Can a Roller Coaster's Loop Height Affect the Force on the Track?

[hoseA]

A roller coaster car of mass 1216 kg slides on
a frictionless track starting at a distance 32 m
above the bottom of a loop 25 m in diameter.
The acceleration of gravity is 9.81 m/s^2

If friction is negligible, what is the magni-
tude of the force of the track on the car when
the car is at the top of the loop? Answer in
units of N.

Fn=(mv^2) /R - Fg

mgh = .5mv^2

v^2= 2gh = 2*9.81*32 = 627.84

Fn = (1216*627.84)/12.5m - (1216*9.81)

= 49147.3152N ?

apparently this is wrong. What did i do wrong?

 

[Doc Al]

Your error was in calculating the KE at the top of the loop:
mg \Delta h = 1/2 m v^2

\Delta h is not 32 m!

 

[mukundpa]

v^2= 2gh = 2*9.81*32 = 627.84 ...

is this the velocity at the top.

 

[mukundpa]

A bit slow in typing then Doc Al.

 

[hoseA]

Your error was in calculating the KE at the top of the loop:
mg \Delta h = 1/2 m v^2
\Delta h is not 32 m!

I presume delta h is then 32m - 25m ?

Thanks for the help.

 

[Doc Al]

I presume delta h is then 32m - 25m ?

That's correct.