Tag Archives: ideal speed

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*}
Advertisements
Advertisements

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 tan⁡θ=v2rg\displaystyle \tan \theta = \frac{v^2}{rg}. We can solve for rr in terms of all the other variables, and we should come up with

r=v2gtan⁡θr = \frac{v^2}{g \tan \theta}

We are given the following values:

  • ideal banking angle, θ=75.0 ∘\displaystyle \theta = 75.0\ ^\circ
  • linear speed, v=30.0 m/s\displaystyle v=30.0\ \text{m/s}
  • acceleration due to gravity, g=9.81 m/s2\displaystyle g=9.81\ \text{m/s}^2

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

r=v2gtan⁡θr=(30.0 m/s)2(9.81 m/s2)(tan⁡75∘)r=24.5825 mr=24.6 m  (Answer)\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 m24.6\ \text{m}.

Part B

The centripetal acceleration aca_c can be solved using the formula

ac=v2ra_c = \frac{v^2}{r}

Substituting the given values, we have

ac=v2rac=(30.0 m/s)224.5825 mac=36.6114 m/s2ac=36.6 m/s2  (Answer)\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 m/s236.6\ \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.

acg=36.6114 m/s29.81 m/s2=3.73\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 ac=3.73ga_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!

Advertisements
Advertisements

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}.

Advertisements
Advertisements