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College Physics by Openstax Chapter 6 Problem 28

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 θ\theta (as defined in the figure) is related to the speed vv and radius of curvature rr of the turn in the same way as for an ideally banked roadway—that is, θ=tan⁡−1(v2/rg)\theta = \tan ^{-1} \left( v^2/rg \right)

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

Figure 6.33 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 θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

Solution:

Part A

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

The force NN and FcF_c are the vertical and horizontal components of the force FF.

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

∑Fy=0Fcos⁡θ−mg=0Fcos⁡θ=mgF=mgcos⁡θEquation 1\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

∑Fx=macFsin⁡θ=macFsin⁡θ=mv2rEquation 2\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.

Fsin⁡θ=mv2rmgcos⁡θsin⁡θ=mv2rmgsin⁡θcos⁡θ=mv2r\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 mm from both sides, and we can apply the trigonometric identity tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

gtan⁡θ=v2rtan⁡θ=v2rgθ=tan⁡−1(v2rg)  (Answer)\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 m/sv = 12.0\ \text{m/s}
  • radius of curvature, r=30.0 mr=30.0\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g = 9.81\ \text{m/s}^2

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

θ=tan⁡−1(v2rg)θ=tan⁡−1[(12.0 m/s)2(30.0 m)(9.81 m/s2)]θ=26.0723∘θ=26.1∘  (Answer)\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*}
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College Physics by Openstax Chapter 6 Problem 25

The ideal banking angle of a curve on a highway

Problem:

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

Solution:

The ideal banking angle (meaning there is no involved friction) of a car on a curve is given by the formula:

θ=tan⁡−1(v2rg)\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)

We are given the following values:

  • radius of curvature, r=1.20 km×1000 m1 km=1200 m\displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}
  • linear velocity, v=105 km/h×1000 m1 km×1 h3600 s=29.1667 m/s\displaystyle v=105\ \text{km/h}\times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 29.1667\ \text{m/s}
  • acceleration due to gravity, g=9.81 m/s2\displaystyle g = 9.81\ \text{m/s}^2

If we substitute these values into our formula, we come up with

θ=tan⁡−1(v2rg)θ=tan⁡−1[(29.1667 m/s)2(1200 m)(9.81 m/s2)]θ=4.1333∘θ=4.13∘  (Answer)\begin{align*} \theta & = \tan^{-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan^{-1} \left[ \frac{\left( 29.1667\ \text{m/s} \right)^2}{\left( 1200\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 4.1333 ^\circ \\ \\ \theta & = 4.13 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal banking angle for the given highway is about 4.13∘ 4.13 ^\circ.

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