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Q27 PE

Expert-verified
Found in: Page 224

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) What is the radius of a bobsled turn banked at $75.0°$and taken at $30.0\text{m}/\text{s}$, assuming it is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem large to you?

(a) The radius is $24.6m$.

(b) $\text{36.6 m/s}$ is centripetal acceleration.

(c) Yes, the centripetal acceleration is $3.7$times greater than gravity's acceleration.

See the step by step solution

## Step 1: Definition of Centripetal acceleration

The rate of change of transverse velocity is known as centripetal acceleration.

## Step 2: Calculating radius of bobsled turn

(a) The ideal velocity for a banked road is,

$v=\sqrt{Rg\mathrm{tan}\theta }$

Here, R is the radius of the turn, g is the acceleration due to gravity, and $\theta$ is the ideal angle of banking.

Rearranging the above equation in order to get an expression for the radius of the turn,

$R=\frac{{v}^{2}}{g\mathrm{tan}\theta }$

Substitute $\text{30.0 m/s}$ for v,${\text{9.8 m/s}}^{\text{2}}$ for g, and $75.{0}^{0}$ for $\theta$,

$R=\frac{{\left(30.0m/s\right)}^{2}}{\left(9.8m/{s}^{2}\right)×\mathrm{tan}\left(75.{0}^{0}\right)}\phantom{\rule{0ex}{0ex}}=24.6m$

Here, the radius of the bobsled turn is $=24.6m$

## Step 3: Calculating centripetal acceleration

(a) The centripetal acceleration is,

${\text{a}}_{\text{c}}\text{=}\frac{{\text{v}}^{\text{2}}}{\text{r}}$

Substitute $\text{30.0 m/s}$ for v , and $\text{24.6 m}$ for r,

${\text{a}}_{\text{c}}\text{=}\frac{{\left(\text{30 m/s}\right)}^{\text{2}}}{\left(\text{24.6 m}\right)}\phantom{\rule{0ex}{0ex}}{\text{= 36.6 m/s}}^{\text{2}}\phantom{\rule{0ex}{0ex}}$

Hence, the centripetal acceleration is $\text{36.6 m/s}$.

## Step 4: Determining is acceleration seem larger

The proportion of centripetal to gravitational acceleration is,

$\frac{{\text{a}}_{\text{c}}}{\text{g}}\text{=}\frac{{\text{36.6 m/s}}^{\text{2}}}{{\text{9.8 m/s}}^{\text{2}}}\phantom{\rule{0ex}{0ex}}\text{= 3.7}\phantom{\rule{0ex}{0ex}}\text{= 3.7g}\phantom{\rule{0ex}{0ex}}$