Maximum Speed for Successful Roller Coaster Hill Climb

[shell4987]

Homework Statement

A roller-coaster car has a mass of 1050 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 19 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.7 m/s? (b) What is FN if v = 18 m/s?

Homework Equations

fynet= mg-FN=ma
a= v squared/r

put together: FN= mg-mv squared/r

The Attempt at a Solution

I've tried this problem and put g as -9.8m/s squared and got -14,472.9 for part (a) and used the same gravity again for part (b) and got -28,195.26... i also did use the positive g (+9.8) and got the wrong answers again. Is there something I am doing wrong?

 

[Doc Al]

I don't see how you got your answers from those equations. (In what direction does the normal force act?)

 

[dynamicsolo]

fynet= mg-FN=ma
a= v squared/r

put together: FN= mg-mv squared/r

The equation you derived for the normal force (presumably from a free-body diagram for the car) is correct. However, since you have already taken the directions of the forces along the radius of the circle into account, you don't want to put in (-g) here; you only need to use magnitudes at this point.

Your result for FN should be positive then, since it is a magnitude. However, if the car goes fast enough, FN could drop to zero, meaning that the car is just breaking contact with the rails; if you get a negative answer for the magnitude, that is actually nonsense, but it means the car is going too fast to stay on the hill.

[edit: I've tried this out with the values given. It would be a good idea for you to show your calculation for part (a) to start with.]

 

[Doc Al]

Good catch, dynamicsolo! FYI: g just stands for the magnitude of the acceleration due to gravity; g = 9.8 m/s^2.

 

[shell4987]

Your result for FN should be positive then, since it is a magnitude. However, if the car goes fast enough, FN could drop to zero, meaning that the car is just breaking contact with the rails; if you get a negative answer for the magnitude, that is actually nonsense, but it means the car is going too fast to stay on the hill.

[edit: I've tried this out with the values given. It would be a good idea for you to show your calculation for part (a) to start with.]

Thanks for your help... here is what I put in for the values of (a) with the equation FN=mg-m(vsquared/r) FN=1050(9.8)-(1050(8.7)^2/19)... I ended up getting 6107.132 as my answer... I don't know if that sounds correct? Also for part (b) I used the same equation and put FN=1050(9.8)-(1050(18)^2/19) and I ended up getting -7615.26. I feel as if those do not sound right at all? What do you think?

 

[Sourabh N]

If your calculations are correct then they sound perfect (atleast to dynamicsolo) :)

 

[dynamicsolo]

Thanks for your help... here is what I put in for the values of (a) with the equation FN=mg-m(vsquared/r) FN=1050(9.8)-(1050(8.7)^2/19)... I ended up getting 6107.132 as my answer... I don't know if that sounds correct? Also for part (b) I used the same equation and put FN=1050(9.8)-(1050(18)^2/19) and I ended up getting -7615.26. I feel as if those do not sound right at all? What do you think?

The units on those are, of course, Newtons. I get substantially the same results (I used g = 9.81). Your negative result for part (b) tells us that the car would not stay on the rails if it were going that fast (FN cannot be less than zero); instead, it would have flown off the hill before reaching the summit.

A question often asked on exams or homework problems is to find the greatest speed the car could have to still be able to make it over the top without flying off. (That would be when FN at the summit just equals zero.) We now know that the answer is between 8.7 and 18 m/sec. ...