lecturenotes_oct14

lecturenotes_oct14 - Lecture notes for Lecture #4,...

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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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Canceling 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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lecturenotes_oct14 - Lecture notes for Lecture #4,...

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