OpenStax College Physics, Chapter 6, Problem 23 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We begin by writing down some data. The mass of the rider on the merry-go-round is 22 kilograms and her angular speed, in part A anyhow, is 40 revolutions per minute which we convert into radians per second because radians per second are the units we need in the calculations for the formulas. So we multiply 40 revolutions per minute by one minute for every 60 seconds and then times two pi radians per revolution, giving us radians per second. So 40 divided by 60 times two pi gives 4.1888 radians per second. The radius of the merry-go-round in part A is 1.25 meters. So the centripetal force that the child will have to apply in order to stay on the ride is mass times centripetal acceleration, and centripetal acceleration is the radius multiplied by the angular velocity squared. So we have 22 kilograms times 1.25 meters, times 4.1888 radians per second squared, giving 483 newtons. In part B we have a different merry-go-round with an angular speed of three revolutions per minute which we convert into 0.31416 radians per second. Its radius is 8 meters. So the centripetal force is mass times radius, times negative velocity squared. So that's 22 kilograms times 8 meters times 0.31416 meters per second squared, giving 17.4 newtons. Comparing each of these forces, we have centripetal force in part A which I've labeled F a, the centripetal force in part B I've labeled F b, and we're going to divide each of them by gravity in order to compare to the weight. The weight of the girl is going to be mass times acceleration due to gravity. So that's 22 kilograms times 9.8 newtons per kilogram, giving 215.6 newtons. So we divide the answer for part A by that to get 2.24. Then in part B, we divide 17.4 by 215.6 to give 0.0807.