A uniform circular disc with mass $m_1$ and radius $R_1$ is fixed at its midpoint and rotates at angular velocity $w_1$. Another uniform circular disc with mass $m_2$ and radius $r_2$ is carefully lowered without rotating down to disc 1. Disc 2 is prevented from moving in horizontal direction via a friction free groove.
Initially, the two discs will slide against each other, but after a while the friction between the discs will ensure that they roll without sliding. Calculate the angular velocity of disc 1 as the discs roll without sliding.
Let $w_1(t) $ and $w_2(t)$ be the angular velocity of body 1 and body 2 respectively at time $t$. Let $t=0$ be the time when body 2 starts making contact with body 1 and let $t=a$ be the time when the discs stop sliding.
Let $I_{G1}$ and $I_{G2}$ be the moment of inertia for body 1 and body 2 respectively around their center of mass, $G_1$ and $G_2$.
We are trying to find $w_1(a)$ as a function of the known parameters and constants.
At $t=0$, it's easy to realize that the total energy for (body 1 + body 2) is pure rotational energy, because their center of mass is not moving and there are no springs attached to this system. But this is true for any time $t$, so we let $T(t)$ be the total rotational energy for (body 1 + body 2) at a given time $t$: $$T(t) = \frac{1}{2}I_{G1}w_1^2(t) + \frac{1}{2}I_{G2}w_2^2(t) $$
At $t=0$, body 2 is not rotating so we have $$ T(0) = \frac{1}{2}I_{G1}w_1^2(0) $$
At $t=a$, body 1 and body 2 are both spinning, so we have $$ T(a) = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}I_{G2}w_2^2(a) $$
Now, using the work-energy-theorem we get $$ (1):= U= T(a)-T(0) = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}I_{G2}w_2^2(a) - \frac{1}{2}I_{G1}w_1^2(0) $$
where $U$ is the work done by the forces acting on the system (body 1 + body 2) between the time $t=0$ and $t=a$. The only forces acting on the system is the reaction forces at $G_1$ and $G_2$, but these points are not moving so the work done is zero: $$ U = 0$$
$U=0$ in eq. $(1)$ gives us the equation $$(1)':= 0 = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}I_{G2}w_2^2(a) - \frac{1}{2}I_{G1}w_1^2(0) $$
which is one equation with two unknowns, $w_1(a)$ and $w_2(a)$. But we know the relationship between these two, since at time $t=a$ when the discs stop sliding their tangential velocity must be the same: $$w_1(a)R_1 = w_2(a)R_2$$
solving for $w_2(a)$ we get $$w_2(a)= \frac{w_1(a)R_1}{R_2}$$
and putting this into eq. $(1)'$: $$0 = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}\frac{I_{G2}w_1^2(a)R_1^2}{R_2^2} - \frac{1}{2}I_{G1}w_1^2(0) $$
and we solve for $w_1^2(a)$: $$w_1^2(a)= \frac{I_{G1}}{I_{G1}+\frac{I_{G2}R_1^2}{R_2^2}}w_1^2(0)$$
The moment of inertia for the bodies are known as $I_{G1}= \frac{m_1R_1^2}{2}$ and $I_{G2}= \frac{m_2R_2^2}{2}$ respectively, so the final answer can be written as $$(2):= w_1^2(a)= \frac{m_1}{m_1+m_2}w_1^2(0)$$
The correct solution to this problem, however, is $$(3):= w_1^2(a)= \frac{m_1^2}{(m_1+m_2)^2}w_1^2(0)$$
Comparing $(2)$ and $(3)$, we see that $(3) < (2)$ due to the logic that $x^2 < x$ when $x < 1$. Here is $$x= \frac{m_1}{m_1+m_2} = 1 - \frac{m_2}{m_1+m_2} < 1$$ so $(3) < (2)$ must be true. That is, the system should have less rotational energy than my solution implies.
This implies that either
- There is work being done from forces acting on the system (body 1 + body 2) which I have missed.
- The total energy is not solely made of rotational energy.
- Calculation error.
Of these three, 2 and 3 is highly unlikely. This means that 1 is true, work is being done on the system. Is this true, and what forces are doing work on the system?
We only care about the work done from forces acting on the system, not forces inside the system like frictional forces and normal forces between the disks, since these forces inside the system doesn't change the total energy of the system.
