## Friday, 11 June 2010

### Mechanics IV

Circular motion:
2) T: period: time to complete one revolution, SI unit is s.
3) Angular speed (velocity but we don’t care about the direction): ω=2π/T, with unit rad s-1.
4) Linear velocity v = s/t = rθ/t = rω = 2πr/T, the unit is m s-1.
5) (Centripetal) acceleration: Consider the object moves Δθ in a short interval Δt. Then the limit of Δv is equal to the arc of the circular path. That is, Δv=vΔθ. By a = dv/dt = vdθ/dt = vω. Therefore a = vω = v2/r = rω2. The acceleration is pointing towards the center of the circular path.
6) Centripetal force: F = ma = mvω = mv2/r = mrω2 is also pointing towards the center of the path. Moreover it isn’t a new type of force, but it can be provided through different sources like tension and friction.
Under uniform circular motion (v and ω are constant), when F> mv2/r, then the excess F will accelerate more seriously towards the center until F= mv2/r’ where r’ is the new radius smaller than the original one, therefore it will spiral inwards. Oppositely if F < mv2/r, r will increase so that the object will spiral outwards.
Complex case on circular motion:
1) Conical pendulum: A light and inextensible string of length l, making angle θ with the vertical line. In this case the centripetal force is given by the tension of the string T. Now consider the vertical motion (is zero), by ΣF=ma, we have the vertical component is equal to the weight of the pendulum. i.e., Tcosθ = mg. In the horizontal view of the circular path we have the horizontal component of tension as the centripetal force. i.e., Tsinθ=mv2/r. Combining the two equations will give tanθ=v2/rg. With the assistance of v = rω we can find the remaining information. Note that since F = Tsinθ = mrω2 = mlsinθω2, we will have T = mlω2.
2) In the flight of aircraft is actually a conical pendulum model, while the lifting force gives the centripetal force instead of tension. The inclined angle of the aircraft is also equal to the angle made with the vertical line.
3) Cycling in a level road: when the bicycle turns, its c.g. have to be closer to the center of the circular path. Otherwise the friction produced when the bicycle is turning around is unbalanced. If it is suitably stanted, the moment produced by the couple mg and R is offset by the friction. Let the horizontal and vertical distance between c.g. and the wheel be a and h respectively. Now F = mv2/r, R=mg and Fh=Ra by taking moment on G. Combining the equations together gives tanθ=v2/rg and the resultant forces acting on the wheel is given by R+F.
4) Cycling in a banked road of θ: In this case we assume the bicycle is perpendicular to the road. The centripetal force is given by the resultant of R and mg. by Rsinθ = mv2/r and Rcosθ = mg, we will have tanθ=v2/rg.