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Q6.3-28 PE

Expert-verifiedFound in: Page 222

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight). **

**(a) Show that \(\theta \) (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, \(\theta = {\tan ^{ - 1}}\,{v^2}/rg\) **

**(b) Calculate \(\theta \) for a \(12.0{\rm{ m}}/{\rm{s}}\) turn of radius \(30.0{\rm{ m}}\) (as in a race).**

**Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle \(\theta \), the speed \(v\), and the radius of curvature \(r\) of the turn similar to that for the ideal banking of roadways.**

- The ideal banking angle is expressed as \(\theta = {\tan ^{ - 1}}\left( {\frac{{{v^2}}}{{rg}}} \right)\)[sb1] .
- The ideal banking angle is \(26.1^\circ \).

**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 \(26.1^\circ \).

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