General Mechanics/Momentum

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Changing mass[edit]

So far we've assumed that the mass of the objects being considered is constant, which is not always true. Mass is conserved overall, but it can be useful to consider objects, such as rockets, which are losing or gaining mass.

We can work out how to extend Newton's second law to this situation by considering a rocket two ways, as a single object of variable mass, and as two objects of fixed mass which are being pushed apart.

We find that

\mathbf{F} =  \frac{m d(\mathbf{v})}{dt}

Force is the rate of change of a quantity, mv, which we call linear momentum.

Newton's third law[edit]

Newton's third law says that the force, F12 exerted by a mass, m1, on a second mass, m2, is equal and opposite to the force F21 exerted by the second mass on the first.

\frac{d(m_2\mathbf{v}_2)}{dt}= \mathbf{F}_{12}= - \mathbf{F}_{21} 
=  -\frac{d(m_1 \mathbf{v}_1)}{dt}

if there are no external forces on the two bodies. We can add the two momenta together to get,

\frac{d(m_1\mathbf{v}_1)}{dt}+\frac{d(m_2 \mathbf{v}_2)}{dt}

so the total linear momentum is conserved.

Ultimately, this is consequence of space being homogeneous.

Centre of mass[edit]

Suppose two constant masses are subject to external forces, F1, and F2

Then the total force on the system, F, is

\mathbf{F}= \mathbf{F}_{1}+\mathbf{F}_{2} =

because the internal forces cancel out.

If the two masses are considered as one system, F should be the product of the total mass and ā, the average acceleration, which we expect to be related to some kind of average position.

\begin{matrix}(m_1+m_2)\bar{a} & = & \frac{d}{dt}(m_1\mathbf{v}_1+m_2\mathbf{v}_2) \\
& = & \frac{d^2}{dt^2}(m_1\mathbf{r}_1+m_2\mathbf{r}_2)\\
\bar{a}& = & \frac{d^2}{dt^2} \frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}\\

The average acceleration is the second derivative of the average position, weighted by mass.

This average position is called the centre of mass, and accelerates at the same rate as if it had the total mass of the system, and were subject to the total force.

We can extend this to any number of masses under arbitrary external and internal forces.

  • we have n objects, mass m1, m2mn, total mass M
  • each mass mi is at point ri,
  • each mass mi is subject to an external force Fi
  • the internal force exerted by mass mj on mass mi is Fij

then the position of the centre of mass, R, is

\mathbf{R}=\frac{\sum_i m_i \mathbf{r}_i}{M}

We can now take the second derivative of R

M \ddot\mathbf{R} & = & \sum_i m_i \ddot\mathbf{r}_i \\
& = & \sum_i (\mathbf{F}_i +\sum_j \mathbf{F}_{ij}) \end{array}

But the sum of all the internal forces is zero, because Newton's third law makes them cancel in pairs. Thus, the second term in the above equation drops out and we are left with:

M \ddot{\mathbf{R}} = \sum_i \mathbf{F}_i = \mathbf{F}

The centre of mass always moves like a body of the same total mass under the total external force, irrespective of the internal forces.