Circular Motion in a Roller Coaster

[iT95]

Homework Statement

In a certain loop-the-loop roller coaster ride, the coaster is given an intial speed of v ms-1. The coaster then rolls freely under the influence of gravity. The coaster travels down a 10m high slope before arriving at the start of the loop. It then travels the loop which has a diameter of 20m. Calculate the minimum value of v required for the coaster to make the loop.

Homework Equations

mv^2/r = mg
E(kinetic) = 0.5mv^2
E(potential) = mgh

The Attempt at a Solution

I found the increase in speed due to gravity using motion formulae
Assuming up is positive and down is negative,
v^2 = u^2 + 2as
v^2 = 0^2 + 2(-9.8)(10)
v = 14 ms-1 down

When the train reaches the loop it has completely horizontal velocity, therefore its velocity at the bottom of the loop is v+14 ms-1 horizontally

To successfully travel the loop, F(centripetal) must equal F(gravity)
Fc = Fg
mv^2/r = mg
v^2/r = g
v^2 = 10 x 9.8
v = 9.9 ms-1 at the top of the loop

v+14 ms-1 must be able to travel the 20m and still achieve a speed of 9.9ms-1 at the top
E(potential) = mgh
E(potential) = m x 9.8 x 20
E(potential) = 196m

Since Ep at top equals Ek at bottom,
Ep(top) = Ek(bottom)
196m = 0.5mv^2
196/0.5 = v^2
v = 19.8 ms-1

19.8ms-1 is the speed required to reach the top, 9.9ms-1 is the speed it must have at the top, therefore the total speed required is 19.8+9.9 =29.69ms-1

Since 14ms-1 is provided by the travel down the slope,
v = 29.69-14 = 15.7ms-1

Can someone tell me if i am right, and correct me if I am not? Your help is much appreciated =)

 

[Camille]

Since Ep at top equals Ek at bottom,
Ep(top) = Ek(bottom)
196m = 0.5mv^2
196/0.5 = v^2
v = 19.8 ms-1

What about Ek at the top?

 

[iT95]

That was added on after that calculation, 9.9ms-1 is the Ek at the top