In a physics "loop the loop"...?

Question

the height at which it is dropped must be equal to 2.7 times the radius of the loop... from where is this 2.7 derived?

Answer 1

At the highest point h of the loop the upward centrifugal force must be at least equal to the downward force of gravity acting on the ball to keep it on the rail so we must have:
mg = m*v^2/R so v^2 = g*R

note also we have h = 2*R;

At this point the total Energy will be:
E = K(translational) + K(rotational) + V
with
K(translational) = 0.5*m*v^2 = 0.5*m*g*R,
V = m*g*h =m*g*2*R and
K(rotational) = 0.5*I*w^2
with the moment of inertia of the ball I = 0.4*m*r^2 and its angular speed w = v/r with r being the radius of the ball: Note that a point on the surface of the ball is indeed moving at speed v, the speed at which the ball is moving along the rail.
Then K(rotational) = 0.5*(0.4*m*r^2)*(v^2/r^2) = 0.2*m*g*R and

E = 0.5*m*g*R + 0.2*m*g*R + 2*m*g*R
= 2.7*m*g*R
Which must be equal to the initial energy m*g*H where H is the height at which the ball is dropped, so
m*g*H = 2.7*m*g*R
-> H = 2.7*R
CQFD

Answer 2

Say H is the height the car is dropped from. Thus, its total energy before release is TE(H) = PE + KE = mgH, which is just the potential energy; where m = mass of the car and passengers, because KE, kinetic energy, is zero before the car drops.

Because I have no idea what the "radius of the loop" is, I have to assume the radius of curvature axis is at the bottom of the loop so that the height from the bottom to the top (h) of the loop and the radius (R) of curvature at the top are equal. That is, I assume h = R. If that's not the case, work the problem as I've done here but put in the correct relationship between h and R.

To stay upside down on the top of the loop of radius R, the net force acting on the car must be f = ma = 0 = W - C = mg - mv^2/R; thus a, the radial acceleration, = 0 when the loaded weight of the car W is offset by centrifugal force C.

Therefore, g <= v^2/R and v^2 >= gR must be true for the car/passengers to stay on the track upside down at the top of the loop. The required total energy at the top of the loop to keep the car and passengers from falling TE(h) = mgh + 1/2 mv^2 = mgh + 1/2 mgR.

From the conservation of energy law, we know that TE(H) = mgH = mgh + 1/2 mgR = TE(h). That is, as energy is neither created nor destroyed, the total eneries at the two points on the roller coaster path have to be equal. [Note: friction force and consequent work energy losses are ignored as they were not given.]

From above, we have H = h + R/2 = R + R/2 = 1.5 R. Thus, as I have assumed h = R, the height H needs to be 1.5 times the radius of curvature for the top of the loop...not 2.7 times.

[NB: My observation of actual loop the loop roller coasters is that the "loop" is not a circle; so that the axis of the radius of curvature is not the center of the loop. If the height of the top of the loop is h = 2R; H = 2.5 R and still not the 2.7 R you are looking for. Anyway, it should be clear how to work this problem once you have the height at the top of the loop in terms of radius of curvature R.]

Answer 3

You need for the centrifugal acceleration at the top of the loop to equal (or exceed) the gravitational acceleration:

v^2/r > g
v > sqrt (gr)

The kinetic energy at that point equals the potential energy difference between that point and the top of the loop.
KE = 1/2 mv^2 = delta PE = mg(h - 2r)
1/2 mgr < mg(h-2r)
h > 2.5r
The extra bit to get to 2.7 must be to account for friction in the loop (or maybe some kind of rotational KE if the wheels are heavy compared to the cart), which I ignored in my analysis.