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# m2 watch

1. A small bead P is threaded onto a circular wire of radius 0.8 m and centre O which is fixed in a vertical plane. The bead is projected from the point vertically below O with speed u ms^-1 and moves in complete circles about O. During the first complete circle the minimum speed of P is observed to be 60% of its maximum speed.

In an intial model of the situation the wire is assumed to be smooth and the bead is modelled as a particle.

(a) Show that u = 7

I did this:

Greatest speed = u
least speed = 0.6u

Initial k.e = 0.5mu^2
least k.e. = 0.5m(0.4u)^2 = 0.5m*0.16u^2 = 0.08mu^2
loss of k.e. = 0.5mu^2 - 0.08mu^2 = 0.5m(0.34u^2)

gain in g.p.e. = mgh = 1.6mg

Equate these due to conservation of energy:

1.6mg = 0.5m(0.34u^2)
u^2 = (1.6g)/(0.34)
u ~ 6.79 m s^-1

Ok, saw an error staright away... sorry, false alarm. !

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Updated: June 25, 2004
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