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Mechanics

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22 Nov 2008 07:01:32 IST

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645

Minimum time

None

A point P is located above an incined plane. It is possible to reach the plane by sliding under gravity down a straight frictionless wire, joining P to some point P' on the plane. How should P' be chosen so as to minimize the time taken?

Edit: I'll rephrase the problem. See the diagram attached; determine the location of P' in terms of the angle with the vertical .

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VARUN RAJ

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22 Nov 2008 09:12:52 IST

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what does this mean
the point p is above the inclined pane

Avirup Dasgupta

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22 Nov 2008 21:28:18 IST

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Sir,

Correct me if I'm wrong.

Anant Kumar

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22 Nov 2008 21:42:50 IST

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Yeah that's right metal aka abhiroop :) .... the problem is not as such difficult... but can you do that without calculus?

Dipanjan

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22 Nov 2008 23:46:57 IST

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This is without calculus:

Anant Kumar

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30 Nov 2008 21:24:16 IST

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As I said already, this problem is inherently not difficult. Both the solutions given above by metal and jishnu are correct.

However, there is a particularly pleasing solution as follows. We begin by proving first an auxiliary theorem: In a vertical plane, bodies starting simulatneously from a single point $P$ and moving along frictionless slopes in different directions, all lie in a vertical circle with $P$ as the highest point, at any subsequent time.

See Figure (a). Of all the bodies starting from $P$ , the one which moves vertically must travel the largest distance in a given time $t$ , which will be $PA=d=\dfrac{1}{2}gt^2$ . Taking $PA$ as diameter draw a circle. Consider the motion of any other body moving along a slope inclined at $\beta$ to the vertical. If the instantaneous location after time $t$ is not on this circle, let it be somewhere $X$ . Then, $PX=\dfrac{1}{2}(g\cos\beta)t^2=d\cos\beta$ . But that means that $\angle AXP=90^\circ$ . So $X$ must lie on the circle we have constructed. Thus, the above proposition is true.

Now consider the original problem. As the bead slides down the wire, it will always lie on a circle which has $P$ as its point and is growing. In fact, it is a three dimensional situation. The bead will actually lie on a sphere with $P$ as its highest point. But it is sufficient to consider the cross section on a plane that is passing through the line of gratest slope. At some instant this circle will touch the inclined plane. Obviously, the bead will reach the inclined plane in the least possible time if it slides along the chord joining the point $P$ with the point of tangency (say $P^\prime$ ). Now the corresponding angle $\theta$ can be easily found by geometry and it is easy to see (figure (b)) that $\theta = \dfrac{\alpha}{2}$ .

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