Two identical twins, Jena and Jackie, are playing one December on a large merry-

Question

Two identical twins, Jena and Jackie, are playing one December on a large merry-go-round (a disk mounted parallel to the ground on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has a mass of 30.7 kg. The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting a distance 1.88 m from the center of the merry-go-round, must hold on to one of the metal posts attached to the merry-go-round with a horizontal force of 60.0 N to keep from sliding off. Jackie is sitting at the edge, a distance 3.52 m from the center.

* With what horizontal force must Jackie hold on to keep from falling off?


* If Jackie falls off, what will be her horizontal velocity when she becomes airborne?

Explanation / Answer

The 60.0 N that Jena needs to hold on with is the force required to match the centripetal force (angular acceleration * her mass). Angular acceleration can be calculated by velocity^2 / radius, and force = mass * acceleration so Centripetal force = mass * velocity * velocity / radius. Another way of looking at the centripetal force is through angular velocity the linear velocity at any point on the merry-go round is the angular velocity * distance from the center of the merry go round. Let's call the angular velocity w (it is usually the greek letter omega) so V = w * r if F = m * v * v / r, F = m * (w * r) * (w * r) / r so centripetal force = F = m * w * w * r Because they are on the same merry go round, the angular velocity w does not change. The radius (distance from the center) does change, and that will change the force needed to hold on. Looking at this equation, the farther you are from the center of the merry go round, the harder it is to hold on.

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