A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius . 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 , how fast is the roller coaster traveling at the bottom of the dip

2√(gr)

Explanation:

At the top of the dip, the passenger feels the force of twice her weight push against her. This force is the normal force, N. This normal force also equals the centripetal force on her. So,

N = mv₁²/r   where m = mass of passenger, v₁ = velocity of passenger at the top of the dip and r = radius of dip. Since N = 2W where W is the weight of the passenger.

N =mv₁²/r

2W = mv₁²/r      (1)

From work-kinetic energy principles,

work done by gravity on passenger to reach bottom of dip = kinetic energy change of passenger at bottom of dip

Wr = 1/2m(v₂² - v₁²) where v₁ and v₂ are the initial and final velocities of the roller coaster, r is the radius of the dip and m is the mass of the passenger.

From (1) above, v₁² = 2Wr/m. Substituting this value for v₁² above, we have

Wr = 1/2m(v₂² - 2Wr/m)

2Wr/m = v₂² - 2Wr/m

v₂² = 2Wr/m + 2Wr/m

v₂² = 4Wr/m      

v₂ = √(4Wr/m)

v₂ = 2√(Wr/m)   W/m = g

v₂ = 2√(gr)

The roller coaster thus travels with a speed of 2√(gr) at the bottom of the dip