0
$\begingroup$

From Classical Mechanics by Kibble:

Consider a system of three particles, each of mass m, whose motion is described by (1.9). If particles 2 and 3, even though not rigidly bound together, are regarded as forming a composite body of mass 2m located at the mid-point $r=\frac{1}{2}(r_2 +r_3)$, find the equations describing the motion of the two-body system comprising particle 1 and the composite body (2+3). What is the force on the composite body due to particle 1? Show that the equations agree with (1.7). When the masses are unequal, what is the correct definition of the position of the composite (2 + 3) that will make (1.7) still hold?

(1.9) is $$ m_1 a_1 = F_{12} + F_{13}, \\ m_2 a_2 = F_{21} + F_{23}, \\ m_3 a_3 = F_{32} + F_{31}.$$

(1.7) is $$ m_1 a_1 = -m_2 a_2 $$


So I've done the first part, however I don't know how to do the bit in italics. Apparently the answer is $$ r = \frac{m_2 r_2 + m_3 r_3}{m_2 + m_3}, $$ but I don't understand where this answer is coming from.

Any help would be appreciated. Thank you.

$\endgroup$
0
$\begingroup$

First, take a step back and note that this result should be intuitive. The formula given is the weighted average of $r_2,r_3$ with each position contributing proportionally according to its mass. I.e. the total mass is $m_{23}=m_2+m_3,$ and then the position $r_2$ makes up $\frac{m_2}{m_{23}}$ of $r$ and $r_3$ makes up $\frac{m_3}{m_{23}}$ of $r$. The resulting $r$ (from now on I will call it $r_{23}$) is called the center of mass of objects 2 and 3.

You can find this algebraically by positing that the combined mass of the objects should be $m_{23}=m_2+m_3$ and then trying to find the acceleration that multiplies it in the net force. Add (1.9.ii) and (1.9.iii) and you get $$m_2a_2+m_3a_3=-m_1a_1.$$ We want to write the LHS as $m_{23}a_{23}$, so just forcibly factor out $m_2+m_3$ from the existing expression and call what's left $a_{23}.$

$$m_2a_2+m_3a_3=\underbrace{(m_2+m_3)}_{m_{23}}\underbrace{\left(\frac{m_2a_2+m_3a_3}{m_2+m_3}\right)}_{a_{23}}.$$

Integrate $a_{23}$ and you have $r_{23}$ as given.

$\endgroup$
1
  • $\begingroup$ Okay, that make sense. While I do have a good familiarity with calculus, I wasn't expecting it to be needed here. However, I now understand what's going on so thanks for you answer. $\endgroup$ Jul 25 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.