A car moving at 18 m/s drives over the top of a hill...?

Question

A car moving at 18 m/s drives over the top of a hill. The top of the hill forms an arc of a vertical circle 124 meters in diameter.
•What is the centripetal force holding the car in the circle?
•What therefore is the normal force between the car's tires and the road?

Answer 1

This is basically question pertaining to motion on a circular ARC with banked curve.

v =18 m/s, r = radius of arc= 124/2 = 62 m , m= m kg (say)

When on circular ARC of of vertical circle, the gravitation force or its weight (mg) must provide foe the centripetal acceleration needed for car to remain in the circle.

m g = m (v^2 / r) [here force has been asked, but m not given]

(a) centripetal force:
m* (v^2 / r) = m (18*18 / 62) = 5.226*m Newton

(b) on banked curve, it is useful to resolve normal force along and perpendicular to the plane because here acceleration is towards the cent re of circle (horizontal acceleration).

if N is the normal force (perpendicular to plane) and theta is the angle between the road and horizontal, then N will make angle )theta) with line of weight (mg)

[N cos (theta)] = mg (canceled out) -----(A)

the net force:

[N sin (theta)] = m (v^2/r) = [mg /cos (theta)] sin (theta) =
= mg tan (theta) ----------(b)

normal force N = sqrt [ (mg)^2 + (mv2^/r)^2 ]
N = m * sqrt [ g^2 + (v2^/r)^2 ] using above CP force
N = m * sqrt [ ((9.8)^2 + (5.226)^2 ]
N = m * 11.106 Newton (m is not given)