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The analogies between translational and rotational motion
position
Linear
Motion x
Rotational
Motion velocity acceleration mass linear momentum force work v = dx/dt a = dv/dt m p=mv
F = ma angular position angular velocity angular acceleration moment of inertia angular momentum torque work power kinetic energy
P = Fv power kinetic energy
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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 = r
Á
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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 2
¼ rad, then
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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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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
¾
, dm =
¾ dV. For a surface mass density dA. For a line mass density
¸
, dm =
¸ dx. In these cases the moment of inertia can be written: or or
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So the moment of inertia is:
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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 crosssectional 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.
If the axis of rotation is at the center, h = l/2, which also simplifies the equation greatly.
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Moment of Inertia
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
I
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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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The angular momentum with respect to the origin of a particle with position r and momentum
p
= m
v
is defined as because
!
is perpendicular to r , . If the angle between r and p is
µ
, then the magnitude of L is
The time rate of change of the of the angular momentum is
The crossproduct of velocity and momentum is zero, because these vectors are parallel. The magnitude of the torque is
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1.) You push a merrygoround at its edge, perpendicular to the radius. If the merrygoround 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
£
10 8 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
£
10
8 km. How fast is the earth moving at its aphelion?
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Solution
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