Wagon wheel. 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 falls with a downward acceleration of 2.95 m/s2.

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fayemioluwatomisin fayemioluwatomisin · Answer 1 · 2020-04-24T04:50:36+02:00

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically