What Determines the Force at the Bottom of a Circular Loop?

[darksyesider]

[Solved] bottom of hill

Homework Statement

No friction, starts at rest.
What is the force factor at the bottom of the circular loop?
(see attachment)

Homework Equations

Conservation of Energy Equations

The Attempt at a Solution

EDIT: Solved below (incorrect solution was here)

 

[tiny-tim]

hi darksyesider!

Then from here I don't know what to do since the radius is not given. (see diagram)

yes, you certainly need to know the radius

i'll guess it's supposed to be 2h

 

[oneplusone]

I think this problem is missing the radius. I don't see how else to solve it.

 

[tiny-tim]

yes try 2h

 

[darksyesider]

So is the answer 5 times?

I got:

$$F_N = \dfrac{m(2gy)}{r}+mg$$

$$= mg(\dfrac{2y}{r}+1)$$

Since force factor = Fa/Fg

$$ff = \dfrac{mg(\dfrac{2y}{r} + 1 )}{mg}$$

Substituting in y = 2r we get 5 times.

 

[tiny-tim]

if force factor = Fa/Fg, yes

 

[darksyesider]

Sorry, but is that the incorrect definition? :(
I can't find any other definition of it…
And thanks a lot for the help!

 

[tiny-tim]

Sorry, but is that the incorrect definition?

i've never heard of it before

but I'm happy to take your word for it … and if it is, your answer looks fine