Q1

a. To find the equations of motion for the bead as seen by an observer co-rotating with the rod, we can use the relative motion of the bead with respect to the rod. Let x be the distance of the bead from the origin and θ be the angle of the rod with respect to the positive x-axis. Then, the equations of motion are:

m(d^2x/dt^2) = -mgsinθ m(d^2y/dt^2) = mgcosθ

where y is the distance of the bead from the rod. Since the rod is rotating at constant angular speed ω, we have:

θ = ωt

We can use this relation to eliminate θ from the equations of motion and obtain:

d^2x/dt^2 = -gsin(ωt) d^2y/dt^2 = gcos(ωt)

These are the equations of motion for the bead as seen by an observer co-rotating with the rod.

b. To find the time it takes for the bead to reach the end of the rod, we can use the fact that the bead will reach the end of the rod when y = L. From the second equation of motion, we have:

d^2y/dt^2 = g*cos(ωt)

Integrating twice with respect to time, we obtain:

y = (g/ω^2)cos(ωt) + C1t + C2

where C1 and C2 are constants of integration. Since the bead starts from rest at the origin, we have y(0) = 0 and dy/dt(0) = v0. Using these initial conditions, we get:

C1 = 0 C2 = – (g/ω^2)

Substituting these values, we obtain:

y = (g/ω^2)*(1 – cos(ωt))

The bead reaches the end of the rod when y = L. Therefore, we can solve for the time t by setting y = L and solving for t:

L = (g/ω^2)(1 – cos(ωt)) cos(ωt) = 1 – (ω^2L)/g ωt = cos^-1(1 – (ω^2*L)/g) t = (1/ω)cos^-1(1 – (ω^2L)/g)

This is the time it takes for the bead to reach the end of the rod.

c. To find the magnitude of the normal force on the bead at the instant it reaches the end of the rod, we can use the first equation of motion:

m(d^2x/dt^2) = -mg*sinθ

At the end of the rod, x = L and θ = π/2. Therefore, we have:

m(d^2L/dt^2) = -mg

The only force acting on the bead at this instant is the normal force N, which is perpendicular to the surface of the rod. Therefore, we have:

N = mg

This is the magnitude of the normal force on the bead at the instant it reaches the end of the rod.

Q2

a. As seen by an observer co-rotating with the rod, the equations of motion for the bead are:

d^2x/dt^2 = -ω^2x d^2y/dt^2 = -gsin(α)

where x is the coordinate along the rod and y is the vertical coordinate. To derive these equations, we use the fact that the observer is rotating at the same angular speed as the rod, so there is no centrifugal force acting on the bead. The only forces acting on the bead are gravity and the normal force, which are both directed vertically downward and have components along the x and y axes.

b. To find the initial speed uo if the bead just reaches the end of the rod, we can use the fact that the horizontal displacement of the bead when it reaches the end of the rod is L. This occurs when the bead has fallen a vertical distance of h = L*sin(α). Using the equation for the vertical motion of the bead, we can solve for the time t it takes to fall this distance:

h = 1/2gt^2

t = sqrt(2h/gsin(α)) = L/sqrt(2gL*sin(α))

Then, using the equation for the horizontal motion of the bead, we can solve for the initial speed uo that will result in a horizontal displacement of L at time t:

L = v0*t

vo = L/t = sqrt(2gL*sin(α))

So the initial speed required for the bead to just reach the end of the rod is vo = sqrt(2gL*sin(α)).

c. At the instant the bead reaches the end of the rod, it has zero vertical velocity and is moving horizontally with speed v = vo. The only force acting on the bead is the normal force, which must provide the centripetal force needed to keep the bead moving in a circle of radius L.

The magnitude of the centripetal force is given by Fc = mω^2L, where ω is the angular speed of the rod. To find ω, we note that the angular speed of the rod is related to the linear speed of the bead by:

ω = v/L

So, at the instant the bead reaches the end of the rod, we have:

Fc = mv^2/L = m(v/L)^2L = mω^2*L

Solving for the normal force, we get:

Fn = Fc = mv^2/L = m(vo^2)/L = 2mg*sin(α)

So the magnitude of the normal force on the bead at the instant it reaches the end of the rod is 2mg*sin(α).