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?

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

To solve this problem, we can consider the forces acting on the bead as it moves along the rotating rod.

Let's assume the displacement of the bead from the axis of rotation at time t is r(t). Since there is no friction, the only force acting on the bead is the centripetal force, which is provided by the gravitational force.

The gravitational force acting on the bead can be resolved into two components: one along the rod (tangential) and the other perpendicular to the rod (radial). The tangential component doesn't cause any displacement since it acts parallel to the motion. Therefore, we only need to consider the radial component.

The radial component of gravity provides the centripetal force, which is given by the equation:

m * (omega)^2 * r(t) = m * g * sin(theta)

Here, m is the mass of the bead, (omega) is the angular velocity of the rotation (2πf), g is the acceleration due to gravity, and theta is the angle between the rod and the vertical direction.

Since the bead is always at a fixed distance from the axis of rotation, we can assume that the angle theta remains constant. Therefore, we can define a constant k = m * g * sin(theta).

The equation then becomes:

(omega)^2 * r(t) = k

Simplifying this equation, we get:

r(t) = k / (omega)^2

Substituting the value of omega (2πf) and simplifying further, we have:

r(t) = k / (4π^2f^2)

This equation gives the displacement of the bead as a function of time without any arbitrary constants.

To find the expression for the constant b in the displacement equation r(t) = A1(e^bt) + A2(e^–bt), we can rewrite the expression in terms of (omega).

r(t) = A1(e^(bt * (2πf))) + A2(e^(–bt * (2πf)))

Comparing this equation with the previous equation (r(t) = k / (4π^2f^2)), we can see that the angular frequency (b) is related to the frequency (f) by the equation:

b = 2πf

Now, to determine the constants A1 and A2, we need to use initial conditions. We need two known displacement values at different times.

Let's say r(t1) = x1 and r(t2) = x2. Substituting these values in the displacement equation, we have:

x1 = A1(e^(bt1)) + A2(e^(–bt1))
x2 = A1(e^(bt2)) + A2(e^(–bt2))

We have two equations and two unknowns (A1 and A2). We can solve these simultaneous equations to determine the values of A1 and A2. This can be done using various mathematical techniques such as substitution, elimination, or matrix methods.

Once we have the values of A1 and A2, we can substitute them back into the displacement equation r(t) = A1(e^bt) + A2(e^–bt) to obtain the final expression for the displacement of the bead as a function of time.