Question

The fast French train known as the TGV (Train Grande Vitesse) has a scheduled average speed of $216km/h$.
(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 0.050g, 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 1.00km radius to be at the acceleration limit?

Answer 1

Hint: Both the subparts in the given question could be solved using the expression for centripetal acceleration. In the first part, we could substitute the given limit of acceleration and the average speed to get the minimum radius that can be tolerated. Now for the second part, substitute the new value of radius and the same acceleration limit to get the speed of the train. Also, take $g=9.8m{{s}^{-2}}$.

Formula used: Expression for centripetal acceleration,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$

Complete step by step answer:
In the question we are given the fast French train known to be the TGV whose scheduled average speed is given as$216km/h$, that is,
$v=216km{{h}^{-1}}$
$\Rightarrow v=216\times \dfrac{1000}{3600}m{{s}^{-1}}$
$\therefore v=60m{{s}^{-1}}$
This statement is followed by two subparts, so, let us approach them one by one.
(a) Here the maximum acceleration experienced by the passengers is given as 0.050g, where g is the acceleration due to gravity given by,
$g=9.8m{{s}^{-2}}$
Then, we are asked to find the smallest radius of curvature for the track that can be tolerated. Let us recall the expression for centripetal acceleration which given by,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$
Substituting the given values,
$r=\dfrac{{{\left( 60m{{s}^{-1}} \right)}^{2}}}{0.050\times 9.8}$
$\Rightarrow r=7346.9m$
$\therefore r=7.35km$
So we found the smallest radius of curvature for the track that can be tolerated to be 7.35km.
(b) Here, the radius of curve is given as,
$r=1km=1000m$
We are supposed to find the speed of the train at the given acceleration limit.
We could use the same expression as before, that is,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}$
$\Rightarrow v=\sqrt{{{a}_{c}}\times r}$
$\Rightarrow v=\sqrt{0.050\times 9.8\times 1000}$
$\Rightarrow v=\sqrt{490}=22.1m{{s}^{-1}}$
$\therefore v\approx 80km{{h}^{-1}}$
Hence, we found the speed that the train will go around a curve of 1.00km radius within the acceleration limit to be approximately $80km{{h}^{-1}}$.

Note: You may have noted that the values of the quantities are given in different units in the question. So we will have to stick to any one of the given units. In the above solution we have chosen the SI unit system to solve the problem but the final answer is again converted back to $km$ and $km{{h}^{-1}}$ respectively in each part.