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!

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