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Compound Systems#
Now that we have the necessary tools to describe motion in two and three dimensions, we can write Newton’s second law in vector form:
The acceleration \(\mathbf{a}\) always points in the same direction as \(\mathbf{F}\).
In this lecture we will introduce the forces between multiple bodies and the centre of mass. Then we will introduce the notion of momentum, and re-express Newton’s second law in terms of changes in momentum (Newton’s original formulation).
Newton’s third law#
In the example of the block on the ramp in the previous section, the two bodies exert a force on each other: the weight of the block is a force on the ramp, while the ramp exerts an equal and opposite force on the block, namely the normal force. This is a consequence of Newton’s third law that forces always appear in pairs that are equal and opposite, but acting on different objects. This last bit is important to remember, and often gets left out of the slogan “for each action there is an equal and opposite reaction”. What would happen if the reaction acts on the same body as the action?
Example: Sketch a ladder at rest, leaning against a wall (make it an abstract drawing, consisting of only three lines—use a ruler!). Draw the forces on the ladder, the wall, and the ground. Take care with the points of engagement where the forces act on the three bodies. If the wall is frictionless but the floor is not, what is the shallowest angle at which the ladder can rest?
Imagine that object 1 exerts a force \(\mathbf{F}_{12}\) on object 2. The point of engagement of \(\mathbf{F}_{12}\) is on object 2.
The third law states that object 2 also exerts a force on object 1, \(\mathbf{F}_{21}\), that is opposite and equal in magnitude to \(\mathbf{F}_{12}\):
The two objects do not need to be touching, as is the case with the block and the ramp. The gravitational force on the Earth by the Sun is a vector \(\mathbf{F}_{\rm g}\) acting on the Earth and pointing towards the Sun. Similarly, the Earth exerts an equal force on the Sun in opposite direction, \(-\mathbf{F}_{\rm g}\).
The acceleration of the Earth \(\mathbf{a}_{\rm E}\) due to the Sun’s gravity can be written as
where \(m_{\rm E}\) is the mass of the Earth.
Similarly, the acceleration of the Sun \(\mathbf{a}_{\odot}\) due to the Earth’s gravity can be written as
where \(m_{\odot}\) is the mass of the Sun. Even though the forces are the same in magnitude, the acceleration of the Earth is much larger than that of the Sun due to the great difference in masses \(m_{\rm E}\) and \(m_{\odot}\).
Often we want to consider a system of \(N\) masses \(m_1, m_2, \ldots, m_N\). This can be a galaxy of stars, the solar system, an atomic nucleus with orbiting electrons, or a formation of flying drones, to name a few examples. There may be forces between these objects. If there is a force \(\mathbf{F}_{jk}\) on mass \(m_k\) due to mass \(m_j\), then there is also a force \(\mathbf{F}_{kj} = -\mathbf{F}_{jk}\) on mass \(m_j\) due to mass \(m_k\), as required by Newton’s third law.
Momentum#
At this point we introduce the momentum \(\mathbf{p}\) of a body with mass \(m\) and velocity \(\mathbf{v}\) as
Momentum is a very useful concept when we want to quantify “how much motion” there is in a body. It plays a more fundamental role than acceleration in Newton’s second law, and we can write instead:
The change in motion of a body is equal to the force exerted on that body. This equation is equivalent to \(\mathbf{F} = m\mathbf{a}\), but only when the mass is constant. There are, however, situations in which this is not the case.
Worked Example: Consider a rocket travelling with (increasing) velocity \(\mathbf{v}\) relative to some rest frame (e.g., the Earth), which provides our coordinate system. The rocket expels hot gas with a velocity \(\mathbf{v}'\) relative to the same rest frame, such that the exhaust velocity relative to the rocket is \(\mathbf{v}_e = \mathbf{v}' - \mathbf{v}\). Assume furthermore that \(\mathbf{v}_e\) is constant. We are interested in the force \(\mathbf{F}\) on the rocket by the exhaust gas, and the velocity of the rocket after some time \(t\).
The momentum of the rocket at time \(t\) is \(\mathbf{p}(t) = m \mathbf{v}(t)\), and a short time interval \(dt\) later, the momentum has become
\[ \mathbf{p}(t+dt) = (m - dm)(\mathbf{v}+d\mathbf{v}) + dm \, \mathbf{v}' \, , \]where \(dm\) is the (positive) mass of hot gas expelled by the rocket.
The first term on the right-hand side is the new momentum of the rocket, while the second term, \(dm \, \mathbf{v}'\), is the momentum of the expelled hot gas. Expanding:\[ \mathbf{p}(t+dt) = m\mathbf{v} + m d\mathbf{v} - \mathbf{v} dm + dm \, \mathbf{v}' = m\mathbf{v} + m d\mathbf{v} + \mathbf{v}_e dm \, , \]where we dropped the product of infinitesimals \(dm \, d\mathbf{v}\).
The change in momentum is
\[ d\mathbf{p} = \mathbf{p}(t+dt) - \mathbf{p}(t) = m d\mathbf{v} + \mathbf{v}_e dm \, . \]Hence,
\[ \frac{d\mathbf{p}}{dt} = m \frac{d\mathbf{v}}{dt} + \mathbf{v}_e \frac{dm}{dt} \, . \]If there is an external force \(\mathbf{F}\) on the rocket, such as Earth’s gravity, then
\[ \mathbf{F} = \frac{d\mathbf{p}}{dt} = m \frac{d\mathbf{v}}{dt} + \mathbf{v}_e \frac{dm}{dt} \, . \]If the rocket accelerates in free space without external forces, then
\[ \mathbf{F} = 0 \quad \longrightarrow \quad m \frac{d\mathbf{v}}{dt} + \mathbf{v}_e \frac{dm}{dt} = 0 \, , \]and the momentum imparted on the hot gas balances the momentum gained by the rocket and its remaining fuel.
Let \(\mathbf{F} = m \mathbf{g}\), the gravitational force close to the Earth’s surface. Our equation of motion becomes
\[ \frac{d\mathbf{v}}{dt} + \frac{\mathbf{v}_e}{m} \frac{dm}{dt} = \mathbf{g} \, . \]Here \(\mathbf{v}\) and \(\mathbf{v}_e\) point upward, while \(\mathbf{g}\) points downward.
This becomes a one-dimensional problem:\[ \frac{dv}{dt} + \frac{v_e}{m} \frac{dm}{dt} = -g \, , \]where the positive direction is upward. This is the equation of motion in terms of the rocket’s velocity.
Integrating:
\[ \int_{v_0}^v dv + v_e \int_{m_0}^m \frac{dm'}{m'} = -g \int_0^t dt \, , \]or
\[ v = v_0 + v_e \ln \frac{m}{m_0} - g t \, . \]If \(m\) is the final mass and \(t\) is the burn time, then \(v\) is the final speed of the rocket.
We assumed that \(g\) does not change with altitude and ignored air resistance.
Centre of mass#
The centre of mass position \(\mathbf{R}\) of \(N\) particles is the mass-weighted average of their positions:
The positions of the particles relative to the centre of mass are \(\mathbf{r}_j' = \mathbf{r}_j - \mathbf{R}\). Using these, the centre of mass always lies at the origin, which is often very convenient.
The velocity of the centre of mass is obtained by differentiating \(\mathbf{R}\):
where \(M = \sum_j m_j\) is the total mass.
Defining the total momentum \(\mathbf{P} = \sum_j \mathbf{p}_j\), we have
Thus the total momentum equals the total mass times the velocity of the centre of mass.
When two constituent masses exchange momentum, one gains \(\Delta \mathbf{p}\) while the other loses \(-\Delta \mathbf{p}\), ensured by Newton’s third law.
This means the overall momentum \(\mathbf{P}\) remains constant, even if the internal motion is complicated.
Only an external force \(\mathbf{F}_{\rm ext}\) can change \(\mathbf{P}\):
This is Newton’s second law again. It means that we can solve the motion of complex bodies if we consider the position and momentum of the centre of mass. It is also why we can treat planets and stars as point objects.