91Ó°ÊÓ

The object’s angular velocity may be expressed as the rate of change of the object’s angular position. All parts of a rigid particle revolving about a fixed axis have an identical angular velocity at any instant.

(a)

The Earth makes one orbit around the Sun in one year. Therefore, the angular velocity of the Earth in its orbit around the Sun can be calculated as follows:

\begin{aligned}{l}{\omega _{{\rm{orbit}}}} &= \frac{{\Delta \theta }}{{\Delta t}}\\{\omega _{{\rm{orbit}}}} &= \frac{{\left( {2\pi \;{\rm{rad}}} \right)}}{{\left( {1\;{\rm{yr}}} \right)}}\left( {\frac{{1\;{\rm{yr}}}}{{3.16 \times {{10}^7}\;{\rm{s}}}}} \right)\\{\omega _{{\rm{orbit}}}} &= 1.99 \times {10^{ - 7}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}

Thus, the angular velocity of the Earth in its orbit around the Sun is \(1.99 \times {10^{ - 7}}\;{{{\rm{rad}}}\mathord{\left/{\vphantom{{{\rm{rad}}}{\rm{s}}}} \right.} {\rm{s}}}\).

(b)

The Earth makes one revolution about its axis in one day. Therefore, the angular velocity of the Earth about its axis can be calculated as follows:

\(\begin{aligned}{l}{\omega_{{\rm{rotation}}}}&=\frac{{\Delta\theta}}{{\Delta t}}\\{\omega _{{\rm{rotation}}}}&=\frac{{\left( {2\pi \;{\rm{rad}}} \right)}}{{\left( {1\;{\rm{day}}}\right)}}\left({\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}}\right)\\{\omega _{{\rm{rotation}}}}&=7.27 \times {10^{ - 5}}\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the angular velocity of the Earth about its axis is \(7.27 \times {10^{ - 5}}\;{{{\rm{rad}}}\mathord{\left/{\vphantom{{{\rm{rad}}}{\rm{s}}}} \right.} {\rm{s}}}\).