A certain roller coaster has a circular, vertical loop of radius 10 meters, such that at its top the riders are upside down. How fast does the roller coaster move at this point if the riders feel perfectly weightless when passing the top of the loop?

To find the speed of the roller coaster at the top of the loop, we can use the concept of Centripetal Force.

At the topmost point of the vertical loop, the riders feel weightless. This means that the net force acting on them is zero, since weight is balanced by the normal force.

The only force acting on the riders at the top of the loop is the normal force (N). At the very top, the normal force provides the centripetal force required to keep the riders moving in a circular path.

The centripetal force is given by the equation:

Fc = m * ac

Where Fc is the centripetal force, m is the mass of the riders, and ac is the centripetal acceleration.

At the very top of the loop, the normal force is equal to the weight of the riders (mg), where g is the acceleration due to gravity.

So, Fc = mg

At the top of the loop, the centripetal acceleration (ac) is given by:

ac = v^2 / R

Where v is the velocity of the roller coaster at the top of the loop, and R is the radius of the loop.

Since the centripetal force equals the weight:

mg = mv^2 / R

Simplifying and canceling out the mass:

g = v^2 / R

Rearranging the equation:

v^2 = g * R

Now, we can solve for the velocity (v):

v = sqrt(g * R)

Plugging in the values, where the radius (R) is 10 meters and the acceleration due to gravity (g) is approximately 9.8 m/s^2:

v = sqrt(9.8 * 10)
v = sqrt(98)
v ≈ 9.90 m/s

So, the roller coaster must move at approximately 9.90 meters per second at the top of the loop for the riders to feel weightless.

I will be glad to critique your thinking.

Pass on the message to the other aliases that zach uses.