- #1

PascalPanther

- 23

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Here is what I did initially:While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argument with your co-author regarding the moment of inertia of an actual wooden wagon wheel. The 70-kg wheel is 120-cm in diameter and has heavy spokes connecting the rim to the axle. Your co-author claims that you can approximate using I = MR2 (like for a hoop) but you anticipate I will be significantly less than that because of the mass located in the spokes. To find I experimentally, you mount the wheel on a low-friction bearing then wrap a light cord around the outside of the rim to which you attach a 20-kg bag of sand. When the bag is released from rest, it drops 3.77 m in 1.6 s.

v = 3.77m/1.6s

mgh = (1/2)mv^2 + (1/2)*I*omega^2

omega = v/R

(20kg)(9.8)(3.77m) = (1/2)(20kg)(2.36 m/s)^2 + (1/2)I((2.36)^2/(0.60)^2)

I = 88.3 ... which is more than I = MR^2 = 25.2 ...

So this is entirely wrong, because I know it should be less, since the mass is more evenly distributed.

Second approach, I believe the drops 3.77 m in 1.6 s is still due to gravity, is not constant velocity. There must also be tension then since it is less than gravitational acceleration.

mg - T = ma

y=(1/2)a*t^2

(3.77m) = (1/2)a(1.6)^2

a= 2.95 m/s^2

(20 kg)(9.8) - T = (20kg)(2.95)

T = 137N

torque = F*R

Only T does something in relation to the rotation of the axis? So:

torque = T*R = I * alpha

alpha = a/R = 2.95m/s^2 / 0.60m = 4.92 rad/s^2

137N * 0.60m = I * (4.92 rad/s^2)

I = 16.6 N s

Well I got the less inertia part now. But was that the right way to do it?