A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m. The acceleration of gravity is 9.8 m/s 2. 1.1 m 0.2 m B D C A 3.7 g Find the normal force pressing on the track at A, where A is at the same level as the center of the loop. Answer in units of N.

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 * 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g (h-r) = 1/2v²

v² = 2g (h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N = mv²/r

replacing the value for v² = 2g (h-r) in the above relation; we have:

Normal force = 2mg (h-r) / r

Normal force = 2 * 3.7 * 10⁻³ * 9.8 (1.1 - 0.2) / 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N