A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If r = 19.0 m, how fast is the roller coaster traveling at the bottom of the dip?
that's an interesting question. Let me ask my professor, and I'll get back to you.
To find the speed of the roller coaster at the bottom of the dip, we need to use the concept of centripetal force and the relationship between force, mass, and acceleration.
First, let's break down the problem and identify the given information:
- Radius of the vertical circle (r) = 19.0 m
- Force felt by the passenger (F) = 2 times her weight
- Mass of the passenger (m)
From the problem, we know that the force felt by the passenger (F) is equal to twice her weight. In other words, F = 2mg, where m is the mass of the passenger and g is the acceleration due to gravity (9.8 m/s^2).
We can equate this force to the centripetal force required to keep the roller coaster moving in a circle at the bottom of the dip. The centripetal force is given by the formula Fc = mv^2/r, where v is the speed of the roller coaster at the bottom of the dip.
So, we have: 2mg = mv^2/r
Now, let's solve for v:
Divide both sides of the equation by m:
2g = v^2/r
Multiply both sides of the equation by r:
2gr = v^2
Take the square root of both sides:
√(2gr) = v
Now, substitute the given value of r = 19.0 m and the acceleration due to gravity g = 9.8 m/s^2 into the equation to find the speed (v) of the roller coaster:
v = √(2 * 9.8 * 19.0)
Calculating the value, we find:
v ≈ 19.85 m/s
Therefore, the roller coaster is traveling at approximately 19.85 m/s at the bottom of the dip.