Solved thermodynamics exercise

Translation, rotation and thermal agitation

Exercise 14 · Lesson 4 — The First Law of Thermodynamics

König's theorem
rotation
kinetic energy
internal energy
friction

Statement

A system of $N$ particles of masses $m_{i}$ has total mass $M$ and a centre of mass with position $R$ and velocity $V$ . Let $ρ_{i} = r_{i} - R$ , and decompose the velocities as

v_{i} = V + ω \times ρ_{i} + u_{i}^{'} .

The overall angular velocity $ω$ is chosen so that the residual motion carries no overall angular momentum:

i \sum m_{i} u_{i}^{'} = 0, i \sum m_{i} ρ_{i} \times u_{i}^{'} = 0 .

We take as given the identity $\sum_{i} m_{i} ∣ ω \times ρ_{i} ∣^{2} = ω \cdot I_{CM} ω$ , where $I_{CM}$ is the inertia tensor about the centre of mass.

Expand $E_{kin} = \frac{1}{2} \sum_{i} m_{i} v_{i}^{2}$ and show that the three cross terms vanish. Deduce the decomposition into translation, overall rotation and residual motion.
In the convention where overall translation and rotation are counted as macroscopic motions, which kinetic energy is included in the internal energy? Is it enough to work in the centre-of-mass frame to retain only thermal agitation?
A uniform disc of mass $M$ and radius $R$ , whose axis is fixed, initially rotates with angular velocity $ω_{0}$ . Its moment of inertia is $I = M R^{2} /2$ . A brake brings it to rest. The disc-and-brake system is thermally insulated; its stationary supports do no work. All the dissipated energy remains in this system, whose total heat capacity $C$ is constant. Derive its change in internal energy and its rise in temperature.
Calculate the energy converted and the temperature rise for $M = 2, 0 kg$ , $R = 0, 10 m$ , $ω_{0} = 100 rad s^{- 1}$ and $C = 500 J K^{- 1}$ . Has this energy crossed the boundary of the system as heat?

Hint

Use

\sum_{i} m_{i} ρ_{i} = 0

and the scalar triple product identity

(ω \times ρ_{i}) \cdot u_{i}^{'} = ω \cdot (ρ_{i} \times u_{i}^{'})

. During braking,

U + E_{kin}^{macro}

is conserved.

Detailed solution

Solution

Question 1. The definition of the centre of mass gives

\sum_{i} m_{i} r_{i} = M R

. Consequently,

i \sum m_{i} ρ_{i} = i \sum m_{i} r_{i} - M R = 0 .

Expanding each square gives

E_{kin} = \frac{1}{2} M V^{2} + \frac{1}{2} i \sum m_{i} ∣ ω \times ρ_{i} ∣^{2} + \frac{1}{2} i \sum m_{i} u_{i}^{'}^{2} + V \cdot (ω \times i \sum m_{i} ρ_{i}) + V \cdot i \sum m_{i} u_{i}^{'} + i \sum m_{i} (ω \times ρ_{i}) \cdot u_{i}^{'} .

The first cross term vanishes by the definition of the centre of mass. The second vanishes because the residual momentum is zero. For the third, the scalar triple product gives

i \sum m_{i} (ω \times ρ_{i}) \cdot u_{i}^{'} = ω \cdot i \sum m_{i} (ρ_{i} \times u_{i}^{'}) = 0.

With the identity given for the inertia tensor, we are left with

E_{kin} = \frac{1}{2} M V^{2} + \frac{1}{2} ω \cdot I_{CM} ω + \frac{1}{2} i \sum m_{i} u_{i}^{'}^{2}

Question 2. In this convention, the kinetic contribution to the internal energy is the last term, $\frac{1}{2} \sum_{i} m_{i} u_{i}^{'}^{2}$ . The internal energy also includes the internal potential energies and any other internal degrees of freedom. Working in the centre-of-mass frame removes the translation term but leaves the overall rotation term. A disc rotating about a fixed axis thus has macroscopic kinetic energy even though its centre of mass is at rest.

Question 3. The heat and work crossing the boundary of the disc-and-brake system are zero. The forces exerted by the supports can transmit angular momentum, but since their points of application are stationary, they transmit no work. Neglecting any other change in macroscopic energy, conservation gives

Δ U + Δ E_{rotation} = 0.

The disc comes to rest, so $Δ E_{rotation} = 0 - \frac{1}{2} I ω_{0}^{2}$ . Consequently,

Δ U = \frac{1}{2} I ω_{0}^{2} = \frac{1}{4} M R^{2} ω_{0}^{2} > 0.

Once internal thermal equilibrium is reached, the constant heat capacity model gives $Δ U = C (T_{f} - T_{i})$ , that is,

T_{f} - T_{i} = \frac{M R ^{2} ω _{0}^{2}}{4 C} .

Friction converts the energy of ordered motion into internal energy within the system.

Question 4. The moment of inertia is

I = 2, 0 (0, 10)^{2} /2 = 0, 010 kg m^{2}

. We obtain

Δ U = \frac{1}{2} (0, 010) (100)^{2} = 50 J, T_{f} - T_{i} = \frac{50}{500} = 0, 10 K .

The system warms up even though $Q = 0$ : this energy did not cross its boundary as heat. It was initially present as rotational kinetic energy.