Nonuniform circular motion

1. Sep 17, 2003

Jonathan

How does one analyse nonuniform circular motion of an unbalanced wheel where the instantaneous acceleration of the anomalous point mass (the one that makes it unbalanced) depends on the position in rotation? In this case, the acceleration depends on the cosine of the angle relative to right-hand side of a horizontal line (0 = 3 o'clock, &pi;/2 = 12 o'clock, etc. as usual).

Last edited: Sep 17, 2003
2. Sep 19, 2003

HallsofIvy

Staff Emeritus
I wanted to think about this for a while but as soon as I started actually working on it I noticed something: Since the weight is concentrated at a single point, we can ignore the disk and think of the weight as attached to the center of the wheel by a rod: this is the classic "pendulum problem"!

Drawing a force diagram and, of course, using "F= ma", we get
m r d2&theta;/dt2= -mg sin&theta; where &theta; is 0 when the weight is directly below the center of the wheel and r is the distance from the center of the wheel to the weght.

This is a (very) non-linear equation so there is no general method of solution. If &theta; is small, we can approximate sin&theta; by &theta; and get r d2&theta;/dt2 = - g &theta; or d2&theta;/dt2+ g/r&theta;= 0.

That's a linear homogeneous equation with constant coefficients and its general solution is &theta;(t)= C1 cos([squ](g/r)&theta;)+ C2 sin([squ](g/r)&theta;). In particular, if we hold the wheel so that the weight makes initial angle &Theta; with the vertical and release it, &theta;(t)= &Theta;cos([squ](g/r)&theta;). The weight moves through the vertical and to an equal height on the other side then repeats periodically.

More generally, we can use "quadrature". If we let &omega;= d&theta;/dt, we have d2&theta;/dt2= d&omega;/dt and then, using the chain rule, d&theta;/dt d&omega;/d&theta;= &omega; d&omega;/d&theta;.

The equation becomes &omega;d&omega;/d&theta;= -g/r sin&theta; so &omega;d&omega;= (-g/r) sin&theta;d&theta; and
(1/2)&omega;2= (g/r)cos&theta;+ C.

Theoretically, one could solve for &omega;= d&theta;/dt and then integrate that but it gives an "elliptic integral" which cannot be done in closed form. What we can do is draw the "phase plane diagram". For a number of different values of C, graph &omega; against &theta;. For some values of C you get "circular" graphs (periodic solutions- the wheel swings back and forth). For other values it's not: the wheel just keeps going around in the same direction.

edit: fixed &theta;s and &omega;s

Last edited by a moderator: Sep 19, 2003
3. Sep 19, 2003

Jonathan

Finally a post!

Well, just as I thought, no (simple) solution. I hate it when that happens, and it always does because I don't work with simple problems. Thanks!

4. Sep 27, 2003