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Chapter 8 Rotational Equilibrium and Dynamics

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Chapter 8 Rotational Equilibrium and Dynamics Powered By Docstoc
					  Chapter 8
  Rotational
Equilibrium and
  Dynamics
8-1 Torque
     Torque (t)
A quantity that measures
 the ability of a force to
 rotate an object around
 some axis.
t  Greek letter ‘tau’
   Net Torque
produces rotation.
Just like a net force
        causes
motion/acceleration.
Torque depends
upon a force and
the length of the
   lever arm.
How easily an object
 rotates depends not
 only on ‘how much’
 force is applied, but
 also where the force
 is applied.
      Lever Arm
The perpendicular
 distance from the axis
 of rotation to a line
 drawn along the
 direction of the force.
Torque also depends
  upon the angle
between a force and
   the lever arm.
l
  t = Fd(sinq)
Metric Units of
torque are Newton-
meters (N-m)
        Torque, like
displacement and force,
   is a vector quantity.
Since torque is a
 rotational motion it has
 two directions:
1.Counterclockwise
 (CCW), which is
 positive.
2.Clockwise (CW),
 which is negative.
   Ex 1: Sean has a flat and is
  changing his tire, to remove
 the lug nuts he applies a force
  of 25 N on the tire iron at an
    angle of 87 o. What is the

torque produced if the tire iron
         is 0.6 m long?
G: d = 0.6 m, F= 25 N,
   q = 87 o

U: t = ?
E: t = Fd(sinq)
S: t = (25 N)(0.6 m)(sin   87 o)

S: t =14.98 N-m
Ex 2: What angle
 produces a torque of
 400 N-m, if the force
 applied is 505 N at a
 distance from the axis
 of rotation of 0.82 m?
G: t = 400 N-m,
  F = 505 N, d =0.82 m
U: q = ?
G: t = 400 N-m,
   F = 505 N, d =0.82 m
U: q = ?
E:         t
  sin  
            Fd
G: t = 400 N-m,
   F = 505 N, d =0.82 m
U: q = ?
E:              1 t 
          sin  
                 Fd 
           1     400      
S :   sin                
               (505)(0.82) 
           1     400      
S :   sin                
               (505)(0.82) 


S :   75     o
8-2 Rotation and
     Inertia
    Point mass
Where all the mass is
assumed to be located
in one point.
   Center of Mass
The point at which all the
 mass of a body can be
 considered to be
 concentrated when
 analyzing translational
 motion.
     Rotational and
translational motion can
      be combined.
 We use the the center of
 mass, as a reference, to
  analyze its translation
         motion.
 Center of Gravity
The position at which the
 gravitational force acts
 on an extended object as
 if it were a point mass.
        Toppling
• If the Center of Gravity (CG)
  is above the support area,
  then the object will remain
  upright.
• If the CG extends outside the
  support area, the object will
  topple.
Unstable Equilibrium
    Is when any
     movement/
  displacement of a
   balanced object
   lowers the CG.
Stable Equilibrium
•Is when any
 motion/displacement
 of a balanced object
 raises its CG.
Neutral Equilibrium
    • Is when any
motion/displacement of
  a balanced object
  neither raises nor
    lowers its CG.
   Moment of Inertia
         (MOI)
The tendency of a body
 to rotate freely about a
 fixed axis to resist a
 change in rotational
 motion.
 MOI is the rotational
  analog of mass.
Very similar to mass, but
 MOI is not an intrinsic
 property of an object.
 It depends upon the
object’s mass and the
 distribution of mass
  around the axis of
       rotation.
The farther the mass is, on
  average from the axis of
  rotation, the greater the
 the object’s MOI and the
    more difficult it is to
     rotate the object.
Calculating the MOI
Pg 285: Table 8-1 has
 equations/formulas for a
 few common shapes.
M – mass in kilograms
R – radius in meters
l – length in meters
Units for MOI are
      kg-m 2
For an object to be in
      complete
     equilibrium,
   requires zero net
  force and zero net
        torque.
If the net force is zero
     the object is in
      translational
    equilibrium. (1  st

      Condition of
      Equilibrium)
  If the net torque is
   zero, than it is in
rotational equilibrium.
  (2 nd Condition for

      Equilibrium)
8-3 Rotational
  Dynamics
Newton’s    2 nd
            Law for
     Rotation
         F = ma
We know in a rotational
system, torque is a
function of force, also:
1. MOI replaces the
mass.
2. Acceleration is
replaced with angular
acceleration.
Net torque = MOI x
            Angular acel.

   t net  I
Remember that if the
  net torque is zero,
  that the object can
 still be rotating, just
at a constant velocity.
Ex 4: Simon decides to ride
 the ‘Gravitron’, it has a
 radius of 3 m, if his mass is
 79.4 kg. What is the net
 torque produced when his
 angular acceleration is 7
 rads/sec 2?
G: R = 3 m, M= 79.4 kg,
    = 7 rads/sec 2

U: tnet = ?
E: tnet = I
We are going to assume
 he is a point mass.
   MOI, I =    MR2

S:tnet =MR2

S: tnet = (79.4)(3) 2(7)

S: tnet = 5002.2 N-m
Ex 5: In, 1995, a fully functional
 pencil with a mass of 24 kg and
 a length of 2.74 m was made.
 Suppose this pencil is suspended
 at its midpoint and a force of 1.8
 N is applied perpendicular to its
 end, causing it to rotate. What is
 the angular acceleration of the
 pencil?
G: M = 24kg, F = 1.8 N,
   l = 2.74 m
U:  = ?
E: t = I   = t / I
We need to find the values
 for Torque and MOI.
Since the midpoint is the
 pivot point the lever arm is
 ½ l. Also, the force acts
 perpendicular, so the angle
 is 90 degrees.
t = Fd(sinq)
t = ½Fl (sinq)
t = ½(1.8)(2.74)sin90
t = 2.466 N-m
We use the formula for
 MOI on pg 285 for a thin
 rod rotating around its
 midpoint.
I = 1/12 Ml  2

I = 1/12 (24)(2.74)2

I = 15.02 kg-m  2
S:  =2.466 /15.02
S:  =0.164 rads/s 2
Angular Momentum (L)
The product of a rotating
   object’s moment of
 inertia and the angular
  speed about the same
           axis.
       L = Iw
L = angular momentum
I = Moment of Inertia
w = angular speed
Angular Momentum
      Units

    Kg-m 2/s
Law of Conservation
  of Momentum
 When the net external torque
 acting on an object or objects
is zero, the angular momentum
        does not change.
          Li = L f
Ex 6: A figure skater jumps into the
 air to perform a twisting
 maneuver. When she first jumps
 her moment of inertia is 86 kg-m2.
 While she’s in the air she brings
 her arms in and decreases her
 momentum to 77 kg-m2. If her
 initial angular speed was 2 rads/sec
 what is her final angular speed, if
 momentum is conserved?
G: wi = 2 rads/sec, Ii = 86
      2 , I = 77 kg-m2
 kg-m f
U: wf = ?
Since momentum is
 conserved:
E:            Li = Lf
            Iiwi = Ifwf
     wf = Iiwi / If
S: wf = (86)(2) / (77)
S: wf = 2.23 rads/sec
Ex 7: A 65 kg student is spinning on
 a merry-go-round that has a mass
 of 525 kg and a radius of 2 m. She
 walks from the edge of the merry-
 go-round toward the center. If the
 angular speed is initially 0.2 rads/s,
 what is the angular speed when the
 student reaches a point 0.50 m
 from the center?
G: M = 525 kg, Ri =
 0.2m, m =65 kg, Rf =
 0.5 m,    wi = 0.2
 rads/s
U: wf = ?
E: Momentum is
 conserved
          Li = L f
  Lm,i + Ls,i = Lm,f + Ls,f
For MOI, treat the
merry-go-round as a
 solid disc and the
 student as a point
       mass.
I m  MR
    1
    2
           2
I m  MR
    1
    2
           2
               I s,i  mR   2
Im    1
       2
         MR2
               Is,i  mR   2
                               Is,f  mrf
                                        2
Im    1
       2
         MR2
               Is,i  mR   2
                                Is,f  mrf
                                         2




Lm,i  I mw i         1
                       2
                               MR w i
                                    2
Im    1
       2
         MR2
               Is,i  mR   2
                                Is,f  mrf
                                         2




Lm ,i  I mw i        1
                       2
                               MR w i
                                    2


Lm ,i     1
           2
               (525)( 2) (0.2) 2
Im    1
       2
         MR2
               Is,i  mR   2
                                Is,f  mrf
                                         2



Lm,i  I mw i         1
                       2
                               MR w i
                                    2


Lm,i      1
           2
               (525)( 2) (0.2) 2


Lm,i  210 k g  m / s          2
Ls ,i  I sw i  mR w i
                   2
Ls ,i  I sw i  mR w i
                   2


Ls ,i  (65)( 2) (0.2)
               2
Ls ,i  I sw i  mR w i2


Ls ,i  (65)(2) (0.2)
               2


Ls ,i  52k g  m / s
                   2
Li  Lm,i  Ls ,i
Li  Lm ,i  Ls ,i
Li  210  52
Li  Lm,i  Ls ,i
Li  210  52
Li  262 kg  m / s 2
L f  Lm , f  Ls , f
L f  Lm, f  Ls , f
L f  I mw f  I s , f w f
L f  Lm , f  Ls , f
L f  I mw f  I s , f w f
Lf        MR w f  mr f w f
       1      2              2
       2
L f  Lm , f  Ls , f
L f  I mw f  I s , f w f
Lf         MR w f  mr f w f
       1       2                 2
       2

Lf    1
        2
            MR  mr f w f
                2        2
                             
Lf    
       1
       2
         (525)(2) 2
                       (65)(0.5) w f
                                2
                                    
Lf    
       1
       2
         (525)(2) 2
                    (65)(0.5) w
                              2
                                    f

Lf    (1066 .25kg  m )w
                      2
                          f
Li  L f
Li  L f
                (                 )
262 kg  m / s  1066 .25kg  m w f
           2                  2
Li  L f
                (                 )
262 kg  m / s  1066 .25kg  m w f
           2                  2


      262 kg  m / s
                 2
wf 
     1066 .25kg  m 2
Li  L f
                 (                )
262 kg  m / s  1066 .25kg  m w f
           2                  2


       262 kg  m / s
                 2
wf 
      1066 .25kg  m 2


w f  0.25rads / sec
  Rotational Kinetic
   Energy (KERot)

Energy of an object due
to its rotational motion.
 Calculating Rotational
    Kinetic Energy
      KErot = ½ Iw2

KErot = rotational KE
I = moment of inertia
w = angular speed
EX 8: What is the MOI
     for a 0.75 kg top
 spinning at an angular
 speed of 1.75 rads/sec,
 if the Rotational KE is
        12 Joules?
G: M = 0.75 kg, KErot =
 12 J, w = 1.75 rads/sec
U: I = ?
E: KErot = ½ Iw 2

  I = 2 KErot /w 2
S: I = 2(12)/(1.75)   2

S: I = 7.84 kg-m 2
 Remember Mechanical
Energy may be conserved.
ME is the sum of all types of
 KE and PE.
ME = KEtrans + KErot + PE
ME = ½   mv2   +½   Iw 2   + mgh
If ME is conserved:
MEinitial = MEfinal
   If ME is conserved:
   MEinitial = MEfinal
KEtrans ,i  KE rot ,i  PE i
               
KEtrans , f  KE rot , f  PE f
                  Iw i  mghi
1 mv 2       1           2
 2  i            2

                     
      mv f            Iw f  mgh f
1        2       1           2
    2                2

				
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posted:8/31/2011
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