If you have two rotating disks with a different angular velocity $\omega_0$ and $\omega_1$, different area and material, how can you describe the angular momentum w.r.t. time which seems to be a function of area, friction coefficient etcetera.

My guess it would be something like: $$ L(t) = \omega_0 I(1-e^{-t/C})$$

How can you derive it? What is exactly $C$?

Also my guess would be that the rate of 'flow of angular momentum' can be described by something like: $$ J_A = k A \frac{\Delta L}{\Delta x}$$

Is there any similarity with newton's law of viscosity? $$ u = \mu \frac{\Delta v}{\Delta y}$$

  • $\begingroup$ The two rotating discs are placed on top of one another? $\endgroup$ Apr 29, 2022 at 12:51

2 Answers 2


enter image description here


$$ \underbrace{I\,\frac{d\omega}{dt}}_{\dot L}=\tau\quad \Rightarrow\\ I\,\int_{\omega_i}^{\omega_f}\,d\omega=I\,(\omega_f-\omega_i)=\int \tau\,dt$$

where $\omega_f~$ is the final angular velocity, $~\omega_i~$ the initial angular velocity , $~\tau~$ is the torque , I is the inertia about the rotation axis and L is the angular momentum

your case

the equations immediately after the collision

$$ I_1\,(\omega_1-\Omega_1)=-\tau_\mu\,t\tag 1$$ $$I_2\,(\omega_2-\Omega_2)=+\tau_\mu\,t\tag 2$$

  • $~\Omega_i~$ initial angular velocity disk i
  • $~\omega_i~$ final angular velocity disk i
  • $~\tau_\mu~$ friction torque between disk one and two

friction torque ?

$$d\tau_\mu=r\,dF\quad\text{and}\\ dF=\mu\,g\,dm=\mu\,g\,\rho\,dV=\mu\,g\,\rho\,z\,dA= \mu\,g\,\rho\,z\,r\,d\phi\,dr\quad\Rightarrow\\ \tau_\mu=c\,\int r\,dF=c\,\int_0 ^{2\pi}d\phi\,\int_0^{R/2}r^2\,dr =c\,\frac{\pi\,R^3}{12}\quad,c=\mu\,g\,\rho\,z\\ $$

  • $~\rho~$ disk desity
  • $~z~$ disk thickness
  • $~\mu~$ friction coefficient
  • $~A~$ disk area
  • $~V~$ disk volume
  • $~R~$ disk radius

from equation (1) and (2) you obtain

$$\omega_1=\Omega_1-\frac{\tau_\mu}{I_1}\,t\quad\Rightarrow L_1=-{\tau_\mu}\\ \omega_2=\Omega_2+\frac{\tau_\mu}{I_2}\,t\quad\Rightarrow L_2={\tau_\mu}$$


$$I_1=\frac{m_1}{2}\,\left(\frac{R}{2}\right)^2\\ I_2=\frac{m_2}{2}\,R^2$$

you can ask "how many seconds after the collision it takes, until the angular velocity of disk one is equal to the angular velocity of disk two "


$$\omega_1=\omega_2\quad \Rightarrow\\ t_f=\frac{1}{\tau_\mu}\frac{I_1\,I_2}{I_1+I_2}|\Omega_1-\Omega_2|$$



  • 1
    $\begingroup$ Thank you for your indepth comment $\endgroup$ May 4, 2022 at 20:03

I think you mean they have a coefficient of friction $\mu$ between their surfaces. If that's the case, for instance let us consider the lower disc, the upper disc applies a torque on it trying to slow it down as there is relative motion between the discs.

Make a dr elemental ring on the disc at a distance r from the center enter image description here

So, the mass of the elemental ring is $dm =\sigma 2\pi r dr$, where $\sigma$ is mass per unit area of the disc. Now the upper disc applies a torque $\mu dm g r$. $$d \tau = \mu \sigma g \cdot 2\pi r^2 dr $$

Integrating this we get a constant torque as the friction in this case is independent of velocity.

$$\tau=-\frac{dL}{dt}=\frac{2\mu \sigma \pi g}{3}R^3$$

So the angular momentum of the lower disc would be:

$$I_{lower}\omega_0-\frac{2\mu \sigma \pi g}{3}R^3t$$

And upper disk would be:

$$\frac{2\mu \sigma \pi g}{3}R^3t$$ The angular momenta of the bodies obey this linear dependence until relative motion ceases and then they remain constant. This is an isolated system so the net torque is 0, implying that the angular momentum of the whole system is conserved. If you wanted the time after which the relative motion ceases, follow this: There would be friction until both of them get equal angular velocities, so the ratio of their angular momenta must be $I_{upper}/I_{lower}$.


$$\frac{I_{lower}\omega_0-\frac{2\mu \sigma \pi g}{3}R^3t_{stop}}{\frac{2\mu \sigma \pi g}{3}R^3t_{stop}}=\frac{I_{lower}}{I_{higher}}$$.

Now solve this for your conditions.

The reason your intuition doesn't hold up is that the friction in your case is independent of any velocity gradient unlike in the case of viscosity. If it were like viscosity you would get an exponential function as expected.

For example spin the same disk but on the surface of water.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.