Irodov 1.278: Sticking Together Always Burns Energy – Amit & Deepali Sinha Foundation

Problem Irodov 1.278

Two horizontal discs turn freely, without friction at their bearings, about one and the same vertical axis through both their centers. Their moments of inertia about that axis are $I_1$ and $I_2$ , and they are spinning with angular velocities $\vect\omega_1$ and $\vect\omega_2$ , both directed along the axis. The upper disc, of moment of inertia $I_1$ , then drops onto the lower one. Their faces rub, and after a while the two turn together as one body. Find:

the common angular velocity $\vect\omega$ once the sliding has stopped
the work $A$ done by the friction forces between the two faces over the whole episode

The two discs on their common vertical axis before the drop, the upper one of moment of inertia I_1 turning at ω_1 and the lower one of I_2 at ω_2. — The two discs on their common vertical axis before the drop, the upper one of moment of inertia $I_1$ turning at $\vect\omega_1$ and the lower one of $I_2$ at $\vect\omega_2$ .

The lower disc seen from above while the faces are still sliding, with the friction couple N_fr that the upper face exerts on it. The upper disc feels exactly the opposite couple. — The lower disc seen from above while the faces are still sliding, with the friction couple $\vect N_{\mathrm{fr}}$ that the upper face exerts on it. The upper disc feels exactly the opposite couple.

Given

$I_1$
$I_2$
$\omega_1$
$\omega_2$

Find

(a)

\omega

, (b)

A

Solution

Step 1The axis feels no external moment, so

L_z

carries across

The friction between the two faces acts on both discs at once, with equal and opposite couples about the axis, so it cancels out of the total. Everything else that touches the discs (their weights, the normal load between them, the frictionless bearings) either acts along the axis or is applied at it. The angular momentum of the pair about the axis is therefore the same before the drop and after the sliding has ceased, and each disc contributes its own $I\omega$ .

I_1\vect\omega_1 + I_2\vect\omega_2 = (I_1+I_2)\,\vect\omega \tag{1}

Dividing through by the combined moment of inertia gives the steady state.

\vect\omega = \dfrac{I_1\vect\omega_1+I_2\vect\omega_2}{I_1+I_2}

(2)

The pair once the sliding has died away, resting one on the other and turning as a single body at the common ω. The sense drawn is the case I_1ω_1+I_2ω_2>0. — The pair once the sliding has died away, resting one on the other and turning as a single body at the common $\vect\omega$ . The sense drawn is the case $I_1\omega_1+I_2\omega_2>0$ .

Step 2The energy that the sliding faces removed

All three angular velocities point along the one axis, so from here on $\omega_1$ , $\omega_2$ and $\omega$ stand for their signed components along it. The friction couples act on the two discs through different angles, since the faces are sliding over one another, so their two contributions do not cancel in the work account the way they canceled in the moment account. All of the work of the friction couples goes into the change of rotational kinetic energy, because nothing else exerts a moment about the axis; the fall of the upper disc and the impact that arrests it do their work on the vertical motion, which is uncoupled from the spin and enters neither $K_{\mathrm{i}}$ nor $K_{\mathrm{f}}$ . First the two rotational energies, the initial one a sum over separate discs and the final one a single body turning at $\omega$ .

\begin{align*}K_{\mathrm{i}} &= \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 \\ K_{\mathrm{f}} &= \frac{1}{2}(I_1+I_2)\,\omega^2\end{align*}

Substituting the $\omega$ of (1) into $K_{\mathrm{f}}$ and subtracting gives the work over a common denominator.

A = \dfrac{(I_1\omega_1+I_2\omega_2)^2 - (I_1+I_2)(I_1\omega_1^2+I_2\omega_2^2)}{2(I_1+I_2)}

(3)

Step 3The numerator collapses to a perfect square

Expanding both products in the numerator of (3) is the whole of the remaining work; the squared terms cancel in pairs and what is left is a cross term.

\begin{align*}(I_1\omega_1+I_2\omega_2)^2 &= I_1^2\omega_1^2 + 2I_1I_2\omega_1\omega_2 + I_2^2\omega_2^2 \\ (I_1+I_2)(I_1\omega_1^2+I_2\omega_2^2) &= I_1^2\omega_1^2 + I_1I_2\omega_2^2 + I_1I_2\omega_1^2 + I_2^2\omega_2^2\end{align*}

Subtracting the second line from the first leaves $-I_1I_2$ multiplying $\omega_1^2 - 2\omega_1\omega_2 + \omega_2^2$ , which is a perfect square.

(I_1\omega_1+I_2\omega_2)^2 - (I_1+I_2)(I_1\omega_1^2+I_2\omega_2^2) = -I_1I_2(\omega_1-\omega_2)^2

A = -\dfrac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2

(4)

Check

The work is negative for every choice of the data except $\omega_1=\omega_2$ , where it vanishes, and that is the case in which the faces never slide at all and friction has nothing to rub against. Let the lower disc be enormously massive, $I_2\gg I_1$ : then (a) gives $\omega\to\omega_2$ , the heavy disc is unaffected, and (b) gives $A\to-\frac{1}{2}I_1(\omega_1-\omega_2)^2$ , exactly the kinetic energy the light disc had in the frame turning with the heavy one, all of it destroyed. Dimensionally $I\omega^2$ is an energy and the combination $I_1I_2/(I_1+I_2)$ carries the units of a single $I$ .

Remark

Write $I_{\mathrm{red}} = I_1I_2/(I_1+I_2)$ and the loss is $-\frac{1}{2}I_{\mathrm{red}}(\Delta\omega)^2$ , the rotational twin of the energy lost in a perfectly inelastic head-on collision, $-\frac{1}{2}\left[m_1m_2/(m_1+m_2)\right](\Delta v)^2$ . The correspondence is exact term by term, mass going to moment of inertia and velocity to angular velocity, because both problems are the same two-line argument: one conserved additive quantity, one final common rate. The loss depends only on the relative angular velocity, never on how rough the faces were or how long they took to lock, since roughness fixes the duration of the slipping but not its outcome.