tt 0 tt = ddωω dddd Units: αα = rrrrrrrrrrrrrr ssssssssssss 2 ωω = vv rr αα = ddωω dddd = dd vv rr dddd = 1 rr ddvv dddd = 1 rr aa tt aa tt = αα rr aa rr = vv2 rr = rr ωω2 7

8 Linear Motion Analog to Rotation 8

9 Example 10.2 αα = 0.12 rrrrrrrrrrrrrr/ss 2 ωω oo = 2.2 rrrrrr/ss ωω = 0 ωω 2 = ωω oo αα θθ θθ oo θθ θθ oo = 20 rrrrrrrrrrrrrr = 3.2 rrrrrrrrrrrrrrrrrrrrrr 9

10 Torque ττ FF rr ssssss θθ FF = FF ssssss θθ 10

11 Example 10.3 rr = 45 cccc θθ = 67 ττ = 95 NN mm FF =?? ττ = FF rr ssssss θθ FF = 230 NN 11

12 Newton s Second Law for Rotational Motion aa tt = αα rr FF = mm aa tt = mm αα rr ττ = FF rr = mm αα rr 2 ττ = II αα II mm rr 2 Moment of Inertia or Rotational Inertia 12

13 Moment of Inertia along a rotation axis II ii mm ii rr ii 2 13

14 Example 10.4 mm = 0.64 kkkk LL = 85 cccc II = 5 8 mm LL2 = 0.29 kkkk mm 2 14

15 Moment of Inertia for a continuous distribution of mass II ii mm ii rr ii 2 II rr 2 dddd 15

16 Example 10.5 MM rr dddd dddd = MM LL rr = xx II rr 2 dddd = LL/2 LL/2 rr 2 dddd LL/2 = LL/2 xx 2 MM LL dddd II = 1 MM LL

17 Example 10.5, modified, solution a rr rr = xx dddd dddd = MM LL MM rr = xx + LL 2 II rr 2 dddd = LL/2 LL/2 rr 2 dddd LL/2 = LL/2 xx + LL 2 2 MM LL dddd II = 1 3 MM LL2 17

18 Example 10.5, modified, solution b ddxx dddd dddd = MM LL rr = xx 0 rr xx II rr 2 dddd LL = rr 2 dddd 0 LL = 0 xx 2 MM LL dddd II = 1 3 MM LL2 18

19 Example 10.6 MM II rr 2 dddd = RR 2 dddd = RR 2 dddd II = MM RR 2 19

20 Example 10.7 dddd dddd = MM ππ RR 2 MM dddd = 2 ππ rr dddd dddd = 2 ππ rr dddd MM ππ RR 2 II rr 2 dddd RR = 0 rr 2 2 ππ rr dddd MM ππ RR 2 II = 1 2 MM RR2 20

21 Moment of Inertia: Shapes and Axis of Rotation 21

22 Parallel Axis Theorem II =?? = II CCCC + MM dd 2 Proof: Problem

23 Now that we have learned how to calculate the moment of inertia of an object at an specific axis of rotation, let s apply it. ττ = II αα 23

24 Example RR = 1.4 mm MM = 940 kkkk ωω oo = 10 rrrrrr FF = 20 NN ωω = 0 tt =?? αα = cccccccccccccccc ττ = 2 FF RR sin(90 ) ωω = ωω oo αα tt ττ = II αα αα = ττ II tt = ωω ωω oo αα II = 1 2 MM RR2 = ωω ωω oo 4 FF/ MMMM 24

25 Example 10.9 MM RR II = 1 2 MM RR2 aa =?? mm gg aa = mm MM ττ = TT RR = II αα mmmm TT = mm aa αα = aa RR TT = II αα RR = II aa RR 2 25

26 Kinetic Energy of a Rotating Object: KK rrrrrr = 1 2 dddd vv2 = 1 2 dddd ωω rr 2 = 1 2 ωω2 dddd rr 2 = 1 2 ωω2 II KK rrrrrr = 1 2 II ωω2 26

27 Example Cylindrical Rotor MM = 135 kkkk RR = 30 cccc ωω = 31,000 rrrrrr KK rrrrrr = 1 2 II ωω2 II = 1 2 MM RR2 KK rrrrrr = 32 MMMM 27

28 Work and Kinetic Energy of a Rotating Object: WW rrrrrr = FF dddd dddd = rr dddd FF rr = ττ dddd = ωω dddd = FF rr dddd = ττ dddd ττ = II ddωω dddd = II dddd dddd ωω dddd WW rrrrrr = 1 2 II ωω ff II ωω ii 2 = II ωω dddd = KK rrrrrr,ff KK rrrrrr,ii = KK rrrrrr 28

29 WW rrrrrr = FF dddd = 1 2 II ωω ff II ωω ii 2 = KK rrrrrr 29

30 Example II = 2.7 kkkk mm 2 ωω ii = 0 rrrrrr ωω ff = 700 rrrrrr ττ =?? θθ = 25 rrrrrrrrrrrrrrrrrrrrrr θθ ffττ WW rrrrrr = dddd θθ ii = 1 2 II ωω ff II ωω ii 2 θθ ffdddd ττ θθ = ττ θθ ττ = 1 2 II ωω ff 2 ff θθ ii = ττ = II ωω ff 2 θθ ii 2 θθ = 46 NN mm 30

31 Total Kinetic Energy for objects under rotational and translational motions NN KK = KK CCCC + ii=1 KK ii,rrrrrr KK = KK CCCC + KK rrrrrr 31

32 Rolling Objects rolling without slipping ωω = ππ tt vv CCCC = ππ RR tt vv rrrrrr = ωω RR = ππ RR tt = vv CCCC 32

33 Work done by Friction: xx ffffxx WW ffffffffffffffff = dddd xx ii FF ppppppp vv = 0 WW ffffffffffffffff = 0 static friction FF ffffffffffffffff FF ppppppp vv WW ffffffffffffffff < 0 kinetic friction FF ffffffffffffffff vv WW ffffffffffffffff = 0 static friction vv = 0 rolling without slipping 33

34 Conservation of Mechanical Energy for a rolling object: WW ffffffffffffffff = 0 rolling without slipping KK + UU = cccccccccccccccc 34

35 Example MM RR vv =?? KK CCCC = 1 KK rrrrrr = 2 MM vv CCCC II vv CCCC RR 2 UU = MM gg h II = 2 5 M RR2 KK + UU = cccccccccccccccc UU = MM gg yy KK = KK CCCC + KK rrrrrr KK CCCC = 1 2 MM vv CCCC 2 KK rrrrrr = 1 II ωω2 = II vv CCCC RR 2 vv CCCC = 2 gg h 1 + II MM RR 2 = 10 gg h 7 35

36 Rolling Race 36

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