# Chapter 8 Rotational Equilibrium and Dynamics

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```					  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
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
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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