How Small Can the Track's Radius Be for Safe Train Speeds?

[Oliviam12]

Homework Statement

A train has a speed of v = 66.5 km/h. If the acceleration experienced by the passengers is to be less than 0.35 g, (g = 9.81 m/s2), find the smallest radius of curvature R acceptable for the track.

Homework Equations

A=v^2/r
a=(mv^2)/r

The Attempt at a Solution

I tried putting 10.16 m/s^2 (added 9.81 + .35) for the acceleration and 18.4722 m/s for the velocity and putting it in the first equation to get 33.5848 m, however, this answer is close but not correct... I think the problem is, I don't understand what to do with the gravity? Is their another equation I am suppost to use or what? or am I only suppost to use .35 m/s/s as my acceleration? Please help, I cannot find any other simular examples in the text; the text only has examples involving tension and mass.

 

[D H]

That 0.35 is a unitless scale factor. You can't add it to 9.81 m/s^2. The problem says "0.35 g" which means "0.35 * g", not "0.35 + g". Multiply!

Carrying out this multiplication will give you the limit on the tolerable centripetal acceleration.