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I would like to put this into a differential equation. This is what I have.

$$r \times F = I \ddot\theta + \mu \dot\theta + k \theta$$

What I need verified:

  • $\text{Torque} = I\ddot\theta + \mu \dot\theta + k \theta$
    • $I$ = moment of inertia (constant)
    • $\mu$ = kinetic friction (constant)
    • $k$ = spring constant ( in my application it will be 0 ) (constant)
    • $\theta$ = rotation (function)
  • Torque = $r F$
    • $r$ = radius
    • $F$ = force
  • It's safe to set the two equations for Torque equal to describe the system.
  • That a solution to the differential for $\theta(t)$ will describe the rotation with respect to time of the circle cylinder.

Any insight would be much appreciated, thanks.

  • 2
    $\begingroup$ Patrick, we have the MathJax engine running on the site, which can render more attractive math from a LaTeX-alike syntax. $\endgroup$ Jun 3, 2013 at 1:58
  • $\begingroup$ Note friction is usually applied at a radius, so you need $\mu r \dot{\theta}$. $\endgroup$ Jun 3, 2013 at 5:12

2 Answers 2


Is this the situation?

If yes then the friction force has to be proportional to the applied force $F$ and not $\omega$. With a perfectly relaxed rope, there is no contact and hence friction. What you include as a friction term is actually a rotational damper, which resists $\omega$. So you have a spring damper system with

$$ r\, F - k \theta - \mu \dot{\theta} = I \ddot{\theta} $$



In other words, you're asking whether it is correct to say that $\tau = I \alpha + \mu \omega + k \theta$ (where $\alpha = \theta ''$ and $\omega = \theta'$) in the specific case of us having a rigit object with torsion, kinetic friction and inertia. However, your equation frictional force is wrong: it should be $\tau_{friction} = \mu_k F_N r$ , where $\mu_k$ is the coefficient of kinetic friction, $F_N$ is the normal force. Yes, your general idea is true because $\tau_{effective} = \Sigma \tau$. Just as a side example, consider the case in which we have $n$ torques,

$$\Sigma_{i=1}^n \tau_i = \tau_1+\tau_2+...+\tau_n = (r_1 \times F_1) + (r_2 \times F_2) + ... + (r_N \times F_n)= \Sigma (r_i \times F_i) $$ In the case that $r= r_1 = r_2=...=r_n$, then our equation above simplifies to $\Sigma \tau = \Sigma (r \times F)$. Similarly, in your case we have 3 torques, $\tau_1$,$\tau_2$ and $\tau_3$ cause by three different things. However, the correct equation for your case is $\tau = I \theta'' + \mu_k F_N r + k \theta$. Note that the torque due to friction does not depend on $\theta$, $\theta'$ or $\theta''$.


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