Question

Short answer

The angular momentum of the ball is \(5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\).

Step-by-step solution

Given data

The mass of the ball is\(m = 0.270\;{\rm{kg}}\).

The radius of the circle is\(r = 1.35\;{\rm{m}}\).

The angular speed is \(\omega = 10.4\;{\rm{rad/s}}\).

Understanding angular momentum

In this problem, use the product of the moment of inertia of the ball and its angular speed to evaluate the angular momentum.

Determine the angular momentum of the ball

The relation of angular momentum can be written as:

\(\begin{aligned}{l}L &= I\omega \\L &= \left( {m{r^2}} \right)\omega \end{aligned}\)

Here, Iis the moment of inertia of the ball.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}L &= \left( {0.270\;{\rm{kg}} \times {{\left( {1.35\;{\rm{m}}} \right)}^2}} \right)\left( {10.4\;{\rm{rad/s}}} \right)\\L &= 5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

Thus, \(L = 5.1\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\) is the required angular momentum.