Frictionless Bead Sliding Down A Parabola

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SUMMARY

The discussion focuses on the dynamics of a bead sliding down the parabola defined by the equation y=(1/2)(x-1)[SUP]2 under the influence of gravity. The bead starts from the point (0, 1/2) and the objective is to express its position as a function of time, r = [x(t), y(t)]. Key equations include the relationship between the angle of descent (A), tangential velocity (v), and tangential acceleration (a), with a defined as a = -g sin A. The discussion also raises questions about the conservation of quantities during the bead's motion.

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devon cook
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Homework Statement


It seems simple, doesn't it? A bead starts from (0,1/2) and simply slides down the parabola y=(1/2)(x-1)2 under gravity. The problem is to get its position rel. to origin in terms of time, ie.
r = [x(t),y(t)]. Anyone into this?


Homework Equations


y' = x-1 = tan A where A is angle tangential veloc. (v) (and tang. accel.(a)) makes with x axis.
a = -g sin A .


The Attempt at a Solution


For starters, can I say that sin A = -(x-1)/sqr(1+(x-1)^2) ?
 
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What quantities do you think are conserved? Can you use them?
 

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