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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 27

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve

Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

Solution:

Part A

For ideally banked curved, the ideal banking angle is given by the formula \displaystyle \tan \theta = \frac{v^2}{rg}. We can solve for r in terms of all the other variables, and we should come up with

r = \frac{v^2}{g \tan \theta}

We are given the following values:

  • ideal banking angle, \displaystyle \theta = 75.0\ ^\circ
  • linear speed, \displaystyle v=30.0\ \text{m/s}
  • acceleration due to gravity, \displaystyle g=9.81\ \text{m/s}^2

If we substitute all the given values into our formula for r, we have

\begin{align*}
r & = \frac{v^2}{g \tan \theta} \\ \\
r & = \frac{\left( 30.0\ \text{m/s} \right)^2}{\left( 9.81\ \text{m/s}^2 \right)\left( \tan 75^\circ  \right)} \\ \\
r & = 24.5825\ \text{m} \\ \\
r & = 24.6\ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The radius of the ideally banked curve is approximately 24.6\ \text{m}.

Part B

The centripetal acceleration a_c can be solved using the formula

a_c = \frac{v^2}{r}

Substituting the given values, we have

\begin{align*}
a_c & = \frac{v^2}{r} \\ \\
a_c & = \frac{\left( 30.0\ \text{m/s} \right)^2}{24.5825\ \text{m}} \\ \\
a_c & = 36.6114\ \text{m/s}^2 \\ \\
a_c & = 36.6 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The centripetal acceleration is about 36.6\ \text{m/s}^2.

Part C

To know how large is the computed centripetal acceleration, we can compare it with that of acceleration due to gravity. 

\frac{a_c}{g} = \frac{36.6114\ \text{m/s}^2}{9.81\ \text{m/s}^2} = 3.73

The computed centripetal acceleration is 3.73 times the acceleration due to gravity. That is a_c = 3.73g.

This does not seem too large, but it is clear that bobsledders feel a lot of force on
them going through sharply banked turns!

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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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