Question

The moment of inertia of a sphere of uniform density rotating on its axis is \(\frac{2}{5}M{R^2}\). Use data given at the end of this book to calculate the magnitude of the rotational angular momentum of the Earth.

Short answer

The magnitude of the rotational angular momentum of the Earth is \(7.1 \times {10^{33}}\,\,{\rm{kg}} \cdot {{\rm{m}}^2}/{\rm{s}}\)

Step-by-step solution

Step 1:Definition of Moment of Inertia and Rotational angular Momentum.

In physics, a moment of inertia is a quantitative measure of a body's rotational inertia, or its resistance to having its speed of rotation along an axis changed by the application of a torque.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

Solve the angular momentum of rotation of the Earth

Use the expression for the angular momentum to solve for the angular momentum of rotation of Earth.

The expression for the angular momentum of the rotating object is,

\(L = I\omega \)

Here, \(I\)is the moment of inertia of the object in this case Earth, and\(\omega \) is the angular velocity of rotation of the object in this case Earth.

The expression for the moment of the inertia of the solid sphere is,

\(I = \frac{2}{5}M{R^2}\)

Here, \(M\)is the mass of the Earth, and \(R\) is the radius of the Earth.

Combine \(I = \frac{2}{5}M{R^2}\)and to write the final expression for the angular momentum of rotation of Earth.

\(L = \left( {\frac{2}{5}M{R^2}} \right)\omega \)

Solve the given expression

The expression for the angular velocity of rotation of Earth is,

\(\omega = \frac{{2\pi }}{T}\)

Here,\(T\) is the period of rotation of Earth.

Substitute \(24h\)for \(T\)in \(\omega = \frac{{2\pi }}{T}\)

\(\begin{aligned}{}\omega &= \frac{{2\pi }}{{\left( {24h} \right)\left( {\frac{{3600s}}{{1h}}} \right)}}\\ &= 7.29 \times {10^{ - 5}}\,\,{\rm{rad}} \cdot {{\rm{s}}^{ - 1}}\end{aligned}\)

Therefore, the angular velocity of rotation of Earth is \(7.29 \times {10^{ - 5}}\,{\rm{rad}} \cdot {{\rm{s}}^{ - 1}}\).

Substitute \(5.97 \times {10^{24}}{\rm{kg}}\)for\(M\), \(6371 \times {10^{23}}{\rm{m}}\)for \(R\) and \(7.29 \times {10^{ - 5}}{\rm{rad}} \cdot {{\rm{s}}^{ - 1}}\) for\(\omega \)in

\(L = \left( {\frac{2}{5}M{R^2}} \right)\omega \)

\(\begin{aligned}{}L& = \frac{2}{5}\left( {5.97 \times {{10}^{24}}\,\,{\rm{kg}}} \right){\left( {6371 \times {{10}^3}\,\,{\rm{m}}} \right)^2}\left( {7.29 \times {{10}^{ - 5}}{\rm{rad}} \cdot {{\rm{s}}^{ - 1}}} \right)\\ &= 7.1 \times {10^{33}}\,\,{\rm{kg}} \cdot {{\rm{m}}^2}/{\rm{s}}\end{aligned}\)

Therefore, the angular momentum of rotation of Earth is\(7.1 \times {10^{33}}\,\,{\rm{kg}} \cdot {{\rm{m}}^2}/{\rm{s}}\).