physics question : about gravitational force?

Question

Describe all the forces acting on a child riding a horse on a merry-go-round. Which of these forces provides the centripetal acceleration of the child?

Answer 1

There are three major forces: gravity, centripetal, and drag.

Gravity determines the weight of the child from W = mg; where m is the kid's mass and g is acceleration due to gravity.

Drag is the push back by the wind as the kid rotates around the MGR's axis of rotation. D = 1/2 rho Cd A v^2; where rho is the air density, Cd is the coefficient of drag, A is the kid's cross sectional area, and v = wR, which is the tangential velocity of the kid R distance from the axis of rotation for the MGR rotating at angular velocity w.

Centripetal force is best calculated by C = mv^2/R, which is really the equation for centrifugal force, but centripetal force is the same value even though it points inward along the R radial instead of outward like centrifugal. In your case, v is the same tangential velocity v = wR cited above. Thus, we can also write C = m(wR)^2/R = mwRw = mw^2R.

Note centripetal acceleration inward along the radial R is a = v^2/R = w^2R; so in fact we have C = ma, which is clearly a force as f = ma is well known.

Answer 2

Gravity acts down.
The horse pushes up and in on the child.

The up part of the horse's push cancels gravity. The in part provides the centripetal acceleration. (Or in the spinning reference frame of the child, you'd say that the in part cancels the centrifugal acceleration).

Answer 3

ok, now, pay interest...weight is the call we supply to a tension derived from gravity. once you step on your bathing room scale and examine (gasp) your weight, you're analyzing how puzzling Earth is pulling on you with its tension of gravity. As W = mg = GmM/R^2 = 26 N; the place m = .seventy two kg, G = 6.67E-eleven, and R = 4945007 m, you have each little thing you may desire to remedy for M = ? the mass of the planet.