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Chapter 8
Rotational Equilibrium
and
Rotational Dynamics
Torque
 Torque, , is the tendency of a force to
rotate an object about some axis

 is the torque
– symbol is the Greek tau
 F is the force
 d is the lever arm (or moment arm)
Fd


Direction of Torque
 Torque is a vector quantity
 The direction is perpendicular to the plane
determined by the lever arm and the force
 For two dimensional problems, into or out
of the plane of the paper will be sufficient
 If the turning tendency of the force is
counterclockwise, the torque will be
positive
 If the turning tendency is clockwise, the
torque will be negative
Units of Torque
 SI
 Newton x meter = Nm
 US Customary
 foot x pound = ft lb
Lever Arm
 The lever arm, d, is
the perpendicular
distance from the
axis of rotation to a
line drawn from the
axis of rotation to a
line drawn along the
the direction of the
force
 d = L sin Φ
An Alternative Look at Torque
 The force could also
be resolved into its
x- and y-
components
 The x-component,
F cos Φ, produces 0
torque
 The y-component,
F sin Φ, produces a
non-zero torque
Torque, final
 From the components of the force or
from the lever arm,
 F is the force
 L is the distance along the object
 Φ is the angle between the force and the
object
 sinFL
Net Torque
 The net torque is the sum of all the
torques produced by all the forces
 Remember to account for the direction of
the tendency for rotation
 Counterclockwise torques are positive
 Clockwise torques are negative
Torque and Equilibrium
 First Condition of Equilibrium
 The net external force must be zero
 This is a necessary, but not sufficient, condition to
ensure that an object is in complete mechanical
equilibrium
 This is a statement of translational equilibrium
0Fand0F
0F
yx 

Torque and Equilibrium, cont
 To ensure mechanical equilibrium, you
need to ensure rotational equilibrium as
well as translational
 The Second Condition of Equilibrium
states
 The net external torque must be zero
0
Mechanical Equilibrium
 In this case, the First
Condition of Equilibrium
is satisfied
 The Second Condition is
not satisfied
 Both forces would
produce clockwise
rotations
N500N5000F 
0Nm500 
Axis of Rotation
 If the object is in equilibrium, it does not
matter where you put the axis of rotation for
calculating the net torque
 The location of the axis of rotation is completely
arbitrary
 Often the nature of the problem will suggest a
convenient location for the axis
 When solving a problem, you must specify an axis
of rotation
 Once you have chosen an axis, you must maintain that
choice consistently throughout the problem
Center of Gravity
 The force of gravity acting on an object
must be considered
 In finding the torque produced by the
force of gravity, all of the weight of the
object can be considered to be
concentrated at one point
Calculating the Center of
Gravity
 The object is divided
up into a large
number of very
small particles of
weight (mg)
 Each particle will
have a set of
coordinates
indicating its
location (x,y)
Calculating the Center of
Gravity, cont.
 The torque
produced by each
particle about the
axis of rotation is
equal to its weight
times its lever arm
Calculating the Center of
Gravity, cont.
 We wish to locate the point of
application of the single force , whose
magnitude is equal to the weight of the
object, and whose effect on the rotation
is the same as all the individual
particles.
 This point is called the center of gravity
of the object
Coordinates of the Center of
Gravity
 The coordinates of the center of gravity
can be found from the sum of the
torques acting on the individual
particles being set equal to the torque
produced by the weight of the object
i
ii
cg
i
ii
cg
m
ym
yand
m
xm
x






Center of Gravity of a Uniform
Object
 The center of gravity of a homogenous,
symmetric body must lie on the axis of
symmetry.
 Often, the center of gravity of such an
object is the geometric center of the
object.
Experimentally Determining
the Center of Gravity
 The wrench is hung
freely from two
different pivots
 The intersection of the
lines indicates the
center of gravity
 A rigid object can be
balanced by a single
force equal in
magnitude to its weight
as long as the force is
acting upward through
the object’s center of
gravity
Notes About Equilibrium
 A zero net torque does not mean the
absence of rotational motion
 An object that rotates at uniform angular
velocity can be under the influence of a
zero net torque
 This is analogous to the translational situation
where a zero net force does not mean the
object is not in motion
Solving Equilibrium Problems
 Draw a diagram of the system
 Isolate the object being analyzed and
draw a free body diagram showing all
the external forces acting on the object
 For systems containing more than one
object, draw a separate free body diagram
for each object
Problem Solving, cont.
 Establish convenient coordinate axes for
each object. Apply the First Condition
of Equilibrium
 Choose a convenient rotational axis for
calculating the net torque on the object.
Apply the Second Condition of
Equilibrium
 Solve the resulting simultaneous
equations for all of the unknowns
Example of a
Free Body Diagram
 Isolate the object to
be analyzed
 Draw the free body
diagram for that
object
 Include all the
external forces
acting on the object
Example of a
Free Body Diagram
 The free body
diagram includes the
directions of the
forces
 The weights act
through the centers
of gravity of their
objects
Fig 8.12, p.228
Slide 17
Torque and Angular
Acceleration
 When a rigid object is subject to a net
torque (≠0), it undergoes an angular
acceleration
 The angular acceleration is directly
proportional to the net torque
 The relationship is analogous to ∑F = ma
 Newton’s Second Law
Moment of Inertia
 The angular acceleration is inversely
proportional to the analogy of the mass
in a rotating system
 This mass analog is called the moment
of inertia, I, of the object
 SI units are kg m2
2
mrI 
Newton’s Second Law for a
Rotating Object
 The angular acceleration is directly
proportional to the net torque
 The angular acceleration is inversely
proportional to the moment of inertia of
the object
 I
More About Moment of Inertia
 There is a major difference between
moment of inertia and mass: the
moment of inertia depends on the
quantity of matter and its distribution in
the rigid object.
 The moment of inertia also depends
upon the location of the axis of rotation
Moment of Inertia of a
Uniform Ring
 Image the hoop is
divided into a
number of small
segments, m1 …
 These segments are
equidistant from the
axis
22
ii MRrmI 
Other Moments of Inertia
Rotational Kinetic Energy
 An object rotating about some axis with
an angular speed, ω, has rotational
kinetic energy ½Iω2
 Energy concepts can be useful for
simplifying the analysis of rotational
motion
Total Energy of a System
 Conservation of Mechanical Energy
 Remember, this is for conservative forces,
no dissipative forces such as friction can be
present
fgrtigrt )PEKEKE()PEKEKE( 
Angular Momentum
 Similarly to the relationship between force
and momentum in a linear system, we can
show the relationship between torque and
angular momentum
 Angular momentum is defined as
 L = I ω
 and
t
L



Angular Momentum, cont
 If the net torque is zero, the angular
momentum remains constant
 Conservation of Linear Momentum
states: The angular momentum of a
system is conserved when the net
external torque acting on the systems is
zero.
 That is, when ffiifi IIorLL,0 
Problem Solving Hints
 The same basic techniques that were
used in linear motion can be applied to
rotational motion.
 Analogies: F becomes , m becomes I and
a becomes , v becomes ω and x
becomes θ


More Problem Solving Hints
 Techniques for conservation of energy
are the same as for linear systems, as
long as you include the rotational
kinetic energy
 Problems involving angular momentum
are essentially the same technique as
those with linear momentum
 The moment of inertia may change,
leading to a change in angular momentum

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4773390

  • 2. Torque  Torque, , is the tendency of a force to rotate an object about some axis   is the torque – symbol is the Greek tau  F is the force  d is the lever arm (or moment arm) Fd  
  • 3. Direction of Torque  Torque is a vector quantity  The direction is perpendicular to the plane determined by the lever arm and the force  For two dimensional problems, into or out of the plane of the paper will be sufficient  If the turning tendency of the force is counterclockwise, the torque will be positive  If the turning tendency is clockwise, the torque will be negative
  • 4. Units of Torque  SI  Newton x meter = Nm  US Customary  foot x pound = ft lb
  • 5. Lever Arm  The lever arm, d, is the perpendicular distance from the axis of rotation to a line drawn from the axis of rotation to a line drawn along the the direction of the force  d = L sin Φ
  • 6. An Alternative Look at Torque  The force could also be resolved into its x- and y- components  The x-component, F cos Φ, produces 0 torque  The y-component, F sin Φ, produces a non-zero torque
  • 7. Torque, final  From the components of the force or from the lever arm,  F is the force  L is the distance along the object  Φ is the angle between the force and the object  sinFL
  • 8. Net Torque  The net torque is the sum of all the torques produced by all the forces  Remember to account for the direction of the tendency for rotation  Counterclockwise torques are positive  Clockwise torques are negative
  • 9. Torque and Equilibrium  First Condition of Equilibrium  The net external force must be zero  This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium  This is a statement of translational equilibrium 0Fand0F 0F yx  
  • 10. Torque and Equilibrium, cont  To ensure mechanical equilibrium, you need to ensure rotational equilibrium as well as translational  The Second Condition of Equilibrium states  The net external torque must be zero 0
  • 11. Mechanical Equilibrium  In this case, the First Condition of Equilibrium is satisfied  The Second Condition is not satisfied  Both forces would produce clockwise rotations N500N5000F  0Nm500 
  • 12. Axis of Rotation  If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque  The location of the axis of rotation is completely arbitrary  Often the nature of the problem will suggest a convenient location for the axis  When solving a problem, you must specify an axis of rotation  Once you have chosen an axis, you must maintain that choice consistently throughout the problem
  • 13. Center of Gravity  The force of gravity acting on an object must be considered  In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point
  • 14. Calculating the Center of Gravity  The object is divided up into a large number of very small particles of weight (mg)  Each particle will have a set of coordinates indicating its location (x,y)
  • 15. Calculating the Center of Gravity, cont.  The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm
  • 16. Calculating the Center of Gravity, cont.  We wish to locate the point of application of the single force , whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles.  This point is called the center of gravity of the object
  • 17. Coordinates of the Center of Gravity  The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object i ii cg i ii cg m ym yand m xm x      
  • 18. Center of Gravity of a Uniform Object  The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry.  Often, the center of gravity of such an object is the geometric center of the object.
  • 19. Experimentally Determining the Center of Gravity  The wrench is hung freely from two different pivots  The intersection of the lines indicates the center of gravity  A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object’s center of gravity
  • 20. Notes About Equilibrium  A zero net torque does not mean the absence of rotational motion  An object that rotates at uniform angular velocity can be under the influence of a zero net torque  This is analogous to the translational situation where a zero net force does not mean the object is not in motion
  • 21. Solving Equilibrium Problems  Draw a diagram of the system  Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object  For systems containing more than one object, draw a separate free body diagram for each object
  • 22. Problem Solving, cont.  Establish convenient coordinate axes for each object. Apply the First Condition of Equilibrium  Choose a convenient rotational axis for calculating the net torque on the object. Apply the Second Condition of Equilibrium  Solve the resulting simultaneous equations for all of the unknowns
  • 23. Example of a Free Body Diagram  Isolate the object to be analyzed  Draw the free body diagram for that object  Include all the external forces acting on the object
  • 24. Example of a Free Body Diagram  The free body diagram includes the directions of the forces  The weights act through the centers of gravity of their objects Fig 8.12, p.228 Slide 17
  • 25. Torque and Angular Acceleration  When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration  The angular acceleration is directly proportional to the net torque  The relationship is analogous to ∑F = ma  Newton’s Second Law
  • 26. Moment of Inertia  The angular acceleration is inversely proportional to the analogy of the mass in a rotating system  This mass analog is called the moment of inertia, I, of the object  SI units are kg m2 2 mrI 
  • 27. Newton’s Second Law for a Rotating Object  The angular acceleration is directly proportional to the net torque  The angular acceleration is inversely proportional to the moment of inertia of the object  I
  • 28. More About Moment of Inertia  There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object.  The moment of inertia also depends upon the location of the axis of rotation
  • 29. Moment of Inertia of a Uniform Ring  Image the hoop is divided into a number of small segments, m1 …  These segments are equidistant from the axis 22 ii MRrmI 
  • 30. Other Moments of Inertia
  • 31. Rotational Kinetic Energy  An object rotating about some axis with an angular speed, ω, has rotational kinetic energy ½Iω2  Energy concepts can be useful for simplifying the analysis of rotational motion
  • 32. Total Energy of a System  Conservation of Mechanical Energy  Remember, this is for conservative forces, no dissipative forces such as friction can be present fgrtigrt )PEKEKE()PEKEKE( 
  • 33. Angular Momentum  Similarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum  Angular momentum is defined as  L = I ω  and t L   
  • 34. Angular Momentum, cont  If the net torque is zero, the angular momentum remains constant  Conservation of Linear Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero.  That is, when ffiifi IIorLL,0 
  • 35. Problem Solving Hints  The same basic techniques that were used in linear motion can be applied to rotational motion.  Analogies: F becomes , m becomes I and a becomes , v becomes ω and x becomes θ  
  • 36. More Problem Solving Hints  Techniques for conservation of energy are the same as for linear systems, as long as you include the rotational kinetic energy  Problems involving angular momentum are essentially the same technique as those with linear momentum  The moment of inertia may change, leading to a change in angular momentum