Controls System experiment

Question

We performed an experiment in which we turned a bicycle on its side and placed a piece of green tap on one of the spokes. We spun the wheel and recorded the wheel spinning until it came to a complete stop. A stop watch was in the frame of the video so we could note each time that the wheel completed one revolution, so we could calculate the angular velocity.

For example, the green tape made on complete revolution in .6 seconds, and 2 revolutions in 1.31 seconds, 3 revolutions in 2.19 seconds, etc.

So my question is, how do I go about graphing this to determine if the wheel is undamped, under damped, or over damped. I am having a hard time thinking of a way to plot the data to replicate some sort of oscillation.

I may be wrong, but I calculated the arc that the wheel traveled as a function of time, synonyms to a position vs time graph, in order to view the response of my systems but clearly that doesn’t produce an oscillating graph. As time goes on, the arc distance traveled (s) is always getting larger until the wheel stops.

Examples I see, shows the displacement vs time graph of a spinning wheel but I don’t understand how can the displacement initially goes up and then come down, and then rise again and produce an isolating motion. That is easy for me to imagine for a spring mass, because the motion is literally swaying back and forth but in the case of a wheel, the distance is always getting larger.

Answer 1

Foreword/disclaimer: This is my derivation (as I mentioned I've never seen this procedure written ), so I would appreciate any constructive feedback.

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I am assuming that the viscous "damping" coefficient for this system produces a torque which is proportional to the angular velocity.

$$M(\omega(t) ) = c_t \omega(t) \tag{eq.1}$$

(this is an analogy to the viscous damping $F=c\cdot \dot{x}$.)

At any given point the rotational motion of the wheel is given by:

$$\sum M = I\cdot \alpha \tag{eq.2}$$

where:

• $\sum M = - M_t$ is the sum to torsional moments on the wheel (I am using the minus because the torque decelerates the wheel).
• $I$ is mass moment of inertia of the wheel
• $\alpha = \frac{d\omega}{dt}$ is the angular acceleration of the wheel.

From eq.1, eq.2 we obtain that:

$$-c_t \omega(t) = I\cdot \frac{d\omega}{dt} $$ $$c_t dt = -I\cdot \frac{d\omega}{\omega(t) } $$

by integrating both parts we obtain:

$$\int_{t_0}^{t_1} c_t dt = -\int_{\omega_0}^{\omega_1} I\cdot \frac{d\omega}{\omega(t) } $$

$$c_t (t_1 -t_0) = -I\left[\ln \omega\right]_{\omega_0}^{\omega_1} $$ $$c_t (t_1 -t_0) = -I\ln \left(\frac{\omega_1}{\omega_0}\right)$$

Now, provided you know the mass moment of area A, you can make an estimation of the $c_t$.

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numerical example

revs	elapsed time	lap period	$ω$	$-\ln\frac{ω_ι}{ω_0}$	$c_t$
1	0.6	0.6	10.47	0	NA
2	1.31	0.71	4.79	0.780	0.39
3	2.19	0.88	2.87	1.294	0.43
...	...	...	...	...	...

assuming you have a few more numbers you can get a values for $c_t$.

Notes:

• This assumes that $c_t$ is constant, so you should get the mean value of all the values in the $c_t$ column. (in a more advanced analysis you could assume that $c_t$ is a function of $\omega$)
• the above values are assuming $I= 1 kg.m^2$. The actual mass moment of inertia is required to obtain proper units and values.
• ... (probably something else I am forgetting).