Motorcycles- Dynamics of Uniform Circular Motion! please help?

Question

A motorcycle has a constant speed of 25m/s as it passes over the top of a hill whose radius of curvature is 126m. The mass of the motorcycle and driver is 342kg. Find the magnitude of
(a) The centripetal force
(b)the normal force that acts on the cycle

A motorcycle is traveling up the sude of a hill and down the other side. the crest is a circular arc with a radius of 45m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.

Thank You!

Answer 1

Centripetal force is m*(v^2/r) = 342*(625/126)N =1696.43N

I would think that the normal force would be force due to gravity minus the centripetal force, which would be:

1655.17N

For the second gravity would have to be more than the centripetal force:

g > v^2/r (I factored out mass)

9.8m/s^2 > v^2/45m
441m^2/s^2 > v^2

v > + or - 21m/s

Answer 2

The centripetal is
m*v^2/r
=342*25^2/126
=1696 N

the normal force is
m*g-m*v^2/r
=1659 N

for the second one
m*g=m*v^2/r
v=sqrt(r*g)
=sqrt(45*9.81)
21 m/s
j