0
$\begingroup$

To be more specific on the problem, a 50kg astronaut, an 80kg astronaut and a 20kg rock are tied together by a light rope during a space walk. I am asked to find the center of mass.

Now, I've chosen my reference point as the 50kg astronaut and I've used the formula $(m_1r_1+m_2r_2+m_3r_3)/(m_1+m_2+m_3)$ to find the position of the center of mass ($r$) of the system. I am uncertain if this is the correct method used.

If someone could just inform me if i am on the right track. Thanks.

Improve this question
$\endgroup$
2

1 Answer 1

Reset to default
1
$\begingroup$

You are on the right track. Mathematically, the center of mass $\vec{c}$ is defined as

$$ \vec{c} \sum_i m_i = \sum_i m_i \vec{r}_i $$

where $m_i$ are the individual masses and $\vec{r}_i$ the individual position vectors.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Or more clearly $$ \vec{c} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} $$ $\endgroup$
    – Floris
    Commented Mar 11, 2015 at 19:16
  • $\begingroup$ The reason I stated it as I did because it describes the fact that the net mass at $\vec{c}$ equals all the masses at their own $\vec{r}$ $\endgroup$
    – John Alexiou
    Commented Mar 11, 2015 at 21:08
  • $\begingroup$ I think that both are good. Your definition leads to my calculation. $\endgroup$
    – Floris
    Commented Mar 11, 2015 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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