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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⁡θ=v2rg\tan \theta = \frac{v^2}{rg}

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

v=rgtan⁡θv= \sqrt{rg \tan \theta}

For this problem, we are given

  • radius, r=100 mr=100\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g=9.81\ \text{m/s}^2
  • banking angle, θ=15.0∘\theta = 15.0^\circ

Substituting all these values in the formula, we have

v=rgtan⁡θv=(100 m)(9.81 m/s2)tan⁡15.0∘v=16.2129 m/sv=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*}

Part B

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

Summing forces in the vertical direction, we have

Ncos⁡θ+fsin⁡θ−w=0Equation 1N \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

Nsin⁡θ−fcos⁡θ=FcEquation 2N \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 metersr=100\ \text{meters}
  • banking angle, θ=15.0∘\theta = 15.0^\circ
  • velocity, v=20 km/h=5.5556 m/sv=20\ \text{km/h} = 5.5556\ \text{m/s}
  • We also know that the friction, f=μsNf=\mu_{s} N and weight, w=mgw=mg

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

Ncos⁡θ+fsin⁡θ−w=0Ncos⁡θ+μsNsin⁡θ=mgN(cos⁡θ+μssin⁡θ)=mgN=mgcos⁡θ+μssin⁡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*}

We also solve for N in equation 2.

Nsin⁡θ−μsNcos⁡θ=mv2rN(sin⁡θ−μscos⁡θ)=mv2rN=mv2r(sin⁡θ−μscos⁡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*}

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

mgcos⁡θ+μssin⁡θ=mv2r(sin⁡θ−μscos⁡θ)\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 μs\mu_s.

mgcos⁡θ+μssin⁡θ=mv2r(sin⁡θ−μscos⁡θ)rg(sin⁡θ−μscos⁡θ)=v2(cos⁡θ+μssin⁡θ)rgsin⁡θ−μsrgcos⁡θ=v2cos⁡θ+μsv2sin⁡θμsv2sin⁡θ+μsrgcos⁡θ=rgsin⁡θ−v2cos⁡θμs(v2sin⁡θ+rgcos⁡θ)=rgsin⁡θ−v2cos⁡θμs=rgsin⁡θ−v2cos⁡θv2sin⁡θ+rgcos⁡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*}

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

μs=rgsin⁡θ−v2cos⁡θv2sin⁡θ+rgcos⁡θμs=100 m(9.81 m/s2)sin⁡15.0∘−(5.5556 m/s)2cos⁡15.0∘(5.5556 m/s)2sin⁡15.0∘+100 m(9.81 m/s2)cos⁡15.0∘μs=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*}
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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 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⁡θ=v2rg\tan \theta = \frac{v^2}{rg}, we can solve for vv.

v=rgtan⁡θv = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

  • radius of curvature, r=100 mr=100\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g=9.81\ \text{m/s}^2
  • banking angle, θ=20.0∘\theta = 20.0 ^\circ

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

v=rgtan⁡θv=(100 m)(9.81 m/s2)(tan⁡20.0∘)v=18.8959 m/sv=18.9 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) \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 m/s18.9\ \text{m/s}.

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