Have a physics problem of Circular Motion in Two Dimensions I need help with?

Question

A carnival clown rides a motorcycle down a ramp and around a vertical loop. IF theloop has a radius of 18m, what is the slowest speed the rider can have at the top of the loop to avoid falling?

Answer 1

x^2 / 18 = 9.81
solve for x

Answer 2

To "avoid falling" means that the motorcycle has to stay connected to the track. This means that the normal force (Fn, the perpendicular force that the track exerts on the motorcycle) must always be greater than zero (because if it drops to zero, that means we've lost contact with the track).

Use the "Fnet = ma" equation, and also the equation that describes acceleration for circular motion. Both equations have "a" in them; you can use that to combine the equations.

At the top of the circle:

Fnet = ma
(Weight + Normal Force) = ma
mg + Fn = ma

Acceleration in a circle:
a = v²/r

Combine the two equations:

mg + Fn = m(v²/r)

Solve for Fn:

Fn = m(v²/r − g)

That's the normal force at the top of the circle. We said this must be greater than or equal to zero, so:

m(v²/r − g) >= 0

Divide by m:

v²/r − g >= 0

Solve for v:

v >= sqrt(gr)

Answer 3

merely remember your kinematics equations, and remember that action interior the x direction could be separated from the action interior the y direction. (do no longer ignore to split your preliminary speed into x and y components, additionally)