Thread: rotation of a weighted pulley

A light wheel of radius a has a uniform semicircular rim of mass M, and my rotate freely in a vertical plane about a horizontal axis through its center. A light string passes around the wheel and suspends a mass m.

The system is governed by the equation:
(M+m)a^2 (d^2(x)/dt^2)= a*g*(m -(2Msin(x))/(pi))
where x is the angle between the downward vertical and the diameter through the center of mass of the heavy rim.

Find the equilibrium points and the conditions for their existence and stability.
Let k = m/M. how does the wheel behave when k is large?

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I have some thoughts on the problem, but I am not sure if they are right:

i think when k is large, d^2(x)/dt^2 tends to g/a, which means the wheel behaves like a simple pendulum??

and one of the equilibrium points is, of course, (k,x) = (0,0). then (0, pi) and (0,2*pi) are equilibrium points as well.

another observation is when k = 2/pi, sin (x) = 1. i think bifurcation occurs here, but i am not sure what would happen when k>(2/pi), becuase it seems like that means sin(x) >1...

Originally Posted by totoro

A light wheel of radius a has a uniform semicircular rim of mass M, and my rotate freely in a vertical plane about a horizontal axis through its center. A light string passes around the wheel and suspends a mass m.

The system is governed by the equation:
(M+m)a^2 (d^2(x)/dt^2)= a*g*(m -(2Msin(x))/(pi))
where x is the angle between the downward vertical and the diameter through the center of mass of the heavy rim.

Find the equilibrium points and the conditions for their existence and stability.
Let k = m/M. how does the wheel behave when k is large?

-----------------

I have some thoughts on the problem, but I am not sure if they are right:

i think when k is large, d^2(x)/dt^2 tends to g/a, which means the wheel behaves like a simple pendulum??

and one of the equilibrium points is, of course, (k,x) = (0,0). then (0, pi) and (0,2*pi) are equilibrium points as well.

another observation is when k = 2/pi, sin (x) = 1. i think bifurcation occurs here, but i am not sure what would happen when k>(2/pi), becuase it seems like that means sin(x) >1...

At the equilibrium points all the derivatives of x are zero, so:

m = 2Msin(x)/(pi)

or:

sin(x) = (pi/2) m/M.

Now m and M are both positive, and we want x in the range [0, 2 pi), so

x = arcsin((pi/2) m/M), and x = pi - arcsin((pi/2) m/M).

and there are no real solutions unless (pi/2) (m/M)<=1.

Or in terms of k, for equlibrium points to exist k <= 2/pi, and they are:

x = arcsin((pi/2) k), and x = pi - arcsin((pi/2) k).

RonL

thanks!
so given these fixed points, i have yet to determine their stability. with only the equation of the second derivative how can i determine the stability of the fixed points?

also i need to find out where the bifurcation occurs and what type they are. i don't know if there is a point, ie a value of k, such that the two fixed points found would coalesce into 1 (tangent bifurcation)? or is it another type of bifurcation?

Originally Posted by totoro

thanks!
so given these fixed points, i have yet to determine their stability. with only the equation of the second derivative how can i determine the stability of the fixed points?

If x0 is one of the equilibrium points we consider a solution which starts
nearby, say x(t)=x0+epsilon(t) x'(0)=0, where epsilon(t) is "small"

(M+m)a^2 (d^2(x)/dt^2) ~= a*g*(m -(2M [sin(x0)-epsilon(t)cos(x0)]))/(pi))

which may be written:

A d^2epsilon(t)/dt^2 ~= K1 + K2 epsilon(t) .............. ...(1),

where:

K1=a*g*(m - 2Msin(x0))/pi

K2 = a*g*(2M)*cos(x0)/pi

A=(M+m)a^2

Then the equilibrium x0 is stable if the solution of (1) is bounded (by some
multiple of epsilon(0)) for all t>0, for sufficently small epsilon(0)>0.

also i need to find out where the bifurcation occurs and what type they are. i don't know if there is a point, ie a value of k, such that the two fixed points found would coalesce into 1 (tangent bifurcation)? or is it another type of bifurcation?

They coalesce when k=1, as then (pi/2) m/M = pi/2 and the two solutions
for the equilibrium x's are equal.

RonL