A small bead of mass m is free to slide along a long, thin rod without any friction. The rod rotates in a horizontal plane about a vertical axis passing through its end at a constant rate of f revolutions per second. Show that the displacement of the bead as a function of time is given by r(t)=A1(e^bt) +A2(e^–bt) , where r is measured from the axis of rotation. Find the expression for the constant b. Also, how would you determine the constants A1 and A2?

Answer 1

F = m a = m w^2 r
where w = 2 pi f

let u = tangential velocity
let v = radial velocity = w r

dv/dt = w dr/dt

at = du/dt = d/dt(wr) = w dr/dt
ar = d^2r/dt^2 = dv/dt= (w^2)r

so
w dr/dt = w^2 r
dr/dt = w r
or
dr/r = w dt
ln r = w dt + c'
r = e^(wt + e^c') = C e^wt
that is the radius versus time and C is the initial radius at t = 0

theta = theta at t =0 +wt = To + w t
for simplicity call To = 0
so
theta = w t9
or R = C e^wt (A1 cos wt + A2 sin wt)
where A1 and A2 depend on the theta at t = 0