Question

I just need from part 2 and on

A rod of total length L 5m is positioned upside down (that is vertically). Its lower end is at L 0 coordinate and its higher end is at y JL 5m coordinate. The mass of the rod is not uniformly distributed and it has position dependent mass density given by the formula por 2y2 1 ml where y is the coordinate of a rod point essentially the distance of the point from its lower end (where it was assigned the y 30 coordinate). Clearly the rod is thinner at the bottom but it becomes thicker and denser as we move upward. We want to calculate the gravitational potential energy of the rod. The problem with this is that the only formula we have learned in class is how to calculate the potential energy of point like masses and not of extended objects or objects who have height dependent mass density. The formula for the potential energy of a point like mass is mgy Where m is the mass of the point like object and y is essentially its "height" (or y coordinate) Critical Thinking and Solution Methodology: We can observe though, that if the rod is sufficiently small we can assume that the mass density does not appreciably vary across its length, it is more or less constant and it is more or less equally to an average value (the mass density in the middle of the bar). We can therefore approximate the bar as a point like particle located at the center of the bar with total mass equal to the "average" mass density times its length. We can then use the formula U mgy that we learned in class to approximate the gravitational potential energy of the rod Some may argue that this approximation is a very crude one and the error is very large. The length of the bar is not short enough to get an accurate result, so we can do better by dividing the bar into two equal parts and assume that the mass density at each segment does not appreciably vary across its length, and therefore can be assumed constant equal to the mass density at the center of each segment. Therefore each of the two segments can be replaced by a point-like mass located at the center of the piece. The mass of the segment will be equal to the approximate (average) mass density of the segment

Answer 1

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