Lecture notes for Lecture #4, October14th, 2010
Today we talked more about mechanical energy systems and spent much time looking at
how certain quantities, like speed and displacement, were invariant to a coordinate
transformation. This is just fancy language that means the following:
these quantities
have the same values regardless of the reference frame in which we choose to measure
them.
Now recall that speed is directly related to kinetic energy. We can watch an object
get tossed into the air, and we know that the speed of this object will change (decrease) as
it reaches its maximum height above the ground. We can choose an arbitrary point in the
object’s path as it rises and falls, and measure its speed and therefore its kinetic energy.
The value that we measure for the object
’
s speed and/or kinetic energy is independent of
the coordinate system in which we choose to make this measurement.
Kinetic Energy, Speed and Coordinate System Dependence
In the diagram above, we are going to show that the speed of the object, say 1 meters
before it reaches its maximum height above the ground is the same whether we measure
this speed in a coordinate system fixed to the ground or a coordinate system that is fixed
somewhere above the ground.
Let’s see if we can prove this.
Coordinate system #1
–
here we will explicitly use “little y” for our y
coordinate.
We know that if the system is closed we can write the following
∆
KE +
∆
PE = 0
½ mv
f
2
½ mv
i
2
+mgy
f
–
mgy
i
=0
Let us just examine the interval defined by the object moving from its initial speed of v
i
to it final speed of v
f
where the final speed that we are interested in is the speed of the
object when it is 1 meter below its maximum height in coordinate system #1. So here are
our following values for v and y:
y
i
= y
i
, y
f
= y
i
+ 1meter, v
f
is what we are looking for, and v
i
is just the initial speed.
Then:
½ mv
f
2
½ mv
i
2
=mg (y
i
+ 1meter)
–
mgy
i
= mgy
i
+
mg (1)
–
mgy
i
Y
i+2 meters
Y
i+1 meter
Y
i
y
i+2 meters
y
i+1 meter
y
i
Coordinate
system
#1(fixed at the ground)
Coordinate
System
#2(fixed somewhere above the ground)
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View Full DocumentCanceling the masses, multiplying both sides by 2, and noting that all the y dependence
vanishes we get:
v
f
2
=
v
i
2
+
g(2)
Now let’s do the same analysis for coordinate system 2, where we will explicitly use “big
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 Fall '08
 PARDINI
 Physics, Energy, Kinetic Energy, Potential Energy, Yi, Coordinate system, coordinate

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