Transcript PPT - Tensors for Tots

Rotational Kinematics
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The analogies between translational and rotational motion
Linear
Motion
Rotational
Motion
position
x
angular
position
velocity
v = dx/dt
angular
velocity
acceleration
a = dv/dt
angular
acceleration
mass
m
moment of
inertia
linear
momentum
p=mv
angular
momentum
force
F = ma
torque
work
power
kinetic energy
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work
P = Fv
power
kinetic energy
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Angular Variables
Consider a planar object rotating about an axis perpendicular to its
plane. The position is described as a point on the object by the
coordinates r and Á, where Á is the angle measured with respect to the
axis. When the object moves through an angle Á, the point moves a
distance s along the arc. We define the angle Á in radians as
s=rÁ
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The linear velocity in meters per second of a point as moves around a circle is
called the tangential velocity
We define the angular velocity ! in radians per second as ! = d Á/dt. Thus
v=r!
If a point is accelarating along its path with tangential acceleration ®, then
We define the angular acceleration ® in radians per second,
Thus
It is common to describe rotating objects by their frequency of revolution in
revolutions per second. Since 2 rev is 2¼ rad, then
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Problems
1.) An electric drill rotates at 1600 rev/min. Through what angle does it
turn in 4 ms? If it reaches this speed from rest in 0.32 s, what is its average
angular acceleration?
2.) The moon goes around the earth in about 27.3 days. What is its angular
velocity?
a.) Convert 27.3 days to seconds
b.) Plug in the numbers
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Moment of Inertia
The moment of inertia represents the effort you need to get
something to turn. Consider a continuous object to be composed
of many small pieces of mass dm. We can write:
If the mass is spread throughout the volume with density ½, the
mass in a volume dV is dm = ½ dV. For a surface mass density
¾, dm = ¾ dA. For a line mass density ¸, dm = ¸ dx. In these
cases the moment of inertia can be written:
or
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Derivation of the Moment of Inertia for a uniform
round disk of thickness b and radius R.
So the moment of inertia is:
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Derivation of the Moment of Inertia for a
uniform rod of dimensions of length l and radius
r
axis
x
0
l–h
h
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We have a rod of uniform composition, same density everywhere, and it
has a mass m and length l. We place the axis of rotation at O, a distance
h from one of the ends. Simple enough, now we pick an element of
volume of a short segment of length dx and cross-sectional area A, a
distance x from O. That means:
Since we need to find the total rotational inertia of the entire rod, we
need to integrate from x = -h to x = l - h:
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If the axis of rotation is at the end of the rod, h =
0, which simplifies the equation greatly.
Thin Rod About Axis Through One End
Perpendicular to Length
If the axis of rotation is at the center, h = l/2,
which also simplifies the equation greatly.
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Thin Rod About Axis Through Center Perpendicular
to Length
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Moment of Inertia
I
hoop or cylindrical shell about its axis
solid cylinder or disk
rod about perpendicular axis through end
rectangular plate
about
perpendicular axis through center
solid sphere
spherical shell
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Problems
1.) Calculate the moment of inertia of a solid cylinder of mass M and radius
R about its axis.
Solution
and
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Angular Momentum and Torque
The angular momentum with respect to the origin of a particle with
position r and momentum p = mv is defined as
because ! is perpendicular to r ,
p is µ, then the magnitude of L is
. If the angle between r and
The time rate of change of the of the angular momentum is
The cross-product of velocity and momentum is zero, because these vectors are
parallel. The magnitude of the torque is
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Problems
1.) You push a merry-go-round at its edge, perpendicular to the radius. If the
merry-go-round has a diameter of 4.0 m and you push with a force of 160 N,
what torque are you applying?
Solution
2.) You apply force on a wrench to loosen a pipe. If the wrench is 22 cm long
and you apply 120 N at an angle of
with respect to the wrench, what
torque are you applying?
Solution
3.) Because the force of gravity is directed along the line joining the centers
of the sun and the earth, the force is negligible. The earth revolves in a slightly
elliptical orbit around the sun. It is 1.47 £ 108 km away from and travelling at
a speed of 30.3 km/s when it is nearest to the sun (perihelion). The earth’s farthest
distance from the sun (aphelion) is 1.52 £ 108 km. How fast is the earth moving at
its aphelion?
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Solution
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