Transcript Chapter 8

Chapter 8
Rotational Equilibrium and Rotational
Dynamics
Rotational kinetic energy
• We consider a system of particles participating in
rotational motion
• Kinetic energy of this system is
2
i i
mv
K 
2
i
• Then
mv
mi (i ri )
K 

2
2
i
i
2
i i
2


2
2
 m (r )
i
i
i
2
Moment of inertia
• From the previous slide
K

2
2
 m (r )
i
i
i
• Defining moment of inertia (rotational inertia) as
I   mi (ri )
2
i
• We obtain for rotational kinetic energy
I
K
2
2
2
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
Moment of inertia: rigid body
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
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
I   mi (ri )   mi R
2
i
R
2
i
m  R M
2
i
i
2
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.
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
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 φ of the
force will be able to cause rotation
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):
  ( Ft )(r )  ( F sin  )(r )
• SI unit: N*m (don’t confuse with J)
Torque
• Torque:
  ( Ft )(r )  ( F sin  )(r )  ( F )( r sin  )
• Moment arm (lever arm): r┴=
• Torque can be redefined as:
force times moment arm
τ = F r┴
r sinφ
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)
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 object’s rate of rotation
doesn’t change
• Forces cause accelerations, whereas torques cause
angular accelerations
Newton’s Second Law for rotation
• Consider a particle rotating under the influence of a
force
• For tangential components
  Ft r  mat r  m(r )r  (mr 2 )  I
  I
• Similar derivation for rigid body
Newton’s Second Law for rotation
  I
   i
i
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?
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
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:

rCM 

 mi ri
i
m
i
i


 mi ri
i
M
Center of mass of two particles
• A system consists of two particles on the x axis
• Then the center of mass is

rCM 

 mi ri
i
m
i


 mi ri
i
M
i
xCM
m1 x1  m2 x2

m1  m2
yCM
m1  0  m2  0

m1  m2
0
Center of mass of a rigid body
• For a system of individual particles we have

rCM 

 mi ri
i
m
i
i
• For a rigid body (continuous assembly of matter)
the sum is calculated over all the elements of the
volume
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.
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
Angular momentum
 f  i

 net  I  I lim
 I lim
t 0 t
t 0
t
I f  Ii
L
I 
 lim
 lim
 lim
t 0
t 0 t
t 0
t
t
• Angular momentum: L  I
L
• SI unit: kg*m2/s
 net  lim


t 0 t
• Recall: p  mv


p
Fnet  lim
t  0  t
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 rod’s mass.)
Conservation of angular momentum
• Newton’s Second Law for rotational motion
 net
L
 lim
t 0 t
• If the net torque acting on a system is zero, then
L
lim
0
t  0 t
L  const
• If no net external torque acts on a system of
particles, the total angular momentum of the system
is conserved (constant)
Conservation of angular momentum
L  I  const
I ii  I f  f
Conservation of angular momentum

L  const
Equilibrium
• Equilibrium:


P  const ; L  const
• Static equilibrium:


P  0; L  0
• 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
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
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
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
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
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

P  const ;

Fnet  0
• Balance of torques: the vector sum of all the
external torques that act on the body, measured
about any possible point, is zero
L  const ;
 net  0
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
Fnet , x  0; Fnet , y  0
• Balance of torques: only the z-component is
considered (the only one perpendicular to the xy
plane)
 net , z  0
Examples of static equilibrium
Examples of static equilibrium
Examples of static equilibrium
Examples of static equilibrium
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.
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
Total energy of a system
• Conservation of mechanical energy (no nonconservative forces present)
Kt  Kr  U i  Kt  Kr  U  f
• In the case where there are non-conservative forces
such as friction, we use the generalized work-energy
theorem instead of conservation of energy:
W  Kt  K r  U
Questions?
Answers to the even-numbered problems
Chapter 8
Problem 20
(b) T = 343 N, H = 171 N, V = 683 N ;
(c) 5.14 m
Answers to the even-numbered problems
Chapter 8
Problem 40
10.9 rads
Answers to the even-numbered problems
Chapter 8
Problem 54
12.3 m/s2