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					       8 Rotational Dynamics
• describe/predict rotational behavior:
• motion, energy, momentum.
• with concepts of: rotational inertia & torque

• Homework:
• 30, 33, 39, 60, 65, 83, 89, 111.


                                              1
       Translation and Rotation
•   Translation: up, down, left, right, in, out
•   Rotation:
•   clockwise (cw),
•   counterclockwise (ccw)
•   //




                                                  2
   Rotational Dynamics:
Newton’s 2nd Law for Rotation


    net rotational action
 
     rotational inertia


                                3
Mass-Distribution Effect
Larger       Larger          Larger
radius      Speed          Effort




                Rotational
                Inertia       ~   MR2


                                        4
     Rotational Inertia ( I )
I  m r  m r 
          1 1
             2
                    2 2
                       2
                                 kg(m)2



Example

I  (4kg)(3m) 2  (5kg)( 2m) 2
   36  20  56 kg  m    2




                                          5
  
I  m r  m r 
       1 1
          2
                  2 2
                     2
                                 
      Example: Four identical masses. Axis
      passes symmetrically through center
      of mass.

  I  m1a  m2a  m3a  m4a
              2          2           2           2



              I  4ma        2




                                             6
  
I  m r  m r 
       1 1
          2
                 2 2
                    2
                           
      Example: Four identical masses. Axis
      passes through one end.

  I  m02  m02  m(2a)2  m(2a)2
      I  4ma  4ma  8ma
                 2          2          2




                                             7
               You Try It

Calculate the Rotational
Inertia of this system:




                            8
 Calculated Rot. Inertias, p.273
• rotational inertias of solid objects can be
  calculated
• need to know: a, b, c, d, e, g, h, j, k.
• omit: f, i




                                                9
     Rotational Kinetic Energy
• Rotational K = ½(I)w2.
• Example: Constant Power Source has
  100kg, 20cm radius, solid disk rotating at
  7000 rad/s.
• I = ½MR2 = ½(100kg)(0.2m)2 = 2kgm2.

• Rot K = ½ (2kgm2)(7000/s)2 = 49 MJ

                                               10
Torque       (t)        [N·m]
     F                      F



                   


         t  F
     = lever-arm               11
           Torque Concept
• a small force can cause a large torque
• a large force can create no torque




                                           12
Rotationally Effective Application of Forces




                                               13
Rotationally Ineffective Application of Forces




                                                 14
• torque = Fsin(f) (r)




                         15
1. A pair of forces with equal magnitudes and
opposite directions is acts as shown. Calculate
the torque on the wrench.




                                                  16
             Example
Rod is 3m Long.

  (3m) cos40        40


   t  F  (5N )(2.3m)  11.5Nm



                                   17
18
Newton’s 2nd Law (Rotation)


      t Net  I

        FNet  ma

                              19
           Concept Review
• Torque: rotational action
• Rotational Inertia: resistance to change in
  rotational motion.
• Torque ~ force x lever-arm




                                                20
               Equilibrium
• Translational: net-force = 0
• Rotational: net-torque = 0




                                 21
The drawing shows a person
whose weight is 584N.
Calculate the net force with
which the floor pushes on each
end of his body.




                                 22
 Rotational Work-Energy Theorem
• (Work)rot = tDq.
• Example: torque of 50 Nm is applied for
  one revolution.
• rotational work = (50Nm)(2prad) = 314 J
• (Rotational Work)net = DKrot.
• Krot = ½Iw2.


                                            23
      Angular Momentum (L)
• analog of translational momentum
• L = Iw [kgm2/s]

• Example: Disk R = 1m, M = 2kg, w = 10/s
• I = ½MR2 = ½(1)(1)2 = 0.5
• L = Iw = (0.5kgm2)(10/s) = 5kgm2/s


                                            24
     Conservation of Angular
          Momentum
• For an isolated system
• (Iw)before = (Iw)after
• Example: Stationary disk M,R is dropped
  on rotating disk M, R, wi.
• (Iw)before = (Iw)after
• (½MR2)(wi) = (½MR2 + ½MR2)(wf)
• (wf) = ½ (wi)
                                            25
                8 Summary
•   All TRANSLATIONAL quantities;
•   speed, velocity, acceleration,
•   force, inertia, energy, and momentum,
•   have ROTATIONAL analogs.




                                            26
                 Problem 33
•   Pivot at left joint, Fj = ?, but torque = 0.
•   ccw (Fm)sin15(18) = mg(26) = cw
•   ccw (Fm)sin15(18) = (3)g(26) = cw
•   (Fm) = (3)g(26)/sin15(18) = 160N
•   Note: any point of arm can be considered
    the pivot (since arm is at rest)


                                                   27
                     #39
•   Left force = mg = 30g, Right = 25g
•   mg = 30g + 25g m = 55kg
•   ccw mg(xcg) = cw 30g(1.6)
•   (55)g(xcg) = 30g(1.6)
•   (55)(xcg) = 30(1.6)
•   Xcg = (30/55)(1.6)


                                         28
              #60, z-axis
• Each mass has r2 = 1.52 + 2.52.
• I = sum mr2 = (2+3+1+4)(1.52 + 2.52)




                                         29
                   #65
• First with no frictional torque, then with
  frictional torque as specified in problem.
• M = 0.2kg, R = 0.15m, m1 = 0.4, m2 = 0.8




                                           30
                    #83
• Pulley M, R. what torque causes it to
  reach ang. Speed. 25/s in 3rev?
• Alpha: use v-squared analog eqn.
• Torque = I = (½MR2)()




                                          31
     #89, uniform sphere part
• Rolling at v = 5m/s, M = 2kg, R = 0.1m
• K-total = ½mv2 + ½Iw2.
• = ½(2)(5x5) + ½[(2/5)(2)(0.1x0.1)](5/0.1)2.
• = 25 + 10 = 35J
• Roll w/o slipping, no heat created, mech
  energy is conserved, goes all to Mgh.
• 35 = Mgh h = 35/Mg = 35(19.6) = 1.79m

                                            32
                    #111
•   Ice skater, approximate isolated system
•   Therefore:
•   (Iw)before = (Iw)after
•   (100)(wi) = (92.5)(wf)
•   (wf) = (100/92.5)(wi)
•   K-rot increases by this factor squared
    times new rot. Inertia x ½.

                                              33
Example: Thin rod formulas.




                              34
K = Ktranslation + Krotation




                               35
            Angular Momentum
•   Symbol: L      Unit: kg·m2/s
•   L = mvr = m(rw)r = mr2w = Iw.
•   v is perpendicular to axis
•   r is perpendicular distance from axis to line
    containing v.




                                                36
            Angular Momentum
•   Symbol: L      Unit: kg·m2/s
•   L = mvr = m(rw)r = mr2w = Iw.
•   v is perpendicular to axis
•   r is perpendicular distance from axis to line
    containing v.




                                                37
13) Consider a bus designed to obtain its motive power from a large
rotating flywheel (1400. kg of diameter 1.5 m) that is periodically
brought up to its maximum speed of 3600. rpm by an electric motor
at the terminal. If the bus requires an average power of 12. kilowatts,
how long will it operate between recharges?
Answer: 39. minutes
Diff: 2 Var: 1 Page Ref: Sec. 8.4




                                                                    38
6) A 82.0 kg painter stands on a long horizontal board 1.55 m from one
end. The 15.5 kg board is 5.50 m long. The board is supported at each
end.
(a) What is the total force provided by both supports?
(b) With what force does the support, closest to the painter, push
upward?




                                                                  39
28) A 4.0 kg mass is hung from a string which is wrapped around a
cylindrical pulley (a cylindrical shell). If the mass accelerates
downward at 4.90 m/s2, what is the mass of the pulley?
A) 10.0 kg
B) 4.0 kg
C) 8.0 kg
D) 2.0 kg
E) 6.0 kg




                                                                40
 19) A solid disk with diameter 2.00 meters and mass 4.0 kg freely rotates about a
 vertical axis at 36. rpm. A 0.50 kg hunk of bubblegum is dropped onto the disk and
 sticks to the disk at a distance d = 80. cm from the axis of rotation.
 (a) What was the moment of inertia before the gum fell?
 (b) What was the moment of inertia after the gum stuck?
 (c) What is the angular velocity after the gum fell onto the disk?




(a) 2.0 kg-m2
(b) 2.3 kg-m2                                                                  41
(c) 31. rpm
42
3. The drawing shows the top view of
two doors. The doors are uniform and
identical. The mass of each door is M
and width as shown below is L. How do
their rotational accelerations compare?




                                     43
A Ring, a Solid-Disk, and a Solid-Sphere are released from
rest from the top of an incline. Each has the same mass and
radius. Which will reach the bottom first?




                                                              44
5. The device shown below is spinning with rotational rate wi when the
movable rods are out. Each moveable rod has length L and mass M. The central
rod is length 2L and mass 2M.
Calculate the factor by which the angular velocity
is increased by pulling up the arms as shown.




                                                                           45
How large is the height h in the drawing? The ball’s initial speed is
4.0 m/s. Does it depend on M and R?




                     solid ball
                     rolls without
                     slipping




                                                                   46
      Rotational Review   (angles in radians)



         s  rq                             Dq
                                  wavg    
                                            Dt
        vt  rw
                                            Dw
                                   avg   
        at  r                             Dt

        a c  rw                    
                                  a  a c  at
                   2




+ 4 kinematic equations



                                                 47
                Graph showing mt2 is more constant than mt

                           Rotation Disk-Hanging Mass

                2500

                2000
mt and mt2/10




                1500
                                                               mt
                                                               mt2
                1000

                 500

                   0
                       0          200        400         600
                                   mass (grams)
                                                                     48
      Angular Momentum Calculation

                L = Iw

Example: Solid Disk M = 2kg R = 25cm
Spins about its center-of-mass at 35 rev/s



L  Iw  1 (2kg)(.25m) 2 )(35rev / s)(2prad / rev) )
          2

 13.7kgm2 / s

                                                   49
4. A one-meter-stick has a mass of 480grams.
a) Calculate its rotational inertia about an axis perpendicular to the
stick and through one of its ends.
b) Calculate its rotational inertia about an axis perpendicular to the
stick and through its center-of-mass.
c) Calculate its angular momentum if spinning on axis (b) at a rate of
57rad/s.




                                                                         50
     Conservation of Angular
          Momentum
• Example: 50 grams of putty shot at 3m/s
  at end of 200 gram thin 80cm long rod free
  to rotate about its center.
• Li = mvr = (0.050kg)(3m/s)(0.4m)
• Lf = Iw = {(1/12)(0.200kg)(0.8m)2 +
  (0.050kg)(0.4m)2}(w)
• final rotational speed of rod&putty =
                                           51
          Dq
wavg    
          Dt

          Dw
 avg   
          Dt


               52
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55
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57
58

				
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