Download 9.1 Impulse and Momentum Ancient Babylonians described

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CHAPTER 9
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9.1 Impulse and Momentum
Ancient Babylonians described fractions of a circle
in terms of 360 divisions that are now called
degrees. A more modern method is based on the
length of the radius and the distance around the
circumference of the circle.
If the object is spinning faster and faster, or
spinning slower and slower, in a certain period of
time, the object has an angular acceleration, . The
relationship between linear and angular acceleration
is a =  r.
In one spin, a point on the edge of a circle
travels a distance equal to 2  times the radius of
the circle. This defines the modern angular
measurement of 2 radians = 360°. The unit of
radians is added for “cosmetic purposes” to provide
a unit analogous to the degree. The unit of one
radian is equivalent to 360° / 2, or 1 rad = 57.3°.
Although the point’s linear velocity works out
to be its tangential speed, the point’s linear
acceleration is not the centripetal acceleration in
this case, since centripetal acceleration applies only
to uniform circular motion, where the spin is not
increasing or decreasing.
As an object rotates, the change in the angle that
it rotates is the angular displacement, . How far
the point on the edge of the circle moves while the
object spins through the angular displacement is
given by the relationship d =  r.
The time it takes to rotate through that angular
displacement is the angular speed, . Angular speed
is defined by the rate change of angular
displacement, or  = /t. The relationship
between the point’s tangential velocity and angular
speed is vt =  r.
The linear speed of the point works out to be the
tangential velocity, because its velocity vector can
only be tangent to the circle.
v=r
 = 
t
vt =2 r
T
Given.
Given.
Substitution
If  is one complete
spin, 2 rad, then t
is the period T.
EQUATION SUMMATION
Displacement

d = r 
Velocity
 = 
vt = r 
t
Acceleration
 = 
a=r
t
vt =2 r
ac = 42 r
ac = vt2
2
T
T
r
ANALOGOUS RELATIONSIPS
LINEAR
ANGULAR
v f = vi + a t
 f = i + a t
d = ½ (v f + vi) t
 = ½ ( f + i) t
2
d = vi t + ½ a t
 = i t + ½  t2
v f2 = vi2 + 2 a d
 f2 = i2 + 2  



(1)
(2)
(3)
(4)
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9.2 Conservation of Momentum
If linear displacements, speeds, and accelerations
have analogous angular quantities, then it makes
sense that force does as well. Torque is a force that
tends to cause rotation around a point called the
fulcrum.
The amount of applied torque depends on the
amount of force, the distance the force is from the
fulcrum, and the angle in which the force is applied
compared to the distance. There are two ways to
express torque.
  = F r sin and
 = r F sin
The quantity r sin is called the lever arm. It is the
perpendicular distance from the fulcrum to the
force. The quantity F sin is the component of the
force perpendicular to the applied object.
This is a subtle difference in the application
alone. The preferred way to express and think of
torques is  = r F sin, since it is the perpendicular
component of the force that tends to cause the
rotation.
The mathematics is still the same. F is the force,
r is the distance from the fulcrum to the force, and 
is the angle between the two.
The direction of the torque can also be
confusing. The simplest description is to call the
torque positive if it would tend to cause counterclockwise motion around the fulcrum. The torque is
negative if it would tend to cause clockwise motion
around the fulcrum.
The net torque is found by adding all the
torques, clockwise and counterclockwise (positive
and negative). If the object is not spinning, the net
torque will be equal to zero. A more convenient
way to express that concept is to equate the sum of
the clockwise torques to the sum of the
counterclockwise torques
 net = 0
or
cw net = ccw net
If the object is spinning, then the net torque will
be equal to the product of the object’s rotational
inertia and the object’s rotational acceleration.
 net = I 
The object’s rotational inertia, also called the
moment of inertia, indicates how the object
responds to changes in rotational motion. This is the
object’s rotation equivalent to its linear inertia. The
higher the moment of inertia, the more difficult is is
to start the object spinning, to spin it faster or
slower, and to stop the object spinning.
The amount of moment of inertia an object has
depends on the distribution of the mass around the
axis of rotation. If the axis of rotation changes, the
moment of inertia changes. If the mass gets closer
to or farther from the axis of rotation, the moment
of inertia changes. Obviously, if the amount of mass
changes, the moment of inertia changes.
The moment of inertia for different objects
spinning around different axes is listed below. To
use these in the previous equation for net torque,
select the correct object and axis for the situation. If
the object or axis is not listed, the situation cannot
be solved (Well, it can. The correct equation for the
moment of inertia needs to be found, then proceed
as needed.)
Thin ring; axis through center:
I = m r2
Thick ring; axis through center:
I = ½ m (r12 + r22)
Disk; axis through center:
I = ½ m r2
Disk; axis tangent to edge:
I = ¼ m r2
Solid ball; axis through center:
(i.e. it is rotating):
I = ⅖ m r2
Solid ball; axis at a distance, r, from object
(i.e. it is revolving):
I = m r2
Thin rod; length l; axis through center:
I =  m l2
Thin rod; length l; axis at one end:
I = ⅓ m l2
Rectangular sheet; axis through center:
I = ½ m (a2 + b2)
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9.2 continued
Newton’s Law of Inertia describes two states of
equilibrium: static equilibrium (without motion) and
dynamic equilibrium (with motion). If an object is
in dynamic equilibrium, it can either move in a
straight line at a constant speed (Fnet = 0) or it can
spin at a uniform rate (cw net = ccw net).
In addition to two states of equilibrium there
exists two conditions of equilibrium. The first
condition is translational equilibrium, in which the
object is moving in a straight line at a constant
speed (Fnet = 0). The second condition is rotational
equilibrium, in which the object is spinning at a
uniform rate (cw net = ccw net).
An object in static equilibrium must meet both
conditions of equilibrium. An object in motion can
move in a straight line at a constant speed, or it can
it can spin at a uniform rate, or both, or neither.
If an object is spinning while in equilibrium, its
axis of rotation will go through the center of mass.
The center of mass is the point at which all the mass
of the object can be considered to be concentrated
without changing the object’s linear inertia. A net
force applied through the center of mass from any
direction will cause the object to accelerate without
changing the rotational motion.
Some objects in equilibrium are more stable
than others. An object is stable if a net external
force is needed to tip the object. The degree of
stability depends upon the orientation of the object.
The object remains stable if its center of mass
remains either above or below the object’s base.
When the tipping force is removed, the object
returns to its stable position. If the object can be
positioned to lower its center of mass or to have a
wider base, the object will have greater stability.
An object is neutrally stable if it may be
overturned without changing raising or lowering its
center of mass. Like a ball, tire, or tin can rolling
across the floor.
An object is unstable if it may be overturned
easily, usually in response to a gravitational force,
since the center of mass does not remain precisely
over the base, allowing the center of mass to freely
move to a lower position of greater stability.
The center of mass for any object may be found
by hanging the object by a string with one end of
the string acting as a plumb line. The center of mass
will lie along that line. Hanging the object from a
second point on the object will mark the center of
mass at the intersections of the two plumb lines.
The object can be supported from either above or
below its center of mass and not rotate.
A person’s center of mass lies in the intersection
of three planes: a vertical plane bisecting the body
right from left; another vertical plane intersecting
the ankles dividing the body front from back; a third
plane horizontally dividing the body just above the
male’s hip bone or the female’s hip joint.
The path of the projectile is parabolic in shape.
If the projectile is spinning, it will spin around its
center of mass while its center of mass follows the
parabolic path of the projectile.