So this appears in a problem which looks simple enough in its context; It's something like this:

Two discs, A and B, are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2I$ respectively about the common axis. Disc A is imparted an initial angular velocity $2\omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc B is imparted an angular velocity $\omega$ by the same spring compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.

When disc B was brought in contact with A, they acquire a common angular velocity in time $t$. The average frictional torque by the other during this period is: ?

The answer ($2I\omega/3t$) was obtained by applying angular momentum conservation and is exactly what confused me. How can we apply angular momentum conservation when friction is present?


The angular momentum of each disk individually is not conserved, however the total angular momentum of both disks is conserved because there are no external torques acting.

Start by calculating the total angular momentum of both disks (I'm going to replace "w" by "v" since "w" is confusingly close to "$\omega$"):

$$\begin{align} L_{total} &= I_a \omega_a + I_b \omega_b \\ &= I.2v + 2I.v \\ &= 4Iv \end{align}$$

Now we bring the disks into contact and they settle down to a constant speed $\omega_{final}$. The total angular momentum is now:

$$\begin{align} L_{total} &= I_a \omega_{final} + I_b \omega_{final} \\ &= 3I\omega_{final} \end{align}$$

Because angular momentum is conserved we just equate our two expressions fo $L_{total}$:

$$ 4Iv = 3I\omega_{final} $$


$$ \omega_{final} = \frac{4}{3}v $$

Now, torque $\times$ time is the angular impulse, and we know that the impulse is equal to the change of momentum. So if we calculate the change in momentum of disk A this is equal to the torque times the time i.e. $Tt$, and we know the initial angular momentum of disk A is $2Iv$ so:

$$\begin{align} Tt &= I_a 2v - I_a \omega_{final} \\ &= 2Iv - \frac{4}{3}Iv \\ &= \frac{2}{3}Iv \end{align}$$

And dividing both sides by $t$ gives the answer:

$$ T = \frac{2}{3} \frac{Iv}{t} $$

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  • $\begingroup$ Even though friction is an internal force here, rotational energy is lost in the form of heat. So, the final rotational energy must be lesser than the initial. Here, initial and final rotational energies are $3I\omega ^2$ and $\frac{8}{3}I\omega ^2$ respectively. Clearly the final energy is less than the initial. But how is rotational energy not conserved but the angular momentum is conserved? Is this analogous to inelastic collisions where linear momentum is conserved but not the kinetic energies? $\endgroup$ – Guru Vishnu Nov 20 '19 at 16:26

How can we apply angular momentum conservation when friction is present?

Why not? If we have a closed system, momentum and angular momentum are conserved. In this case, the full system is disk A and disk B, and there are no external forces, so the system is closed. There are internal forces, namely in this case, friction, but that doesn't matter.

You might be mixing this up with conservation of mechanical energy, which is not conserved if there is friction. (but total energy still is - if you include heat loss etc.)

(Note that this implies that the exact form of the frictional force is irrelevant.)

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