# Find Speed in Centripetal Motion

1. Nov 25, 2007

### petern

The radius of curvature of a loop-to-loop roller coaster is 12 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.4mg. Find the speed of the roller coaster at the top of the loop. Answer: v = 12.83 m/s.

I assume .4 mg is the torque, right? I have no clue what to do. Please help.

Last edited: Nov 25, 2007
2. Nov 25, 2007

### rock.freak667

At the top of the loop...what forces provide the centripetal force? the weight and the force that the seat exerts....and so....you can find v....

3. Nov 25, 2007

### petern

What do I do with the mg?

4. Nov 25, 2007

### rock.freak667

Well mg is just the formula to find weight...so that the force exerted by the seat is 0.4*(the weight of the seat)

so that weight of seat+Force exerted by seat, provides the centripetal force

5. Nov 25, 2007

### petern

So I would use the equation F = ma and a = v^2/r. I wouldn't use v = (2*pi*r)/(T) because T = period and that info is not provided. But I still don't understand what I'm suppose to do with the m and .4mg.

6. Nov 25, 2007

### rock.freak667

ok..seems you aren't getting this...

Eq'n: Centripetal Force,$F_c=\frac{mv^2}{r}$

The weight of the seat(mg) + The force that the seat exerts on a passenger(0.4mg) provides the centripetal force.

This means that

$$F_c=mg+.4mg$$

since $F_c=\frac{mv^2}{r}$

then
$$\frac{mv^2}{r}=mg+.04mg$$

What cancels out...and you have the value of 'r' and 'g'

7. Nov 25, 2007

### petern

Solved! Thank you sooo much.