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Q27 PE

Expert-verifiedFound in: Page 224

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 $75.0\xb0$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.

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

(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 $,

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

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

(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 $\text{36.6 m/s}$.

The proportion of centripetal to gravitational acceleration is,

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

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