Question

Question: (II) A uniform horizontal rod of mass M and length l rotates with angular velocity \(\omega \) about a vertical axis through its center. Attached to each end of the rod is a small mass m. Determine the angular momentum of the system about the axis.

Short answer

The angular momentum of the system about the axis is \(\left( {\frac{1}{2}\left( {\frac{1}{6}M + m} \right){l^2}} \right)\omega \).

Step-by-step solution

Formula of angular momentum

The equation for the angular momentum of a rotating object about a fixed rotation axis is as follows:

\(L = I\omega \)

Here, \(I\) is the moment of inertia of the object and \(\omega \) is the angular velocity.

Given information

Given data:

The mass of the rod is\(M\).

The length of the rod is\(l\).

Mass \(m\) is attached to each end of the rod.

Write the expression for the moment of inertia of each component of the system

The expression for the moment of inertia of the rod is as follows:

\({I_1} = \frac{1}{{12}}M{l^2}\)

The expression for the moment of inertia of the mass on one end of the rod is as follows:

\({I_2} = m{R^2}\)

The expression for the moment of inertia of the mass on another end of the rod is as follows:

\({I_3} = m{R^2}\)

Calculate the angular momentum of the system

The distance between the masses on either end of the rotation axis is equal to half its length. That is:

\(R = \frac{l}{2}\)

The moment of inertia of a system is equal to the sum of the moments of inertia of the individual components of the system. Therefore, we can write:

\(\begin{aligned}{c}I &= {I_1} + {I_2} + {I_3}\\I &= \frac{1}{{12}}M{l^2} + m{R^2} + m{R^2}\\I &= \frac{1}{{12}}M{l^2} + 2m{R^2}\\I &= \frac{1}{{12}}M{l^2} + 2m{\left( {\frac{l}{2}} \right)^2}\end{aligned}\)

Solve further as:

\(\begin{aligned}{l}I &= \frac{1}{{12}}M{l^2} + \frac{1}{2}m{l^2}\\I &= \frac{1}{{12}}\left( {M + 6m} \right){l^2}\\I &= \frac{1}{2}\left( {\frac{1}{6}M + m} \right){l^2}\end{aligned}\)

The angular momentum of the system about the axis can be calculated as:

\(\begin{aligned}{l}L &= I\omega \\L &= \left( {\frac{1}{2}\left( {\frac{1}{6}M + m} \right){l^2}} \right)\omega \end{aligned}\)

Thus, the angular momentum of the system about the axis is \(\left( {\frac{1}{2}\left( {\frac{1}{6}M + m} \right){l^2}} \right)\omega \).