Question

The distance between a carbon atom \(\left( {m = 12\;{\rm{u}}} \right)\) and an oxygen atom \(\left( {m = 16\;{\rm{u}}} \right)\) in the CO molecule is \({\bf{1}}{\bf{.13 \times 1}}{{\bf{0}}^{{\bf{10}}}}\;{\bf{m}}\) How far from the carbon atom is the center of mass of the molecule?

FIGURE 7-37Problem 50.

Short answer

The center of mass of the mass system is at \(0.44\;{\rm{m}}\) from the \(1.00\;{\rm{kg}}\) mass.

Step-by-step solution

Given data

To find the center of mass, you have to find the relative positions of all particles for that particle, relative to which you measure the distance of the center of mass.

The given three masses are \({m_1} = 1.00\;{\rm{kg}}\), \({m_2} = 1.50\;{\rm{kg}}\), and \({m_3} = 1.10\;{\rm{kg}}\).

Now, the relative distance of \(1.50\;{\rm{kg}}\) from the \(1.00\;{\rm{kg}}\) mass is \({x_2} = 0.50\;{\rm{m}}\), and the relative distance of the \(1.10\;{\rm{kg}}\) from the \(1.00\;{\rm{kg}}\) mass is:

\(\begin{array}{c}{x_3} = \left( {0.50\;{\rm{m}} + 0.25\;{\rm{m}}} \right)\\ = 0.75\;{\rm{m}}\end{array}\)

Calculation of the center of mass

Now, the center of mass of the mass system is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {x_1}} \right) + \left( {{m_2} \times {x_2}} \right) + \left( {{m_2} \times {x_2}} \right)}}{{{m_1} + {m_2} + {m_3}}}\\ = \frac{{\left( {1.00\;{\rm{kg}} \times 0} \right) + \left( {1.50\;{\rm{kg}} \times 0.50\;{\rm{m}}} \right) + \left( {1.10\;{\rm{kg}} \times 0.75\;{\rm{m}}} \right)}}{{1.00\;{\rm{kg}} + 1.50\;{\rm{kg}} + 1.10\;{\rm{kg}}}}\\ = 0.44\;{\rm{m}}\end{array}\)

Hence, the center of mass of the mass system is at \(0.44\;{\rm{m}}\) from the \(1.00\;{\rm{kg}}\) mass.