Equations of Motion for a Variable Mass System 📂Classical Mechanics

Equations of Motion for a Variable Mass System

Overview¹

In physics, mass is often treated as a fixed constant in many situations. However, there are many cases where this is not true. For example, a falling raindrop absorbs smaller droplets in the atmosphere, increasing its mass. A rocket accelerates by expelling gases from burning fuel, thereby reducing its mass as the fuel is consumed.

Formulas

An Object Moving with Increasing Mass

Consider a case where an object moves and its mass increases. Assume that a large object collides with smaller particles, causing the small particles to stick to the object as shown in the figure below.

Let’s denote the object’s mass and velocity at time $t$ as $m(t), \mathbf{v}(t)$, respectively. Let’s call the velocity of the small particles $\mathbf{u}(t)$. Let’s denote the mass gained by the object after a small time interval $\Delta t$ as $\Delta m$. Then, the mass of the object after time $\Delta t$ is $m(t + \Delta t) = m(t) + \Delta m$, and its velocity is $\mathbf{v}(t + \Delta t) = v(t) + \Delta \mathbf{v}$. Let’s denote the linear momentum of the system as $\mathbf{p}$. Then,

$$ \begin{align*} \mathbf{p}(t) &= m \mathbf{v} + \Delta m \mathbf{u} \\ \mathbf{p}(t + \Delta t) &= m(t + \Delta t) \mathbf{v}(t + \Delta t) = \left( m(t) + \Delta m \right) \left( \mathbf{v}(t) + \Delta \mathbf{v} \right) \end{align*} $$

The change in momentum during $\Delta t$ is as follows.

$$ \begin{align*} \Delta \mathbf{p} &= \left( m + \Delta m \right) \left( \mathbf{v} + \Delta \mathbf{v} \right) - \left( m \mathbf{v} + \Delta m \mathbf{u} \right) \\ &= m \mathbf{v} + m\Delta \mathbf{v} + \Delta m \mathbf{v} + \Delta m \Delta \mathbf{v} - m \mathbf{v} - \Delta m \mathbf{u} \\ &= (m + \Delta m) \Delta \mathbf{v} + \Delta m (\mathbf{v} - \mathbf{u}) \\ \end{align*} $$

Here, if we denote the relative velocity of the small particles to the object as $\mathbf{V} = \mathbf{u} - \mathbf{v}$,

$$ \Delta \mathbf{p} = (m + \Delta m) \Delta \mathbf{v} - \mathbf{V}\Delta m $$

Now, by dividing both sides by $\Delta t$ and taking the limit as $\Delta t \to 0$,

$$ \mathbf{F}_{\text{ext}} = \dfrac{d \mathbf{p}}{dt} = m \dot{\mathbf{v}} - \mathbf{V}\dot{m} $$

Here, $\mathbf{F}_{\text{ext}}$ represents external forces such as gravity and air resistance.

Consider the case where the small particles are stationary. For example, assume that the water droplets in the fog are stationary when some object moves through the fog. This is generally a good approximation. Then, $\mathbf{V} = - \mathbf{v}$, and the equation of motion is as follows.

$$ \mathbf{F}_{\text{ext}} = m \dot{\mathbf{v}} + \mathbf{v}\dot{m} $$

An Object Moving with Decreasing Mass

The result is not different from the above. However, in this case, since the mass is decreasing, the change should be negative.

$$ \Delta m \lt 0 \quad \text{and} \quad \dot{m} \lt 0 $$