Rotational Equilibrium and Rotational Dynamics

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Chapter 8 Rotational Equilibrium and Rotational
Dynamics
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• Rotational kinetic energy
• We consider a system of particles participating
  in rotational motion
• Kinetic energy of this system is
• Then

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• Moment of inertia
• From the previous slide
• Defining moment of inertia (rotational inertia)
  as
• We obtain for rotational kinetic energy

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• Moment of inertia rigid body
• There is a major difference between moment of
  inertia and mass the moment of inertia depends
  on the quantity of matter, its distribution in
  the rigid object and also depends upon the
  location of the axis of rotation
• For a rigid body the sum is calculated over all
  the elements of the volume
• This sum can be calculated for different shapes
  and density distributions
• For a constant density and the rotation axis
  going through the center of mass the rotational
  inertia for 9 common body shapes is given in
  Table 8-1

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Moment of inertia rigid body
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• Moment of inertia rigid body
• The rotational inertia of a rigid body depends
  on the position and orientation of the axis of
  rotation relative to the body

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• Moment of inertia rigid body
• Example moment of inertia of a uniform ring
• The hoop is divided into a number of small
  segments, m1 , which are equidistant from the
  axis

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Chapter 8 Problem 29
Four objects are held in position at the corners
of a rectangle by light rods. Find the moment of
inertia of the system about (a) the x-axis, (b)
the y-axis, and (c) an axis through O and
perpendicular to the page.
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• Torque
• The door is free to rotate about an axis through
  O
• Three factors that determine the effectiveness
  of the force in opening the door
• 1) The magnitude of the force
• 2) The position of the application of the force
• 3) The angle at which the force is applied

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• Torque
• We apply a force at point P to a rigid body that
  is free to rotate about an axis passing through O
• Only the tangential component Ft F sin f of
  the force will be able to cause rotation

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• Torque
• The ability to rotate will also depend on how
  far from the rotation axis the force is applied
• Torque (turning action of a force)
• SI unit Nm (dont confuse with J)

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• Torque
• Torque
• Moment arm (lever arm) r- r sinf
• Torque can be redefined as
• force times moment arm
• t F r-

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• Torque
• Torque is the tendency of a force to rotate an
  object about some axis
• Torque is a vector
• The direction is perpendicular to the plane
  determined by the position vector and the force
• If the turning tendency of the force is CCW
  (CW), the torque will be positive (negative)

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• Torque
• When two or more torques are acting on an
  object, they are added as vectors
• The net torque is the sum of all the torques
  produced by all the forces
• If the net torque is zero, the objects rate of
  rotation doesnt change
• Forces cause accelerations, whereas torques
  cause angular accelerations

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• Newtons Second Law for rotation
• Consider a particle rotating under the influence
  of a force
• For tangential components
• Similar derivation for rigid body

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Newtons Second Law for rotation
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Chapter 8 Problem 35
A 150-kg merry-go-round in the shape of a
uniform, solid, horizontal disk of radius 1.50 m
is set in motion by wrapping a rope about the rim
of the disk and pulling on the rope. What
constant force must be exerted on the rope to
bring the merry-go-round from rest to an angular
speed of 0.500 rev/s in 2.00 s?
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• Center of mass
• 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 a single point
• 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 mass of the
  object

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• Center of mass
• In a certain reference frame we consider a
  system of particles, each of which can be
  described by a mass and a position vector
• For this system we can define a center of mass

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• Center of mass of two particles
• A system consists of two particles on the x axis
• Then the center of mass is

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• Center of mass of a rigid body
• For a system of individual particles we have
• For a rigid body (continuous assembly of matter)
  the sum is calculated over all the elements of
  the volume

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Chapter 8 Problem 8
A water molecule consists of an oxygen atom with
two hydrogen atoms bound to it. The bonds are
0.100 nm in length, and the angle between the two
bonds is 106. Use the coordinate axes shown, and
determine the location of the center of gravity
of the molecule. Take the mass of an oxygen atom
to be 16 times the mass of a hydrogen atom.
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• Center of gravity
• Gravitational force on a body effectively acts
  on a single point, called the center of gravity
• If g is the same for all elements of a body
  (which is not always so see for example Chapter
  7) then the center of gravity of the body
  coincides with its center of mass

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• Angular momentum
• Angular momentum
• SI unit kgm2/s
• Recall

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Chapter 8 Problem 45
A light rigid rod 1.00 m in length rotates about
an axis perpendicular to its length and through
its center. Two particles of masses 4.00 kg and
3.00 kg are connected to the ends of the rod.
What is the angular momentum of the system if the
speed of each particle is 5.00 m/s? (Neglect the
rods mass.)
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• Conservation of angular momentum
• Newtons Second Law for rotational motion
• If the net torque acting on a system is zero,
  then
• If no net external torque acts on a system of
  particles, the total angular momentum of the
  system is conserved (constant)

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Conservation of angular momentum
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Conservation of angular momentum
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• Equilibrium
• Equilibrium
• Static equilibrium
• Stable equilibrium the body returns to the
  state of static equilibrium after having been
  displaced from that state
• Unstable equilibrium the state of equilibrium
  is lost after a small force displaces the body

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• Center of mass stable equilibrium
• We consider the torque created by the gravity
  force (applied to the CM) and its direction
  relative to the possible point(s) of rotation

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• Center of mass stable equilibrium
• We consider the torque created by the gravity
  force (applied to the CM) and its direction
  relative to the possible point(s) of rotation

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• Center of mass stable equilibrium
• We consider the torque created by the gravity
  force (applied to the CM) and its direction
  relative to the possible point(s) of rotation

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• Center of mass stable equilibrium
• We consider the torque created by the gravity
  force (applied to the CM) and its direction
  relative to the possible point(s) of rotation

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• The requirements of equilibrium
• For an object to be in equilibrium, we should
  have two requirements met
• Balance of forces the vector sum of all the
  external forces that act on the body is zero
• Balance of torques the vector sum of all the
  external torques that act on the body, measured
  about any possible point, is zero

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• Equilibrium 2D case
• If an object can move only in 2D (xy plane) then
  the equilibrium requirements are simplified
• Balance of forces only the x- and y-components
  are considered
• Balance of torques only the z-component is
  considered (the only one perpendicular to the xy
  plane)

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Examples of static equilibrium
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Examples of static equilibrium
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Examples of static equilibrium
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Examples of static equilibrium
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Chapter 8 Problem 28
One end of a uniform 4.00-m-Iong rod of weight Fg
is supported by a cable. The other end rests
against the wall, where it is held by friction.
The coefficient of static friction between the
wall and the rod is µs 0.500. Determine the
minimum distance x from point A at which an
additional object, also with the same weight Fg
can be hung without causing the rod to slip at
point A.
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• Indeterminate structures
• Indeterminate systems cannot be solved by a
  simple application of the equilibrium conditions
• In reality, physical objects are
• not absolutely rigid bodies
• Concept of elasticity is employed

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• Total energy of a system
• Conservation of mechanical energy (no
  non-conservative forces present)
• In the case where there are non-conservative
  forces such as friction, we use the generalized
  work-energy theorem instead of conservation of
  energy

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Questions?
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Answers to the even-numbered problems Chapter 8
Problem 20 (b) T 343 N, H 171 N, V 683 N
(c) 5.14 m
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Answers to the even-numbered problems Chapter 8
Problem 40 10.9 rads
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Answers to the even-numbered problems Chapter 8
Problem 54 12.3 m/s2