Mechanics/Momentum

Changing mass

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

$𝐅=\frac{d(m𝐯)}{dt}$

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

Newton's third law

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.

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

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

$\frac{d(m_1{𝐯}_1)}{dt}+\frac{d(m_2{𝐯}_2)}{dt}=\frac{d}{dt}(m_1{𝐯}_1+m_2{𝐯}_2)=0$

so the total linear momentum is conserved.

Ultimately, this is consequence of space being homogeneous.

Centre of mass

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

Then the total force on the system, F, is

$𝐅={𝐅}_1+{𝐅}_2=\frac{d}{dt}(m_1{𝐯}_1+m_2{𝐯}_2)$

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.

$\begin{array}{ccc}(m_1+m_2)\overline{a}& =& \frac{d}{dt}(m_1{𝐯}_1+m_2{𝐯}_2)\\ & =& \frac{d^2}{d{t}^2}(m_1{𝐫}_1+m_2{𝐫}_2)\\ \overline{a}& =& \frac{d^2}{d{t}^2}\frac{m_1{𝐫}_1+m_2{𝐫}_2}{m_1+m_2}\end{array}$

The average acceleration is the second derivative of the average postion, 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 arbitary external and internal forces.

• we have n objects, mass m₁, m₂ … m_n, total mass M
• each mass m_i is at point r_i,
• each mass m_i is subject to an external force F_i
• the internal force exerted by mass m_j on mass m_i is F_ij

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

$𝐑=\frac{\sum _i{m}_i{𝐫}_i}{M}$

We can now take the second derivative of R

$\begin{array}{ccc}M\ddot{𝐑}& =& \sum _i{m}_i{\ddot{𝐫}}_i\\ & =& \sum _i({𝐅}_i+\sum _j{𝐅}_{ij})\end{array}$

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:

$M\ddot{𝐑}=\sum _i{𝐅}_i=𝐅$

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