Conservation of angular momentum and rotational kinetic energy

[Saptarshi Sarkar]

Homework Statement

A thin uniform disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disk of same dimensions but mass M/4 is placed on the first disk coaxially. What is the angular speed of the system is now

Relevant Equations

$K.E. = \frac 12I\omega^2$
$L= I\omega$

I first tried to get the solution by conserving the rotational kinetic energy and got $\omega'=\frac2{\sqrt5} \omega$.

But, it was not the correct answer. Next I tried by conserving the angular momentum and got $\omega'=\frac 45 \omega$, which is the correct answer.

Why is the rotational kinetic energy not conserved here? Where is the energy being used up?

 

[anuttarasammyak]

Energy used up calculated as
\frac{1}{10}I\omega^2
is used up when two disks was getting the same angular speed in most cases by friction between the surfaces.

 

[haruspex]

The moral here is never to assume work is conserved without a good reason to do so.
It is generally clearer whether linear or angular momentum are conserved. If those laws suffice to find the answer, trust them.
The present problem is an example of coalescence. As in a totally inelastic collision, two bodies meet and arrive at the same velocity through some interaction. As @anuttarasammyak observes, that interaction could be by friction. Equally, it could have been two toothed wheels, the teeth engaging on contact.