91影视

To make a safe turn, the cyclist bends slightly away from the vertical axis, providing the required centripetal force.

The free-body equation along the vertical direction is,

\(F\cos \theta = W\) \({\rm{(1}}{\rm{.1)}}\)

Here, \(W\) is the weight of the cyclist and \(\theta \) is the ideal banking angle.

The weight of the cyclist is,

\(W = mg\)

Here, \(m\) is the mass of the cyclist, and \(g\) is the acceleration due to gravity.

From equation \({\rm{(1}}{\rm{.1)}}\),

\(F\cos \theta = mg\) \({\rm{(1}}{\rm{.2)}}\)

The free-body equation along the horizontal direction is,

\(F\sin \theta = {F_c}\) \({\rm{(1}}{\rm{.3)}}\)

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

The centripetal force is,

\({F_c} = \frac{{m{v^2}}}{r}\)

Here, \(v\) is the velocity of the cyclist, and \(r\) is the radius of the circular path.

From equation \({\rm{(1}}{\rm{.3)}}\),

\(F\sin \theta = \frac{{m{v^2}}}{r}\) \({\rm{(1}}{\rm{.4)}}\)

Dividing equation \({\rm{(1}}{\rm{.4)}}\) by equation \({\rm{(1}}{\rm{.2)}}\),

\(\begin{aligned}{}\frac{{F\sin \theta }}{{F\cos \theta }} = \frac{{\frac{{m{v^2}}}{r}}}{{mg}}\\\tan \theta = \frac{{{v^2}}}{{rg}}\end{aligned}\)

Rearranging the above equation in order to get an expression for the ideal banking angle.

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{v^2}}}{{rg}}} \right)\) \({\rm{(1}}{\rm{.5)}}\)

As a result, the ideal banking angle is expressed as \(\theta = {\tan ^{ - 1}}\left( {\frac{{{v^2}}}{{rg}}} \right)\).

The ideal banking angle of the cyclist can be calculated using equation \({\rm{(1}}{\rm{.5)}}\).

Substitute \(12.0{\rm{ m}}/{\rm{s}}\) for \(v\), \(30.0{\rm{ m}}\) for \(r\), and \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{aligned}{}\theta = {\tan ^{ - 1}}\left( {\frac{{{{\left( {12.0{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{\left( {30.0{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}} \right)\\ = 26.1^\circ \end{aligned}\)

Hence, the ideal banking angle of the cyclist is \(26.1^\circ \).