Question2. Modern roller coasters have vertical loops like the one shown here. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g ?

Studdy Solution
Solve for the speed $v$:
First, calculate $1.50 \times 9.81$:
$$1.50 \times 9.81 = 14.715 \, \text{m/s}^2$$
Now, substitute and solve for $v$:
$$\frac{v^2}{15.0} = 14.715$$
Multiply both sides by $15.0$:
$$v^2 = 14.715 \times 15.0$$
$$v^2 = 220.725$$
Take the square root of both sides to find $v$:
$$v = \sqrt{220.725}$$
$$v \approx 14.85 \, \text{m/s}$$
The speed of the roller coaster at the top of the loop is approximately:
$$\boxed{14.85 \, \text{m/s}}$$