Astunt man is attempting to ride his bike through the loop of a track. the
combined mass of the stunt man and his bike is 101 kg. (a) if the radius of
the loop is 10.1 m and he is traveling at 22.5 m/s, what is the normal force
from the track when he is at the top of the loop?
(b.) what is the minimum speed the stunt man can be traveling and still
make the loop?

(a) 4070 N

(b) 9.95 m/s

Explanation:

(a) Draw a free body diagram of the stunt man when he is at the top of the loop.  There are two forces: normal force N pushing down and weight force mg pulling down.

Sum of forces in the radial (-y) direction:

∑F = ma

N + mg = m v² / r

N = m v² / r − mg

N = (101 kg) (22.5 m/s)² / (10.1 m) − (101 kg) (9.8 m/s²)

N = 4070 N

(b) At the minimum speed, there is no normal force at the top of the loop.

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

v = √(9.8 m/s² × 10.1 m)

v = 9.95 m/s

momentum of soft ball three times less than baskey ball