Question

The position vector of three particles of masses m_1=1kg. m_2=2kg and m_3=3kg are r_1=(hati+4hatj+hatk)m, r_2=(hati+hatj+hatk)m and r_3=(2hatj-hatj-2hatk)m respectivley. Find the position vector of their centre of mass.

Answer 1

Position vector of centre of mass = $\frac{m_1r_1 + m_2r_2+m_3r_3}{m_1+m_2+m_3}$

                   = $\frac{1(i+4j+k)+2(i+j+k)+3(2i-j-2k)}{1+2+3}$

                   = $1.5i + 0.5j - 0.5k$

Answer 2

Position vector of center of mass = $\frac{1}{2} (3i+j-k)$

Given:

The position vector of three particles of masses $m_1$=1 kg. $m_2$=2 kg and $m_3$=3kg are $r_1$=$\hat i+4\hat j+\hat k$m, $r_2$=$\hat i +\hat j+\hat k$m and $r_3$=$2\hat j-\hat j-2\hat k$m respectively.

To find:

Position vector of their center of mass.

Formula used:

$r_{CM}$    =$\frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$

  $r_{CM}$ = Position vector of center of mass.

Explanation:

Center of mass:

It is the average position of all the masses from reference axis.

It is very important from physics point of view because all the calculations are made from center of mass only.

$r_{CM}$    =$\frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$

$r_{CM}$ = $\frac{1(i+4j+k)+2(i+j+k)+3(2 j- j-2 k) }{1+2+3}$

$r_{CM}$ = $\frac{9i+3j - 3k}{6}$ = $\frac{1}{2} (3i+j-k)$

Position vector of center of mass = $\frac{1}{2} (3i+j-k)$

To learn more...

1)What are the resultants of the vector product of two given vectors given by .

A = 4i - 2j + k and B = 5i + 3j - 4k?

https://brainly.in/question/4175836

2)Unit vector in the direction of 4i+4j-2k is.

https://brainly.in/question/6789082