The radius of curvature of a loop-the-loop for a roller coaster is 11.8 m. At the top of the loop (with the car inside the loop), the force that the seat exerts on a passenger of mass m is 0.80mg. Find the speed of the roller coaster at the top of the loop.
(g - v^2/r) = .8
To find the speed of the roller coaster at the top of the loop, we can use the centripetal force and gravitational force acting on the passenger.
At the top of the loop, the centripetal force is provided by the normal force (N) of the seat acting towards the center of the loop, and the gravitational force (mg) acting downward. We can write an equation for the net force acting on the passenger:
Net force = Centripetal force + Gravitational force
The centripetal force can be calculated using the formula:
Centripetal force = mass × (velocity^2 / radius of curvature)
Let's denote the speed of the roller coaster at the top of the loop as v.
Centripetal force = m × (v^2 / r), where r is the radius of curvature.
Now, let's substitute the given values into the formula:
0.80mg = m × (v^2 / 11.8)
mg = (v^2 / 11.8)
Next, we can cancel out the mass (m) from both sides of the equation:
g = v^2 / 11.8
To solve for the speed (v), we multiply both sides of the equation by 11.8:
11.8g = v^2
Finally, we can take the square root of both sides to find the speed (v):
v = √(11.8g)
Now, plug in the value for acceleration due to gravity (g) which is approximately 9.8 m/s^2:
v = √(11.8 × 9.8)
v ≈ √115.64
v ≈ 10.76 m/s
Therefore, the speed of the roller coaster at the top of the loop is approximately 10.76 m/s.