A 4.00-kg mass and a 3.00-kg mass are attached to opposite ends of a very light 42.0-cm-long horizontal rod (Fig. 8–61). The system is rotating at angular speed\(\omega = 5.60\;{\rm{rad/s}}\)about a vertical axle at the center of the rod. Determine (a) the kinetic energy KE of the system, and (b) the net force on each mass.

The two masses are\({m_1} = 4\;{\rm{kg}}\)and\({m_2} = 3\;{\rm{kg}}\).

The length of the horizontal rod is\(l = 42\;{\rm{cm}} = 0.42\;{\rm{m}}\).

The angular speed of the system is \(\omega = 5.60\;{\rm{rad/s}}\).

The moment of inertia is a quantity that expresses a body’s tendency to resist angular acceleration about the axis of rotation.

It is given as the product of the mass of the rigid body and the square of the distance from the axis of rotation.

\(I = m{r^2}\)

The rotational kinetic energy is the energy an object possesses because of its rotation. It can be expressed in terms of the moment of inertia, Iand the angular velocity,\(\omega \).

\(K{E_{{\rm{rot}}}} = \frac{1}{2}I{\omega ^2}\)

The rod joining the two masses is very light, i.e., massless. The kinetic energy of the system will be the sum of the kinetic energy of each mass.

Therefore, the total rotational kinetic energy of the system is given by:

\(\begin{aligned}{c}K{E_{{\rm{rot}}}} = \frac{1}{2}{I_1}{\omega ^2} + \frac{1}{2}{I_2}{\omega ^2}\\ = \frac{1}{2}\left( {{m_1}{r^2}} \right){\omega ^2} + \frac{1}{2}\left( {{m_2}{r^2}} \right){\omega ^2}\\ = \frac{1}{2}\left( {{m_1} + {m_2}} \right){r^2}{\omega ^2}\end{aligned}\)

Here,\(r = \frac{{0.42\;{\rm{m}}}}{2} = 0.21\;{\rm{m}}\).

\(\begin{aligned}{c}K{E_{{\rm{rot}}}} = \frac{1}{2}\left( {{m_1} + {m_2}} \right){r^2}\omega _1^2\\ = \frac{1}{2}\left( {4\;{\rm{kg}} + 3\;{\rm{kg}}} \right){\left( {0.21\;{\rm{m}}} \right)^2}{\left( {5.60\;{\rm{rad/s}}} \right)^2}\\ = 4.84\;{\rm{J}}\end{aligned}\)

Thus, the total kinetic energy of the system is 4.84 J.

The net centripetal force of the rotating system is given by:

\(F = m{\omega ^2}r\)

For mass\({m_1}\), the net force is:

\(\begin{aligned}{c}{F_1} = {m_1}{\omega ^2}r\\ = \left( {4\;{\rm{kg}}} \right){\left( {5.60\;{\rm{rad/s}}} \right)^2}\left( {0.21\;{\rm{m}}} \right)\\ = 26.34\;{\rm{N}}\end{aligned}\)

For mass\({m_2}\),the net force is:

\(\begin{aligned}{c}{F_2} = {m_2}{\omega ^2}r\\ = \left( {3\;{\rm{kg}}} \right){\left( {5.60\;{\rm{rad/s}}} \right)^2}\left( {0.21\;{\rm{m}}} \right)\\ = 19.76\;{\rm{N}}\end{aligned}\)

These forces are unequal; so there will be a net horizontal force on the rod due to the masses.