pallidin said:
Not to be silly, but consider this:
I'm sitting on one of those "wheeled" office chairs, which itself is on slick ice.
My feet are not touching the ground.
I extend one arm with a can of propellant. Nozzle oriented to effect my rotation.
I release the propellant. Do I just spin in place?
No. There is also a lateral component, though slight.
What happens is that I spin, but also will have moved off the starting point.
The continuing lateral component will cause the total system to subscribe a circle while spinning in my chair.
If you keep your arm fixed (relative to the chair), then you exert a constant torque, but and a constant force, but with changing direction. Neglecting air drag, your equations of motion are:
[tex]
m \ddot{x} = F \, \cos{\theta}[/tex]
[tex]
m \ddot{y} = -F \, \sin{\theta}[/tex]
[tex]
I \ddot{\theta} = \tau = l \, F[/tex]
With the initial conditions:
[tex]
x(0) = y(0) = 0[/tex]
[tex]
\dot{x}(0) = \dot{y}(0) = 0[/tex]
[tex]
\theta(0) = 0, \; \dot{\theta}(0) = 0[/tex]
The equation of theta has a simple solution:
[tex]
\theta(t) = \frac{\tau}{2 \, I} \, t^{2}[/tex]
and the equations for
x and
y are given by the integrals:
[tex]
\dot{x}(t) = \frac{F}{m} \int_{0}^{t}{\cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dp}[/tex]
[tex]
x(t) = \int_{0}^{t}{\dot{x}(q) \, dq} = \frac{F}{m} \int_{0}^{t}{\int_{0}^{q}{\cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dp} \, dq}[/tex]
[tex]
x(t) = \frac{F}{m} \int_{0}^{t}{\int_{p}^{t}{\cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dq} \, dp}[/tex]
[tex]
x(t) = \frac{F}{m} \int_{0}^{t}{(t - p) \, \cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dq} \, dp}[/tex]
Similarly, for y we get:
[tex]
y(t) = -\frac{F}{m} \int_{0}^{t}{(t - p) \, \sin{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dq} \, dp}[/tex]
One part of these integrals is expressible in terms of elementary functions by substitution:
[tex]
-\frac{F}{m} \, \int_{0}^{t}{p \, \cos{\left(\frac{\tau \, p^{2}}{2 I}\right)} \, dp} = -\frac{I F}{m \tau} \, \int_{0}^{\frac{\tau t^{2}}{2 I}}{\cos{x} \, dx} = -\frac{I}{m \, l} \, \sin{\left(\frac{\tau \, t^{2}}{2 I}\right)}[/tex]
[tex]
\frac{F}{m} \, \int_{0}^{t}{p \, \sin{\left(\frac{\tau \, p^{2}}{2 I}\right)} \, dp} = \frac{I F}{m \tau} \, \int_{0}^{\frac{\tau t^{2}}{2 I}}{\sin{x} \, dx} = \frac{I}{m \, l} \, \left[1 - \cos{\left(\frac{\tau \, t^{2}}{2 I}\right)}\right][/tex]
The other parts of the integrals are expressible in terms of the http://en.wikipedia.org/wiki/Fresnel_integral" :
[tex]
\frac{F t}{m} \int_{0}^{t}{\cos{\left(\frac{\tau p^{2}}{2 I} \right)} \, dp} = \frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \int_{0}^{t \sqrt{\frac{\tau}{2 I}}}{\cos{x^{2}} \, dx} = \frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, C(t \sqrt{\frac{\tau}{2 I}})[/tex]
[tex]
-\frac{F t}{m} \int_{0}^{t}{\sin{\left(\frac{\tau p^{2}}{2 I} \right)} \, dp} = -\frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \int_{0}^{t \sqrt{\frac{\tau}{2 I}}}{\sin{x^{2}} \, dx} = -\frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, S(t \sqrt{\frac{\tau}{2 I}})[/tex]
Finally, we may write:
[tex]
x(t) = \frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, C(t \sqrt{\frac{\tau}{2 I}}) - \frac{F I}{m \, \tau} \, \sin{\left(\frac{\tau \, t^{2}}{2 I}\right)}[/tex]
[tex]
y(t) = -\frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, S(t \sqrt{\frac{\tau}{2 I}}) + \frac{F I}{m \, \tau} \, \left[1 - \cos{\left(\frac{\tau \, t^{2}}{2 I}\right)}\right][/tex]
Choosing a system of units in which [itex]\tau/2 I = 1[/itex] and [itex]F/m = 1[/itex], the equations of motion become:
[tex]
\theta(t) = t^{2}[/tex]
[tex]
x(t) = t \, C(t) - \frac{1}{2} \, \sin{(t^{2})}[/tex]
[tex]
y(t) = -t \, S(t) + \frac{1}{2} \, \left[1 - \cos(t^{2})\right][/tex]
The plot of the trajectory of the center of mass for 5 revolutions is depicted in the following figure (using Mathematica):
Of course, eventually drag will start having a dominant role and becomes an important factor in the equation.
). But not fixed in range of motion.