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Three calculus problems in one

Imagine a rod of length \(l\) standing vertically against a wall. The rod is kicked off the wall slightly to begin moving. Assuming that the rod is always in contact with the wall, find: a)the trajectories of all points on the rod. b)if the rod, as it falls, leaves behind dark marks, the area those marks cover. c)the equation of the boundary of this area.

It'd be nice if someone could derive the equations of motion for a non-constrained rod, but i know it's a difficult problem.However, i can answer the first three. Have fun!

Note by Kyriakos Grammatikos
4 years, 9 months ago

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9 votes

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Comment deleted Feb 06, 2013

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If the center of the rod is not fixed at (0, L/2), that equation does not hold.

Ramon Vicente Marquez - 4 years, 9 months ago

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it is assumed that the rod is pushed to the right. the center of the rod moves downward. let L be the length of the rod. let (0,Ycm) be the position of the center of the rod relative to the initial position of the bottom of the rod. let (0, r) be the initial position of a point on the rod relative to the initial position of the center of the rod. let (x, y) be the position of a point on the rod relative to the initial position of the bottom of the rod. a) y = Ycm (2r + L) / L, x = r sqrt (1 - 4 (Ycm / L)^2 ). b and c) still solving.

Ramon Vicente Marquez - 4 years, 9 months ago

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That's very correct.The parametrization of the points seems to be the only solution. Another similar solution: let \((x_0,0), (0,y_0)\) be the points the rod intersects the x and y axis respectively. The equation of the line representing the rod is \(yx_{0}+xy_{0}=x_{0}y_{0}\), with \(x_{0}^{2}+y_{0}^{2}=l^2\). Suppose we want to parametrize the points of the rod as following: \(x=ax_{0}, 0\leq{a}\leq{1}\).From the above equation we have that \(y=(1-a)y_{0}\). Since only \(l\) is given, we will eliminate \(x_{0}\) and \(y_{0}\) to obtain an equation for the trajectories in terms of a, which is unique for each point. Finally, \(x_{0}=\frac{x}{a}\) \(,y_{0}=\frac{y}{a}\)

which implies that \(\frac{x^2}{(al)^2}+\frac{y^2}{[l(1-a)]^2}=1\) which is in general an ellipse. for a=1/2, X=Xcm, Y=Ycm and it's trajectory is a circle. For a=0 or 1 the trajectory is a straight line as requested.

Kyriakos Grammatikos - 4 years, 9 months ago

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