91Ó°ÊÓ

Angular velocity may be defined as the speed at which an item rotates or revolves around an axis. The SI unit for angular velocity is radians per second because it is expressed as an angle per unit of time.

The centripetal acceleration is calculated as follows:

\({a_c} = 10g\)
The acceleration owing to gravity is denoted by \(g\).

Substitute \({\rm{9}}{\rm{.8 m/}}{{\rm{s}}^{\rm{2}}}\) for \(g\),

\(\begin{aligned}{}{a_c} = 10 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = 98{\rm{ m}}/{{\rm{s}}^2}\end{aligned}\)

The centripetal acceleration is,

\({a_c} = {\omega ^2}r\)

Here, \({a_c}\) is the centripetal acceleration, \(\omega \) is the angular velocity, and \(r\) is the distance of the rider from the center.

To generate an expression for the angular velocity, rearrange the previous equation.\(\omega = \sqrt {\frac{{{a_c}}}{r}} \)

Substitute \(98{\rm{ m}}/{{\rm{s}}^2}\) for \({a_c}\), and \(15.0{\rm{ m}}\) for \(r\),

\(\begin{aligned}{}\omega = \sqrt {\frac{{98{\rm{ m}}/{{\rm{s}}^2}}}{{15.0{\rm{ m}}}}} \\ = 2.55{\rm{ rad}}/{{\rm{s}}^2}\end{aligned}\)

Hence, the angular velocity is \(2.55{\rm{ rad}}/{\rm{s}}\).

The vertical component of the force on the arm will counter the weight i.e.,

\(\begin{aligned}{}{F_{arm}}\sin \theta = W\\ = mg\end{aligned}\) \({\rm{(1}}{\rm{.1)}}\)

Here, \({F_{arm}}\) is the force on the arm, \(\theta \) is the angle below the horizontal, \(W\) is the weight, \(m\) is the mass, and \(g\) is the acceleration due to gravity.

The horizontal component of the force on the arm will provide necessary centripetal force i.e.,

\(\begin{aligned}{}{F_{arm}}\cos \theta = {F_c}\\ = m{a_c}\end{aligned}\) \({\rm{(1}}{\rm{.2)}}\)

Here, \({F_c}\) is the centripetal force, and \({a_c}\) is the centripetal acceleration.

Dividing equation \({\rm{(1}}{\rm{.1)}}\) by equation \({\rm{(1}}{\rm{.2)}}\),

\(\begin{aligned}{}\frac{{{F_{arm}}\sin \theta }}{{{F_{arm}}\cos \theta }} = \frac{{mg}}{{m{a_c}}}\\\tan \theta = \frac{g}{{{a_c}}}\end{aligned}\)

Rearranging the above equation to get an expression angle below the horizontal.

\(\theta = {\tan ^{ - 1}}\left( {\frac{g}{{{a_c}}}} \right)\)

Substitute \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\), and \(98{\rm{ m}}/{{\rm{s}}^2}\) for \({a_c}\),

\(\begin{aligned}{}\theta = {\tan ^{ - 1}}\left( {\frac{{9.8{\rm{ m}}/{{\rm{s}}^2}}}{{98{\rm{ m}}/{{\rm{s}}^2}}}} \right)\\ = 5.71^\circ \end{aligned}\)

Hence, the angle below the horizontal is \(5.71^\circ \).