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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- Thread Starter
- 25-06-2004 15:26