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 θ = v 2 r g \tan \theta = \frac{v^2}{rg} tan θ = r g v 2 Solving for v v v
v = r g tan θ v= \sqrt{rg \tan \theta} v = r g tan θ For this problem, we are given
radius, r = 100 m r=100\ \text{m} r = 100 m
acceleration due to gravity, g = 9.81 m/s 2 g=9.81\ \text{m/s}^2 g = 9.81 m/s 2
banking angle, θ = 15. 0 ∘ \theta = 15.0^\circ θ = 15. 0 ∘
Substituting all these values in the formula, we have
v = r g tan θ v = ( 100 m ) ( 9.81 m/s 2 ) tan 15. 0 ∘ v = 16.2129 m/s v = 16.2 m/s ( Answer ) \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*} v v v v = r g tan θ = ( 100 m ) ( 9.81 m/s 2 ) tan 15. 0 ∘ = 16.2129 m/s = 16.2 m/s ( Answer ) Part B
Let us draw the free-body diagram of the car.
Summing forces in the vertical direction, we have
N cos θ + f sin θ − w = 0 Equation 1 N \cos \theta + f \sin \theta -w= 0 \quad \quad \quad \color{Blue} \text{Equation 1} N cos θ + f sin θ − w = 0 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 θ − f cos θ = F c Equation 2 N \sin \theta - f \cos \theta = F_c \quad \quad \quad \color{Blue} \text{Equation 2} N sin θ − f cos θ = F c Equation 2 We are given the following quantities:
radius of curvature, r = 100 meters r=100\ \text{meters} r = 100 meters
banking angle, θ = 15. 0 ∘ \theta = 15.0^\circ θ = 15. 0 ∘
velocity, v = 20 km/h = 5.5556 m/s v=20\ \text{km/h} = 5.5556\ \text{m/s} v = 20 km/h = 5.5556 m/s
We also know that the friction, f = μ s N f=\mu_{s} N f = μ s N w = m g w=mg w = m g
We now use equation 1 to solve for N in terms of the other variables.
N cos θ + f sin θ − w = 0 N cos θ + μ s N sin θ = m g N ( cos θ + μ s sin θ ) = m g N = m g cos θ + μ s sin θ ( Equation 3 ) \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*} N cos θ + f sin θ − w N cos θ + μ s N sin θ N ( cos θ + μ s sin θ ) N = 0 = m g = m g = cos θ + μ s sin θ m g ( Equation 3 ) We also solve for N in equation 2.
N sin θ − μ s N cos θ = m v 2 r N ( sin θ − μ s cos θ ) = m v 2 r N = m v 2 r ( sin θ − μ s cos θ ) ( Equation 4 ) \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*} N sin θ − μ s N cos θ N ( sin θ − μ s cos θ ) N = m r v 2 = m r v 2 = r ( sin θ − μ s cos θ ) m v 2 ( Equation 4 ) Now, we have two equations of N N N
m g cos θ + μ s sin θ = m v 2 r ( sin θ − μ s cos θ ) \frac{mg}{\cos \theta + \mu_s \sin\theta} = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} cos θ + μ s sin θ m g = r ( sin θ − μ s cos θ ) m v 2 We can now use this equation to solve for μ s \mu_s μ s
m g cos θ + μ s sin θ = m v 2 r ( sin θ − μ s cos θ ) r g ( sin θ − μ s cos θ ) = v 2 ( cos θ + μ s sin θ ) r g sin θ − μ s r g cos θ = v 2 cos θ + μ s v 2 sin θ μ s v 2 sin θ + μ s r g cos θ = r g sin θ − v 2 cos θ μ s ( v 2 sin θ + r g cos θ ) = r g sin θ − v 2 cos θ μ s = r g sin θ − v 2 cos θ v 2 sin θ + r g cos θ \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*} cos θ + μ s sin θ m g r g ( sin θ − μ s cos θ ) r g sin θ − μ s r g cos θ μ s v 2 sin θ + μ s r g cos θ μ s ( v 2 sin θ + r g cos θ ) μ s = r ( sin θ − μ s cos θ ) m v 2 = v 2 ( cos θ + μ s sin θ ) = v 2 cos θ + μ s v 2 sin θ = r g sin θ − v 2 cos θ = r g sin θ − v 2 cos θ = v 2 sin θ + r g cos θ r g sin θ − v 2 cos θ Now that we have an equation for μ s \mu_s μ s
μ s = r g sin θ − v 2 cos θ v 2 sin θ + r g cos θ μ s = 100 m ( 9.81 m/s 2 ) sin 15. 0 ∘ − ( 5.5556 m/s ) 2 cos 15. 0 ∘ ( 5.5556 m/s ) 2 sin 15. 0 ∘ + 100 m ( 9.81 m/s 2 ) cos 15. 0 ∘ μ s = 0.2345 ( Answer ) \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*} μ s μ s μ s = v 2 sin θ + r g cos θ r g sin θ − v 2 cos θ = ( 5.5556 m/s ) 2 sin 15. 0 ∘ + 100 m ( 9.81 m/s 2 ) cos 15. 0 ∘ 100 m ( 9.81 m/s 2 ) sin 15. 0 ∘ − ( 5.5556 m/s ) 2 cos 15. 0 ∘ = 0.2345 ( Answer )
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