Riding a Bicycle in an Ideally Banked Curve

Problem:

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. 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} \left( v^2/rg \right)

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

Solution:

Part A

Let us redraw the given forces in a free-body diagram with their corresponding components.

The force N and F_c are the vertical and horizontal components of the force F.

If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for F, we have

\begin{align*}
\sum F_y & = 0 \\ \\
F \cos \theta - mg & = 0 \\ \\
F \cos \theta & = mg \\ \\
F & = \frac{mg}{\cos \theta}  \quad \quad  & \color{Blue}  \small \text{Equation 1}
\end{align*}

If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have

\begin{align*}
\sum F_x & = ma_c \\ \\
F \sin \theta  & = m a_c \\ \\
F \sin \theta  & = m \frac{v^2}{r}   \quad \quad  & \color{Blue}  \small \text{Equation 2}
\end{align*}

We substitute Equation 1 to Equation 2.

\begin{align*}
F \sin \theta  & = m \frac{v^2}{r} \\ \\
\frac{mg}{\cos \theta} \sin \theta & = m \frac{v^2}{r} \\ \\
mg \frac{\sin \theta}{\cos \theta} & =m \frac{v^2}{r} \\ \\
\end{align*}

We can cancel m from both sides, and we can apply the trigonometric identity \displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

\begin{align*}
g \tan \theta & = \frac{v^2}{r} \\ \\
\tan \theta & = \frac{v^2}{rg} \\ \\
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following values:

• linear velocity, v = 12.0\ \text{m/s}

• radius of curvature, r=30.0\ \text{m}

• acceleration due to gravity, g = 9.81\ \text{m/s}^2

We substitute the given values to the formula of \theta we solve in Part A.

\begin{align*}
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \\ \\
\theta & = \tan ^ {-1} \left[ \frac{\left( 12.0\ \text{m/s} \right)^2}{\left( 30.0\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\
\theta & = 26.0723 ^\circ \\ \\
\theta & = 26.1 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}