Please show work.
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No way to tell, if you're asking for the horizontal G force along the radius of gyration. And the reason for that is because a = W^2 R = (2pi/T)^2 R is the acceleration radial acceleration and G = a/g is the g-force, where g = 9.8 m/s^2 near Earth's surface. The angular speed W = 2pi F = 2pi/T and T = 5 sec is the period, the time to go once around.
As you can plainly see, a also depends on R, the radius of gyration. And that you have not given us. But for a given period, T = 5 s, a is proportional to R. Which means if that cycle did the 5 seconds once around a small circle r < R, that would result in less Gs than if it did that period over a larger circle R > r.
As you can plainly see, a also depends on R, the radius of gyration. And that you have not given us. But for a given period, T = 5 s, a is proportional to R. Which means if that cycle did the 5 seconds once around a small circle r < R, that would result in less Gs than if it did that period over a larger circle R > r.
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the opposite side of the opposing force