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Calculate the moment of inertia of a bicycle wheel (with 8 spokes (thin rods) evenly spaced around the wheel) rotating around its center. Approximate the tire as a thin ring.

2007-04-17 21:34:44 · 3 answers · asked by playb0ybunny55 1 in Science & Mathematics Physics

3 answers

the mass of the wheel I'll call K (for Ksyrium)
The mass of a single spoke I'll call t (for titanium)

The diameter of the wheel is 0.7m or 70 cm

I of a thin ring is MR^2
so the tire and rim is K*0.35^2

I of a rod supported on one end is
1/3*M*L^2

Each spoke (assuming no hub diameter) is
t*.35^2/3

So the moment of inertia for the whole assembly is
(K+8*t/3)*.35^2

j

2007-04-18 19:02:17 · answer #1 · answered by odu83 7 · 0 1

The tires DO rotate at the same rpm which IS NOT the same as what you are looking at which is the linear velocity not rotational speed. The mark further out from the center of rotation ( the axle) travels further with each revolution of the wheel than th mark on the inside. Let's say the outer mark is 12" from the axle the inner mark 6" from axle. For any given wheel speed revolutions/minute, the outer mark move 12╥ inches, the inner mark moves 6╥. As for the race car example the inner lane has LESS distance to go than the outer lane since the arc that represents any turn in the race course has a longer trace for any increase in turn radius.

2016-04-01 06:49:54 · answer #2 · answered by Anonymous · 0 0

8 Spokes

2016-12-16 13:54:08 · answer #3 · answered by lempicki 4 · 0 0

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