Apparent Weight, Centripetal Force question

Question

An elementary student of mass m = 28 kg is swinging on a swing. The length from the upper bar of the swing set to the seat is L= 4.2 m. The child is attempting to swing all the way around in a full circle.

If the velocity at the bottom is the same as the velocity at the top from part (a) = 6.4m/s, what is the child's apparent weight, in newtons, at the very bottom of the path?

I cannot figure out the apparent weight, I would like some assistance please.. T_T

Answer 1

Cece,

When a student of mass m on a swing of length L passes through the lowest point with a velocity v , the rope, at the lowest point of the circle must provide a two part support for the swing

1. pull the swing upward to compensate the student's downward gravitational pull by Earth,

1. F_g=mg

(the swing, apparently, is assumed to be massless) PLUS

2. provide enough extra upward pulling force to ensure enough centripetal acceleration for the swing to stay on a circle of radius L while moving with the speed v ,

1. a_c=v²/L

so the extra pulling (centripetal) force would be

1. F_c=mv²/L

By the 3-d Newtonian Law, the rope would be pulled down by two forces : F_g=mg (which is the "true" weight of the student at rest) plus an additional force mv²/L. Added together, these two forces make what is called the apparent weight

Dr.Ariel B.