Question

What is the angular momentum of a rod with a mass of ${\text{8kg}}$ and length of ${\text{6m}}$ that is spinning around its center at ${\text{5Hz}}$ ?

Answer 1

The product of a body's rotational inertia and rotational velocity (in radians/sec) around a particular axis is angular momentum, which is a vector quantity (more specifically, a pseudovector). We come across this property on a regular basis, whether consciously or unconsciously. The following are some examples : If we try to balance on a bicycle without a kickstand, we will almost certainly fall off. When we start pedalling, however, these wheels gain angular momentum. They will resist improvement, making juggling more difficult.

Complete step by step answer:
The expression for angular momentum is $\overrightarrow {{\text{L}}\,} \,{\text{ = }}\,\,{{I\omega }}$ in which ${\text{I}}$ is the object's moment of inertia, and ${{\omega }}$ is the object's angular velocity. When a thin, rigid rod rotates around its base, its moment of inertia is given by,
${\text{I}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{{\text{12}}}}{\text{M}}{{\text{L}}^{\text{2}}}$, ${{\omega }}\,{\text{ = }}\,{{2\pi f}}$ gives the angular velocity, where f is the frequency.
We are given that ${\text{f}}\,{\text{ = }}\,{\text{5Hz}}$,
The angular velocity can be calculated as follows:
${{\omega }}\,{\text{ = }}\,{{2\pi f}}\,{\text{ = }}\,{{2\pi }}\left( {{\text{5}}{{\text{s}}^{{\text{ - 1}}}}} \right)\,\,{\text{ = }}\,\,{{10\pi }}\dfrac{{{\text{rad}}}}{{\text{s}}}$

We are given that ${\text{M}}\,{\text{ = }}\,{\text{8kg}}\,{\text{and}}\,{\text{L}}\,{\text{ = }}\,{\text{6m}}$, the moment of inertia can be calculated as follows:
${\text{I}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{{\text{12}}}}{\text{M}}{{\text{L}}^{\text{2}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{{\text{12}}}}\left( {{\text{8kg}}} \right){\left( {{\text{6m}}} \right)^{\text{2}}}\,{\text{ = }}\,{\text{24kg}}{{\text{m}}^{\text{2}}}$
Using the values, we calculated for ${\text{I}}$ and ${{\omega }}$, the angular momentum is calculated as follows:
$\overrightarrow {\text{L}} \,{\text{ = }}\,{{I\omega }}\,{\text{ = }}\,\left( {{{10\pi }}\dfrac{{{\text{rad}}}}{{\text{s}}}} \right)\left( {{\text{24kg}}{{\text{m}}^{\text{2}}}} \right)\,{\text{ = }}\,{{240\pi }}\dfrac{{{\text{kg}}{{\text{m}}^{\text{2}}}}}{{\text{s}}}$
$\therefore \overrightarrow L \, \approx \,754\dfrac{{kg{m^2}}}{s}$

Therefore, the angular momentum of a rod with a mass of ${\text{8kg}}$ and length of ${\text{6m}}$ that is spinning around its center at ${\text{5Hz}}$ is $\approx 754\dfrac{{kg{m^2}}}{s}$.

Additional Information:
We come across the property of angular momentum on a regular basis, whether consciously or unconsciously. The following are some examples : If we try to balance on a bicycle without a kickstand, we will almost certainly fall off. When we start pedalling, however, these wheels gain angular momentum. They will resist improvement, making juggling more difficult.

Note: Azimuthal quantum number or secondary quantum number are synonyms for angular momentum quantum number. It is a quantum number that determines the angular momentum of an atomic orbital and defines its size and shape. The average value is between zero and one.