The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h.

(a) If the train goes around a curve at that speed and the acceleration experienced by the passengers is to be limited to 0.060g, 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 0.50 km radius to be at the acceleration limit?

a) Use the formula a=(v^2)/r

a=.060g
g=9.8 m/(s^2)
so a=.588 m/s^2
v=216km/hr * 1hr/3600s * 1000m/1km = 60m/s
.588=(60^2)/r
solve for r, r = 6122.44898

b)Use the formula a=(v^2)/r
a=.060g
g=9.8 m/(s^2)
so a=.588 m/s^2
r=.50km * 1000m/1km =500m
.588=(v^2)/500
solve for v, v= 17.146 m/s

(a) To find the smallest radius of curvature that can be tolerated, we need to use the centripetal acceleration formula:

a = v^2 / r

where:
a = acceleration
v = velocity
r = radius of curvature

Given:
v = 216 km/h = 60 m/s (converting km/h to m/s)
a = 0.060g = 0.060 * 9.8 m/s^2 (converting g to m/s^2)

Substituting the given values into the formula, we can solve for r:

0.060 * 9.8 = (60^2) / r

0.588 = 3600 / r

r = 3600 / 0.588

r ≈ 6122.45 meters

Therefore, the smallest radius of curvature that can be tolerated is approximately 6122.45 meters.

(b) We can use the same centripetal acceleration formula to find the velocity at the acceleration limit:

a = v^2 / r

Given:
a = 0.060g = 0.060 * 9.8 m/s^2
r = 0.50 km = 500 meters

Substituting the given values into the formula, we can solve for v:

0.060 * 9.8 = v^2 / 500

0.588 = v^2 / 500

500 * 0.588 = v^2

v^2 ≈ 294

v ≈ sqrt(294)

v ≈ 17.15 m/s

Therefore, the train must go around the curve at a speed of approximately 17.15 m/s to be at the acceleration limit.

To solve these problems, we need to apply some principles from physics and use some equations related to circular motion.

(a) The first step is to determine the acceleration experienced by the passengers. We are given that the acceleration is limited to 0.060g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the maximum acceleration allowed is 0.060 * 9.8 = 0.588 m/s^2.

Next, we need to relate the acceleration, speed, and radius of curvature using the centripetal acceleration formula:

Ac = v^2 / r

where Ac is the centripetal acceleration, v is the velocity (speed), and r is the radius of curvature.

We are given the speed of the train, which is 216 km/h. To determine the radius of curvature, we need to convert the speed from km/h to m/s by dividing by 3.6:

v = 216 km/h / 3.6 = 60 m/s

Now, we can rearrange the centripetal acceleration formula to solve for the radius:

r = v^2 / Ac

Substituting the values we have:

r = (60 m/s)^2 / 0.588 m/s^2 ≈ 614 m

Therefore, the smallest radius of curvature for the track that can be tolerated is approximately 614 meters.

(b) In this case, we are given the radius of curvature for the curve, which is 0.50 km. We need to find the speed at which the acceleration is limited to 0.060g.

Using the same centripetal acceleration formula:

Ac = v^2 / r

We know the centripetal acceleration limit is 0.060g. Substituting the values:

0.060g = v^2 / 0.50 km

To convert the radius to meters, we multiply by 1000:

0.060g = v^2 / (0.50 km * 1000 m/km)

Simplifying:

v^2 = 0.060g * 0.50 km * 1000 m/km

v^2 = 0.030g * 1000 m

Now, we can solve for the speed by taking the square root of both sides:

v = √(0.030g * 1000 m)

Substituting the value of g (approximately 9.8 m/s^2):

v ≈ √(0.030 * 9.8 m/s^2 * 1000 m)

v ≈ √(294 m^2/s^2)

v ≈ 17.14 m/s

Therefore, the train must go around the curve at approximately 17.14 m/s (or approximately 61.7 km/h) to be at the acceleration limit.