Design a Roller Coaster: Calculate Track Radius for 3800 N Force

[runningirl]

Homework Statement

You have been hired to design a roller coaster and are working on two sections. The roller coaster rolls on the track, but is not attached.

If the cars will have a speed of 35 m/s at the bottom of the hill, what is the radius that the circular bottom section of the track should have so that the seat exerts a force of 3800 N on an 80 kg rider?

Homework Equations

F=mv^2/r

The Attempt at a Solution

i did:
(80*35^2/R)-9.8(80)=3800
98000/R=4584
R=98000/4584
=21.38 m?

but apparently i did something wrong and got the wrong answer.
can someone help me out??

 

[rock.freak667]

The Attempt at a Solution

i did:
(80*35^2/R)-9.8(80)=3800

At the bottom you would have mv²/R = 3800-9.8(80) (as both the reaction and the weight act in opposite directions).

 

[Dick]

Why do you have a minus sign between the two terms? The seat needs to counter both gravity AND the supply the centripetal acceleration. Shouldn't it be plus?

 

[runningirl]

at the bottom of the roller coaster it should be a minus sign instead of a plus.

 

[Dick]

at the bottom of the roller coaster it should be a minus sign instead of a plus.

Why do you think that?

 

[runningirl]

There was a second part to the question.

b) In the second section the coaster goes over a hill with radius 10 m. You do some calculations and find that the speed at the top of the hill is going to be 12 m/s. Is this speed acceptable? Why or why not?
F=mv^2/r
=144*80/10
=1152
i said no because mg=9.8(80)=784 and mg>1152 to work.
1152=mg-Fn... so if Fn>0?!

 

[Dick]

There was a second part to the question.

b) In the second section the coaster goes over a hill with radius 10 m. You do some calculations and find that the speed at the top of the hill is going to be 12 m/s. Is this speed acceptable? Why or why not?
F=mv^2/r
=144*80/10
=1152
i said no because mg=9.8(80)=784 and mg>1152 to work.
1152=mg-Fn... so if Fn>0?!

Fn must have a different sign at the top of the roller coaster than on the bottom shouldn't it?

 

[runningirl]

yeah.. it should be plus...
but i know it doesn't work since Fn>0 to work...

 

[runningirl]

though my teacher said something like 1152=mg-Fn?

 

[Dick]

yeah.. it should be plus...
but i know it doesn't work since Fn>0 to work...

ma=Fn+Fgrav where you think of the symbols as representing forces with a direction. Put a=v^2/r, Fgrav=-mg and let Fn be the magnitude of the normal force. We'll indicate direction (up or down) by the sign of the quantity. At the bottom the centripetal acceleration is upwards and Fn is upwards. So putting plus signs on those quantities, we get ma=Fn-mg or Fn=ma+mg. At the top the centripetal acceleration is downwards and Fn needs to be downwards if you are going to stay in the seat. So we put minus signs on ma and Fn. So -ma=-Fn-mg. Or Fn=ma-mg.