Curvature/centrifugal force problem

[TroyWeathers]

Hey everyone,

I am stuck on a problem that seems so very simple, but I can not get the correct answer on this. I would appreciate anyone who would solve this and/or tell me what I am doing wrong. Sometimes I make stupid mistakes. I know it is probably simple, but just hit a wall on this one.

Problem:A train is running smootly on a curved track at the rate of 50 m/s. A passenger standing on a set of scales observes that his weight is 10% greater than when the train is at rest. The track is banked so that the force acting on the passenger is normal to the floor of the train. What is the radius of curvature of the track?

I am told the final correct answer is 556m.

My attempt is this : Centrifugal force = mg

So, since he is 10% more weight than at rest I have: mg + .10mg = mv^2/r. where m = mass, g= acceleration due to gravity, and r= the radius of the track.

1.1mg = mv^2/r. The masses cancel giving 1.1g = v^2/r. Solving for r gives:

r= v^2/1.1g, which gives me 231.9m for r which isn't correct. Where have I went wrong? Thanks for the help!

 

[PhanthomJay]

Hey everyone,

I am stuck on a problem that seems so very simple, but I can not get the correct answer on this. I would appreciate anyone who would solve this and/or tell me what I am doing wrong. Sometimes I make stupid mistakes. I know it is probably simple, but just hit a wall on this one.

Problem:A train is running smootly on a curved track at the rate of 50 m/s. A passenger standing on a set of scales observes that his weight is 10% greater than when the train is at rest. The track is banked so that the force acting on the passenger is normal to the floor of the train. What is the radius of curvature of the track?

I am told the final correct answer is 556m.

My attempt is this : Centrifugal force = mg

This is wrong. What you are looking for is the centripetal force which acts horizontally toward the center of the curve. There are just 2 forces acting on the car, the weight, which acts vertically down, and the normal force (the scale reading), which acts perpendicular to the banked rails.

So, since he is 10% more weight than at rest I have: mg + .10mg = mv^2/r. where m = mass, g= acceleration due to gravity, and r= the radius of the track.

1.1mg = mv^2/r. The masses cancel giving 1.1g = v^2/r. Solving for r gives:

r= v^2/1.1g, which gives me 231.9m for r which isn't correct. Where have I went wrong? Thanks for the help!

It is the horizontal component of the Normal force that provides the centripetal force. You should be able to get that with some trig by using Newton 1 in the vertical y direction, and the centripetal force equation (Newton 2) in the horizontal x direction.

 

[TroyWeathers]

Thank you very much. I was making it too simple. That's weird because I usually make things way too hard. I will take a look at it and rework it. Thank you again very much.