Question

Two particles of masses \[100g\] and \[300g\] at a given time have positions \[2\vec i + 5\vec j + 13\vec k\] and \[ - 6\vec i + 4\vec j - 2\vec k\] m respectively and velocities \[10\vec i - 7\vec j - 3\vec k\] and \[7\vec i - 9\vec j + 6\vec k\] m/s respectively. Determine the instantaneous position and velocity of C.M

Answer 1

Hint: The center of mass is the point at which the whole mass of the body is concentrated is called the center of mass. For a system center of mass is the point that can be used to define the motion of the system. So we need to use the formula for the position of the center of mass and velocity of the center mass and need to substitute the given values in the formula.
Formula used:
\[{\vec r_{com}} = \dfrac{{\sum\limits_{i = 1}^n {{m_i}{{\vec r}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}\]
\[{\vec v_{com}} = \dfrac{{\sum\limits_{i = 1}^n {{m_i}{{\vec v}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}\]

Complete answer:
Given two position vectors, let them be named as \[{\vec r_1}\] and \[{\vec r_2}\]. Also given that two velocity vectors, let them be named as \[{\vec v_1}\] and \[{\vec v_2}\].
Given
\[{\vec r_1} = 2\vec i + 5\vec j + 13\vec k\] and \[ - 6\vec i + 4\vec j - 2\vec k\]. Also \[{\vec v_1} = 10\vec i - 7\vec j - 3\vec k\] and \[{\vec v_2} = 7\vec i - 9\vec j + 6\vec k\].
Now first we will find the instantaneous position of the center of mass. The formula for the instantaneous position of the center of mass is given as,
\[{\vec r_{com}} = \dfrac{{\sum\limits_{i = 1}^n {{m_i}{{\vec r}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}\]
\[ \Rightarrow {\vec r_{com}} = \dfrac{{{m_1}{{\vec r}_1} + {m_2}{{\vec r}_2}}}{{{m_1} + {m_2}}}\]
Given two masses as \[{m_1}\] and \[{m_2}\] which is equal to \[100g\] and \[300g\].
Substituting all the known values in the above formula we will get,
\[ \Rightarrow {\vec r_{com}} = \dfrac{{100(2\vec i + 5\vec j + 13\vec k) + 300( - 6\vec i + 4\vec j - 2\vec k)}}{{100 + 300}}\]\[\dfrac{{gm.m}}{{gm}}\]
\[ \Rightarrow {\vec r_{com}} = \dfrac{{100(2\vec i + 5\vec j + 13\vec k) + 300( - 6\vec i + 4\vec j - 2\vec k)}}{{400}}m\]
\[ \Rightarrow {\vec r_{com}} = \dfrac{1}{4}(2\vec i + 5\vec j + 13\vec k - 18\vec i + 12\vec j - 6\vec k)m\]
\[ \Rightarrow {\vec r_{com}} = \dfrac{1}{4}( - 16\vec i + 17\vec j + 7\vec k)\]
This is the required position of the center of mass.
Now we need to find the velocity of the center of mass. The formula for the instantaneous velocity of the center of mass is given by,
\[{\vec v_{com}} = \dfrac{{\sum\limits_{i = 1}^n {{m_i}{{\vec v}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}\]
\[ \Rightarrow {\vec v_{com}} = \dfrac{{{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}}}{{{m_1} + {m_2}}}\]
Now substituting all the known values in the above formula we will get as,
\[{\vec v_{com}} = \dfrac{{100(10\vec i - 7\vec j - 3\vec k) + 300(7\vec i - 9\vec j + 6\vec k)}}{{100 + 300}}\dfrac{{gm.m/s}}{{gm}}\]
\[ \Rightarrow {\vec v_{com}} = \dfrac{{100(10\vec i - 7\vec j - 3\vec k) + 300(7\vec i - 9\vec j + 6\vec k)}}{{100 + 300}}m/s\]
\[ \Rightarrow {\vec v_{com}} = \dfrac{1}{4}(10\vec i - 7\vec j - 3\vec k + 21\vec i - 27\vec j + 18\vec k)m/s\]
\[ \Rightarrow {\vec v_{com}} = \dfrac{1}{4}(31\vec i - 34\vec j + 15\vec k)m/s\]
Therefore this is the required velocity of the center of mass.

Note:
We should always note that the center of mass is a point that is hypothetical and it may either lie inside of the system or outside of the system. It is not necessary that there must be some material particle at the center of mass of the system. But in a homogeneous body that is the body that is having a uniform distribution of mass the center of mass will always coincide with the geometrical center of the body.