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Q108P

Expert-verifiedFound in: Page 91

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of${\mathbf{216}}{\mathbf{km}}{\mathbf{/}}{\mathbf{h}}$ ****. (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to ${\mathbf{0}}{\mathbf{.}}{\mathbf{050}}{\mathbf{g}}$***,*** what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a ${\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathbf{km}}$**** radius to be at the acceleration limit?**

a) Smallest radius of curvature of the trackis $7.3km$.

b) Maximum allowed speed of the train is$80km/h$.

1) Average speed of the train is $216km/h$

2) Acceleration experience by the passengers is limited to $0.050g$

**When an object moves in a circular orbit, it will experience a centripetal acceleration. The magnitude of this acceleration depends on the speed of the object as well as the radius of curvature. Using the formula for centripetal acceleration, we can find the radius of the curvature of the track if the velocity is known, or we can find the velocity if the radius of curvature is known.**

Formula:

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

Where ${a}_{c}$is the centripetal acceleration, $v$is the average speed, and $r$is the radius of curvature.

First, we need to convert the units of given speed from km/h to m/s

So,

$\mathrm{v}=216\frac{\mathrm{km}}{\mathrm{h}}\phantom{\rule{0ex}{0ex}}=216\frac{\mathrm{km}}{\mathrm{h}}\times \frac{1000\mathrm{m}}{3600\mathrm{s}}\phantom{\rule{0ex}{0ex}}=60\mathrm{m}/\mathrm{s}$

To determine the smallest radius of curvature, we need to use formula forcentripetal acceleration.

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

Rearranging,

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

Substitute $60\mathrm{m}/\mathrm{s}$the for *v*, and$9.8\mathrm{m}/{\mathrm{s}}^{2}$ for *g* in the above equation.

$=\frac{{\left(60\mathrm{m}/\mathrm{s}\right)}^{2}}{0.050\times 9.8\mathrm{m}/{\mathrm{s}}^{2}}\phantom{\rule{0ex}{0ex}}=7347\mathrm{m}\phantom{\rule{0ex}{0ex}}=7.3\mathrm{km}$

Therefore, the smallest radius of curvature is$7.3\mathrm{km}$ .

Now, the radius of curvature available is given as,

$\mathrm{r}=1\mathrm{km}\phantom{\rule{0ex}{0ex}}=1000\mathrm{m}$

${a}_{c}=\frac{{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}{v}^{2}={a}_{c}\times r$

Substitute the values,

$=0.050\times 9.8\mathrm{m}/{\mathrm{s}}^{2}\times 1000\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{v}=22.1\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\approx 80\mathrm{km}/\mathrm{h}$

Hence, the speed at the curvature should be approximately $80\mathrm{km}/\mathrm{h}$.

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