1. MSU Physics 231 Fall 2015 1 Physics 231 Topic 8: Rotational Motion Alex Brown October 21-26 2015

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4. 4 Key Concepts: Rotational Motion Rotational Kinematics Equations of Motion for Rotation Tangential Motion Kinetic Energy and Rotational Inertia Moment of Inertia Rotational dynamics Rolling bodies Vector quantities Mechanical Equilibrium Angular Momentum Covers chapter 8 in Rex & Wolfson

5. MSU Physics 231 Fall 2015 5 Review: Radians & Radius  r Circumference= 2  r s s =  r  in radians! 360 o =2  rad = 6.28… rad  (rad) = 2  /360 o  (deg)

6. MSU Physics 231 Fall 2015 6 Review: Angular speed Average angular velocity Instantaneous Angular velocity Angular velocity : rad/s

7. MSU Physics 231 Fall 2015 7 Review: Angular  Linear Velocities linear (tangential) velocity

8. MSU Physics 231 Fall 2015 8 Angular acceleration Definition: The change in angular velocity per time unit Average angular acceleration Instantaneous angular acceleration Unit: rad/s 2

9. MSU Physics 231 Fall 2015 9 Angular  Linear Accelerations velocity change in velocity Change in velocity per time unit linear or tangential acceleration

10. MSU Physics 231 Fall 2015 10 Equations of motion Linear motion Angular motion x(t) = x o + v o t + ½at 2 v(t) = v o + at  (t) =  o +  o t + ½  t 2  (t) =  o +  t Angular motion = Rotational Motion!

11. MSU Physics 231 Fall 2015 11 example  =0 A person is rotating a wheel. The handle is initially at rest at  = 0 o. For 5s the wheel gets a constant angular acceleration of 1 rad/s 2. After that The angular velocity is constant. Through what angle will the wheel have rotated after 10s.  t First 5s:  (5)=  0 +  0 t + ½  t 2 = 0 + (0)(5) + ½1(5) 2 = 12.5  (5) =  0 +  t = 0 + (1)(5) = 5 rad/s Next 5s:  (5) = 12.5 + (5)(5) + 0 = 37.5 37.5 rad = 2150 o = 6.0 rev

12. MSU Physics 231 Fall 2015 12 A Rolling Coin d  0 = 18 rad/s  = -1.8 rad/s a)How long does the coin roll before stopping? b)What is the average angular velocity? c) How far (D) does the coin roll before coming to rest? D a)  (t) =  0 +  t 0 = 18 - 1.8t t=10 s b)  = (  0 +  (10))/2 = (18 + 0)/2 = 18/2 = 9 rad/s d = 0.02 m  = area under  vs t curve = (10)(18)/2 = 90 rad c) D = s =  r = (90) (0.02/2) = 0.9 m

13. MSU Physics 231 Fall 2015 13 Torque It is much easier to swing the door if the force F is applied as far away as possible (d) from the rotation axis (O). Torque: The capability of a force to rotate an object about an axis. Torque  = F d (Nm) Torque is positive if the motion is counterclockwise (angle increases) Torque is negative if the motion is clockwise (angle decreases) Top view F d O

14. MSU Physics 231 Fall 2015 14 Multiple forces causing torque. 0.6 m 0.3 m F1 = 100 N F 2 = 50 N Two persons try to go through a rotating door at the same time, one on the l.h.s. of the rotator and one the r.h.s. of the rotator. If the forces are applied as shown in the drawing, what will happen? Top view  1 = -F 1 d 1 = -(100)0.3 = -30 Nm  2 = F 2 d 2 = (50)0.6 = 30 Nm Nothing will happen! The 2 torques are balanced. 0 Nm 

15. MSU Physics 231 Fall 2015 15 Center of gravity (center of mass) F pull d pull (side view) F gravity d gravity ?   = F pull d pull + F gravity d gravity We can assume that for the calculation of torque due to gravity, all mass is concentrated in one point: The center of gravity: the average position of the mass d cg =(m 1 d 1 +m 2 d 2 +…+m n d n ) (m 1 +m 2 +…+m n ) 1 2 3………………………n

16. MSU Physics 231 Fall 2015 16 Center of mass; more general The center of gravity center of gravity = center if mass Note 2: freely rotating systems (I.e. not held fixed at some point) must rotate around their center of gravity. x 1,y 1 x 2,y 2 x 3,y 3 m1gm1g m2gm2g m3gm3g O x cg,y cg

17. MSU Physics 231 Fall 2015 17 examples: (more in the book) Oxygen molecule relative mass O is 16 H is 1 Where is the center of gravity?

18. MSU Physics 231 Fall 2015 18 Quiz Consider a hoop of mass 1 kg. The center of gravity is located at: a)The edge of the hoop b)The center of the hoop c)Depends on how the hoop is positioned relative to the earth’s surface.

19. MSU Physics 231 Fall 2015 19 Object in equilibrium? CG FpFp FpFp dd Top view Newton’s 2nd law:  F=ma F p +(-F p )=ma=0 No acceleration of the center of mass up or down But the block starts to rotate!   = F p d + F p d = 2F p d (both clockwise) There is rotation!

20. MSU Physics 231 Fall 2015 20 Translational equilibrium:  F = ma = 0 The center of gravity does not move Rotational equilibrium:   =0 The object does not rotate Mechanical equilibrium:  F = ma = 0 &   =0 No movement! Torque:  = Fd Center of Gravity:

21. MSU Physics 231 Fall 2015 21 question 20N Is this “freely rotating” object in mechanical equilibrium? a) Yes b) No 1m COG Freely rotating: can only rotate around its COG: Rotational eq.:   = -20x1 + 20x0 + 20x1 =0 Translational eq.:  F= 20 –20 +20 =+20 No translational equilibrium: no mechanical equilibrium!

22. MSU Physics 231 Fall 2015 22 quiz 0-212 m 20N 40N A wooden bar is initially balanced. Suddenly, 3 forces are applied, as shown in the figure. Assuming that the bar can only rotate, what will happen (what is the sum of torques)? a)the bar will remain in balance b)the bar will rotate counterclockwise c)the bar will rotate clockwise Torque:  =Fd

23. MSU Physics 231 Fall 2015 23 Somethings to keep in mind If an object is in equilibrium, it does not matter where one chooses the axis of rotation: the net torque must always be zero. If it is given that an object is at equilibrium, you can choose the rotation axis at a convenient location to solve the problem.

24. MSU Physics 231 Fall 2015 24 Weight of board: w What is the tension in each of the wires (in terms of w) given that the board is in equilibrium? Translational equilibrium  F = ma = 0 T 1 +T 2 -w = 0 so T 1 =w-T 2 Rotational equilibrium   = 0 T 1 0 – (0.5)w + (0.75)T 2 =0 T 2 =0.5/(0.75 w) = 2/3w T 1 =1/3w T 2 =2/3w Note: I chose the rotation point at T 1, but could have chosen anywhere else. By choosing T 1 the torque due to T 1 Does not contribute to the equation for rotational equilibrium w T1T1 T2T2 0 0.75m 0.25m

25. MSU Physics 231 Fall 2015 25 r F t = ma t Torque and angular acceleration m FtFt Newton 2nd law: F=ma F t r = mra t F t r = mr 2  Used a t = r   = mr 2  = I  (used  = F t r) The angular acceleration goes linear with the torque. I = mr 2 = moment of inertia for one point mass

26. MSU Physics 231 Fall 2015 26  = I  Moment of inertia I: I = (  m i r i 2 ) r i is the distance to the axis of rotation I = mr 2 = when all mass is at the same distance r compare with: F=ma The moment of inertia in rotations is similar to the mass in Newton’s 2 nd law. m1m1 m2m2 m3m3 r1r1 r2r2 r3r3 More General for rotation around a common axis

27. MSU Physics 231 Fall 2015 27 Extended objects - hoop I=(  m i r i 2 ) =(m 1 +m 2 +…+m n )R 2 =MR 2 M F r mimi

28. MSU Physics 231 Fall 2015 28 A homogeneous stick Rotation point m m m m m m m m m m F  =(m 1 r 1 2 +m 2 r 2 2 +…+m n r n 2 )   =(  m i r i 2 )   =I  Moment of inertia I: I=(  m i r i 2 )

29. MSU Physics 231 Fall 2015 29 Two inhomogeneous sticks m m m m m m m m 5m F m m m m m m m m F 18m  =(  m i r i 2 )   118mr 2   =(  m i r i 2 )   310mr 2  r Easy to rotate!Difficult to rotate

30. MSU Physics 231 Fall 2015 30 Demo: fighting sticks

31. MSU Physics 231 Fall 2015 31 A simple example A and B have the same total mass. If the same torque is applied, which one accelerates faster? F F r r Answer: A  =I  Moment of inertia I: I=(  m i r i 2 ) A B

32. MSU Physics 231 Fall 2015 32 The rotation axis matters! I = (  m i r i 2 ) =(0.2) 0.5 2 + (0.3) 0.5 2 + (0.2) 0.5 2 + (0.3) 0.5 2 = 0.5 kg m 2 I = (  m i r i 2 ) = (0.2) 0 + (0.3) 0.5 2 + (0.2) 0 + (0.3) 0.5 2 = 0.3 kg m 2 0.5 m 0.2 kg 0.3 kg

33. MSU Physics 231 Fall 2015 33  =I  (compare to  F=ma ) Moment of inertia I: I=(  m i r i 2 )  : angular acceleration I depends on the choice of rotation axis!!

34. MSU Physics 231 Fall 2015 34 Some common cases R R L L hoop/thin cylindrical shell: I=MR 2 disk/solid cylinder I= 1 / 2 MR 2 Long thin rod with rotation axis through center: I= 1 / 12 ML 2 solid sphere I= 2 / 5 MR 2 thin spherical shell I= 2 / 3 MR 2 Long thin rod with rotation axis through end: I= 1 / 3 ML 2

35. MSU Physics 231 Fall 2015 35 L mass: m F g =mg Compare the angular acceleration for 2 bars of different mass, but same length.  = I  = mL 2  /3 also  = Fd = mg(cos  )L/2 So mg(cos  )L/2 = mL 2  /3 and  = 3g(cos  )/(2L) independent of mass! Compare the angular acceleration for 2 bars of same mass, but different length  = 3g(cos  )/(2L) so if L goes up,  goes down! I bar =mL 2 /3   F p = F g cos  = mg cos  perpendicular component Falling Bars

36. MSU Physics 231 Fall 2015 36 Rotational kinetic energy Consider a object rotating with constant velocity. Each point moves with velocity v i. The total kinetic energy is: KE r = ½I  2 Conservation of energy for rotating object must include: rotational and translational kinetic energy and potential energy [PE+KE t +KE r ] initial = [PE+KE t +KE r ] final mimi riri v

37. MSU Physics 231 Fall 2015 37 Convenient way of writing KE for rotation object Ik ACylindrical shell Mr 2 1 BSolid cylinder(1/2) mr 2 0.5 CThin spherical shell (2/3) mr 2 (2/3) DSolid sphere(2/5) mr 2 (2/5) I = kmr 2

38. MSU Physics 231 Fall 2015 38 rolling motion (no slipping) this means that  =v/r

39. MSU Physics 231 Fall 2015 39 KE for rolling motion (no slipping) r = radius KE = KE (translational) + KE (rotational) = [½mv 2 + ½I  2 ] = [½mv 2 + ½(kmr 2 )v 2 /r 2 ] = [½mv 2 +½kmv 2 ] = (1+k)½mv 2 used: I=kmr 2 and  =v/r KE = (1+k)½mv 2

40. MSU Physics 231 Fall 2015 40 Example. 1m Consider a ball (sphere) and a block going down the same 1m-high slope. The ball rolls and both objects do not feel friction. If both have mass 1kg, what are their velocities at the bottom (I.e. which one arrives first?). The radius of the ball is 0.4 m. Block: [½mv 2 +mgy] initial = [½mv 2 + mgy] final so v = 4.43 m/s 0 + mgh = ½mv 2 + 0 v=  2gh

41. MSU Physics 231 Fall 2015 41 Example. 1m Consider a ball and a block going down the same 1m-high slope. The ball rolls and both objects do not feel friction. If both have mass 1kg, what are their velocities at the bottom (I.e. which one arrives first?). The radius of the ball is 0.4 m. Ball: [½m(1+k)v 2 +mgy] initial = [½m(1+k)v 2 + mgy] final so v = 3.74 m/s slower than the block since part of the energy goes into rotational KE 0 + mgh = ½m(1+k)v 2 + 0 v=  [2gh/(1+k)] with k = 2/5 (No mass dependence)

42. MSU Physics 231 Fall 2015 42 Which one goes fastest (h=1m) objectkvfastest Block sliding04.43 Cylindrical shell13.134 Solid cylinder1/23.612 Thin spherical shell2/33.423 Solid sphere2/53.741 The larger the moment of inertia, the slower it rolls! The available E kin is spread over rotational and translational (linear) E kin Or: besides moving the object, you spend energy rotating it

43. MSU Physics 231 Fall 2015 43 Angular momentum Conservation of angular momentum: If the net torque equals zero, the angular momentum L does not change I i  i = I f  f

44. MSU Physics 231 Fall 2015 44 Conservation laws: In a closed system: Conservation of energy E Conservation of linear momentum p Conservation of angular momentum L

45. MSU Physics 231 Fall 2015 45 Neutron star Sun: radius: R s = (7) 10 5 km Period of rotation: 25 days Supernova explosion Neutron star: radius: R n = 10 km What is frequency of rotation ? I sphere =(2/5)MR 2 s = sun n = neutron star (assume no mass is lost so M s =M n )  s = 2  /(25 days x 24 x 3600) = (2.9) 10 -6 rad/s Conservation of angular momentum: I s  s = I n  n  n = (R s /R n ) 2  s = 1.4E+04 rad/s f n = (  n /2  ) = 2200 Hz !

46. MSU Physics 231 Fall 2015 46 The spinning lecturer… A lecturer (60 kg) is rotating on a platform with  i =2  rad/s (1 rev/s). He is holding two 1 kg masses 0.8 m away from his body. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate (m arm =2.5 kg). 0.4m 0.8m I i = 0.5 M lec R 2 (body) +2(M w R w 2 ) (two weights) +2(0.33M arm 0.8 2 ) (two arms) = 1.2 + 1.3 + 1.0 = 3.5 kg m 2 I f = 0.5 M lec R 2 = 1.2 kg m 2

47. MSU Physics 231 Fall 2015 47 The spinning lecturer… A lecturer (60 kg) is rotating on a platform with  i =2  rad/s (1 rev/s). He is holding two 1 kg masses 0.8 m away from his body. He then puts the masses close to his body (R=0.0 m). Estimate how fast he will rotate (m arm =2.5 kg). 0.4m 0.8m Conservation of angular mom. I i  I = I f  f  f = (I i /I f )  i  f = (3.5/1.2) 2  = 18.3 rad/s (approx 3 rev/s) For details see next slide KE i = KE f ? A=yes B=no

48. MSU Physics 231 Fall 2015 48 Details of spinning lecturer I initial =I body +2I mass +2I arms I body = 0.5M lec R lec 2 (solid cylindrical shell) M=60, R=0.2 I mass =M mass R mass 2 (point-like mass at radius R) M=1 R=0.8 I arms =1/3M arm R arm 2 (arm is like a long rod with rotation axis through one of its ends) M=2.5 kg R=0.8 I initial =0.5M lec R lec 2 +2(M mass R mass 2 )+2(0.33M arm R arm 2 ) =1.2+1.3+1.0= 3.5 kgm 2 I final =0.5M lec R 2 =1.2 kgm 2 (assumed that masses and arms do not contribute to moment of inertia anymore) Conservation of angular mom. I i  i =I f  f 3.5*2  =1.2*  f  i = was given to be 2  rad/s  f =18.3 rad/s (approx 3 rev/s) KE i = (1/2)I i (  i ) 2 = 69 KE f = 201

49. MSU Physics 231 Fall 2015 49 r F t = ma t Orbital motion m FtFt I = mr 2 = moment of inertia one point mass L = I  For a point-like mass orbiting a central axis: I=mr 2 and  =v/r So L = (mr 2 ) (v/r) and thus L=mrv

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55. MSU Physics 231 Fall 2015 55 question A B dAdA dBdB Two people are on a see saw. Person A is lighter than person B. If person A is sitting at a distance d A from the center of the seesaw, is it possible to find a distance d B so that the seesaw is in balance? A) YesB) No If M A =0.5M B and d A =3 m, then to reach balance, d B =… A) 1 m B) 1.5 m C) 2 m D) 2.5 m E) 3 m

56. MSU Physics 231 Fall 2015 56 Translational equilibrium (Hor.)  F x = ma = 0 n-T x = n - Tcos37 o = 0 so n = Tcos37 o Translational equilibruim (vert.)  F y = ma = 0  s n – w – w + T y =0  s n - 2w + Tsin37 o = 0 (use n=Tcos37 o )  s Tcos37 0 - 2w + Tsin37 0 = 0 0.4T - 2w + 0.6T=0 T = 2w Rotational equilibrium:   = 0 xw + 2w - 4Tsin37 0 = 0 (4sin37 o ) = 2.4 T = 2w w (x + 2 - 4.8)=0 x = 2.8 m  s =0.5 coef of friction between the wall and the 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide? w snsn n w T TyTy TxTx (x=0,y=0) 37 o x

57. MSU Physics 231 Fall 2015 57 The direction of the rotation axis The conservation of angular momentum not only holds for the magnitude of the angular momentum, but also for its direction. The rotation in the horizontal plane is reduced to zero: There must have been a large net torque to accomplish this! (this is why you can ride a bike safely; a wheel wants to keep turning in the same direction.) L L

58. MSU Physics 231 Fall 2015 58 Rotating a bike wheel! LL A person on a platform that can freely rotate is holding a spinning wheel and then turns the wheel around. What will happen? Initial: angular momentum: L i = I wheel  wheel Closed system, so L must be conserved. Final: L f = - I wheel  wheel + I person  person = L i = I wheel  wheel  person = 2I wheel  wheel /I person