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

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 $c_t$.

numerical example

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

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 $c_t$.

numerical example

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).