# Could someone check if this is right soon.

1. Sep 27, 2009

### Ion1776

Could someone plz check if this is right soon. plz

What would the bank angle be for a circular racetrack with radius 120 m so that a car can go around the curve safely at a maximum of 25 m/s, without the help of frictional force to keep it on the road?

(25^2)/(120)/(9.80)=.53146

ArcSin(.53146)=32.1 Degrees

is this right, could someone check it for me

Thanks

2. Sep 27, 2009

### Delphi51

Re: Could someone plz check if this is right soon. plz

I would have used arctan rather than arcsin. Maybe you should take time to draw the vector diagram to be sure.

3. Sep 27, 2009

### Ion1776

Re: Could someone plz check if this is right soon. plz

ArcTan doesnt give you the right answer. Could you help me at all, more in depthly.

4. Sep 27, 2009

### Delphi51

Re: Could someone plz check if this is right soon. plz

5. Sep 27, 2009

### Ion1776

Re: Could someone plz check if this is right soon. plz

So is it

Arctan(5.208/9.81)=27.96

6. Sep 27, 2009

### Delphi51

Re: Could someone plz check if this is right soon. plz

Yes, that's what I got.

7. Sep 27, 2009

### Ion1776

Re: Could someone plz check if this is right soon. plz

One more thing, how did you get the 5.208

8. Sep 27, 2009

### Delphi51

Re: Could someone plz check if this is right soon. plz

Fc = mv^2/R = m*25^2/120 = 5.208*m
This is the centripetal or centrifugal force. From the point of view of the outside world, the force is inward needed to hold the car in circular motion. From the point of view of the car in circular motion, the force is outward, tending to push it out of circular motion into straight line motion.