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College Physics by Openstax Chapter 6 Problem 30

The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads

Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Solution:

Part A

The formula for an ideally banked curve is

\tan \theta = \frac{v^2}{rg}

Solving for v in terms of all the other variables, we have

v= \sqrt{rg \tan \theta}

For this problem, we are given

  • radius, r=100\ \text{m}
  • acceleration due to gravity, g=9.81\ \text{m/s}^2
  • banking angle, \theta = 15.0^\circ

Substituting all these values in the formula, we have

\begin{align*}
v & = \sqrt{rg \tan \theta} \\ 
v & = \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)\tan 15.0^\circ }\\
v & = 16.2129\ \text{m/s} \\
v & = 16.2\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Let us draw the free-body diagram of the car.

Summing forces in the vertical direction, we have

N \cos \theta + f \sin \theta -w= 0  \quad  \quad \quad \color{Blue} \text{Equation 1}

Summing forces in the horizontal directions taking to the left as the positive since the centripetal force is directed this way, we have

N \sin \theta - f \cos \theta = F_c \quad \quad \quad \color{Blue} \text{Equation 2}

We are given the following quantities:

  • radius of curvature, r=100\ \text{meters}
  • banking angle, \theta = 15.0^\circ
  • velocity, v=20\ \text{km/h} = 5.5556\ \text{m/s}
  • We also know that the friction, f=\mu_{s} N and weight, w=mg

We now use equation 1 to solve for N in terms of the other variables.

\begin{align*}
N \cos \theta + f \sin \theta -w & = 0 \\
N \cos \theta +\mu_s N\sin \theta  & = mg \\
N \left( \cos \theta + \mu_s \sin\theta \right) & =mg \\
N & = \frac{mg}{\cos \theta + \mu_s \sin\theta} \ \qquad \ \color{Blue} \left( \text{Equation 3} \right)
\end{align*}

We also solve for N in equation 2.

\begin{align*}
N \sin \theta - \mu_s N \cos \theta & = m \frac{v^2}{r} \\
N \left( \sin \theta-\mu_s \cos \theta \right) & = m \frac{v^2}{r} \\
N & = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \ \qquad \ \color{Blue} \left( \text{Equation 4} \right)
\end{align*}

Now, we have two equations of N. We now equate these two equations.

\frac{mg}{\cos \theta + \mu_s \sin\theta}  = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)}

We can now use this equation to solve for \mu_s.

\begin{align*}
\frac{\bcancel{m}g}{\cos \theta + \mu_s \sin\theta}  & = \frac{\bcancel{m}v^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \\
rg\left( \sin \theta-\mu_s \cos \theta \right) & = v^2 \left( \cos \theta + \mu_s \sin\theta \right) \\
rg \sin \theta - \mu_s rg \cos \theta & = v^2 \cos \theta +\mu_s v^2 \sin \theta \\
\mu_s v^2 \sin \theta + \mu _s rg \cos \theta & = rg \sin \theta - v^2 \cos \theta \\
\mu _s \left( v^2 \sin \theta + rg \cos \theta \right) & = rg \sin \theta - v^2 \cos \theta \\
\mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta}
\end{align*}

Now that we have an equation for \mu_s, we can substitute the given values.

\begin{align*}
\mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \\
\mu_s & = \frac{100\ \text{m}(9.81\ \text{m/s}^2) \sin 15.0^\circ -\left( 5.5556\ \text{m/s} \right)^2 \cos 15.0^\circ }{\left( 5.5556\ \text{m/s} \right)^2 \sin 15.0^\circ +100\ \text{m}\left( 9.81\ \text{m/s}^2 \right) \cos 15.0^\circ } \\
\mu_s & = 0.2345 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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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 \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).

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 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*}
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College Physics by Openstax Chapter 6 Problem 26

The Ideal Speed on a Banked Curve

Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

Solution:

The formula for the ideal speed on a banked curve can be derived from the formula of the ideal angle. That is, starting from \tan \theta = \frac{v^2}{rg}, we can solve for v.

v = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

  • radius of curvature, r=100\ \text{m}
  • acceleration due to gravity, g=9.81\ \text{m/s}^2
  • banking angle, \theta = 20.0 ^\circ

If we substitute the given values into our formula, we have

\begin{align*}
v = & \sqrt{rg \tan \theta} \\ \\
v = & \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right) \left( \tan 20.0 ^\circ \right) } \\ \\
v = & 18.8959\ \text{m/s} \\ \\
v = & 18.9\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The ideal speed for the given banked curve is about 18.9\ \text{m/s}.

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