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# The fast French train known as the T.G.V (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 $0.05g$, 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?

Last updated date: 13th Jun 2024
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Hint: This question can be solved by the use of the concept of centripetal force. Using the centripetal acceleration formula first we will evaluate the minimum radius of curvature for the track that can be tolerated and then by using the obtained value of the radius we will find the speed of the train using the same centripetal acceleration formula.
Formula used:
The centripetal acceleration
${a_c} = \dfrac{{{v^2}}}{r}$
Where $v$ is the speed and $r$ is the radius of curvature.

Complete step-by-step solution
In this question, the scheduled average speed of the train is given as $216km/h$, hence
$\Rightarrow v = 216km/h$
Now we have to change the unit into the $m/s$for our convenience, therefore
$\Rightarrow v = 216 \times \dfrac{{1000}}{{3600}}m/s$
$\Rightarrow v = 216 \times \dfrac{5}{{18}}m/s$
$\therefore v = 60m/s$
Now we will try to solve the given subparts one by one.
(A) In this subpart, it is given that the maximum magnitude of the acceleration experienced by the passengers is $0.05g$, where $g = 9.8m/{s^2}$which is the gravitational acceleration.
Now using these given values we will try to find the minimum radius of curvature which can be tolerated by the track. Hence we will consider the formula of centripetal acceleration ${a_c}$.
${a_c} = \dfrac{{{v^2}}}{{{r_{\min }}}}$
$\Rightarrow {r_{\min }} = \dfrac{{{v^2}}}{{{a_c}}}$
Now substituting the values of ${a_c} = 0.05 \times g$and the velocity given.
$\Rightarrow {r_{\min }} = \dfrac{{{{\left( {60m/s} \right)}^2}}}{{\left( {0.025 \times 9.8m/{s^2}} \right)}}$
$\Rightarrow {r_{\min }} = 7346.9m$
Hence the minimum radius of curvature that can be tolerated by the track is
$\therefore {r_{\min }} = 7.35km$.

(B) In the second subpart, it is given that the radius of curvature is $1km$ and we have to calculate the speed of the train at this radius, hence using the same formula of centripetal acceleration which is given as
${a_c} = \dfrac{{{v^2}}}{{{r_{\min }}}}$
Which in terms of speed is given as
${v^2} = {a_c} \times r$
$\Rightarrow v = \sqrt {{a_c} \times r}$
Now substituting the value of ${a_c} = 0.05 \times g$and $r = 1km = 1000m$, hence
$\Rightarrow v = \sqrt {0.05 \times 9.8 \times 1000}$
$\Rightarrow v = \sqrt {490} = 22.135m/s$
$\therefore v \approx 80km/h$
Hence the speed at which the train will go around the given curve of the radius $1km$ is given $v = 80km/h$.

Note: To solve this type of question one should be ensured to use the proper units. In this given question we first convert the given units in the CGS system which is in $m/s$ then again for the answer we again convert it into the MKS system $km/h$.