Question

given is a picture of a typical windmill, where the geometry and center of mass of one of the vanes is illustrated. the mass of each vane is 105 kg. the distance from the center of mass of the vane to axis b is ( k_1 = 3.15 ) m. the distance from the center of mass of the vane to the center of the windmill hub is ( k_2 = 4.88 ) m.
if the moment of inertia of a vane about axis a is ( 241 \text{ kg}cdot\text{m}^2 ) and about axis b is ( 5860 \text{ kg}cdot\text{m}^2 ), calculate the moment of inertia ( i_{\text{total}} ) of the entire assembly about the axis that passes through the windmills hub and is perpendicular to the screen.
ignore the hub and assume the vanes are flat.

Explanation:

Step1: Use the parallel - axis theorem

The parallel - axis theorem is \(I = I_{cm}+md^{2}\), where \(I\) is the moment of inertia about an axis parallel to the axis through the center of mass, \(I_{cm}\) is the moment of inertia about the axis through the center of mass, \(m\) is the mass of the object, and \(d\) is the distance between the two axes.

For one vane, if \(I_{A}\) is the moment of inertia about axis \(A\) (axis through the center of mass of the vane) and \(I_{B1}\) is the moment of inertia about axis \(B\) (axis through the hub) for one vane. Then \(I_{B1}=I_{A}+m(k_{1} + k_{2})^{2}\)

Given \(m = 105\space kg\), \(k_{1}=3.15\space m\), \(k_{2}=4.88\space m\), \(I_{A}=241\space kg\cdot m^{2}\)

First, calculate \(d=k_{1}+k_{2}=3.15 + 4.88=8.03\space m\)

Then \(I_{B1}=241+105\times(8.03)^{2}\)

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Step2: Calculate the moment of inertia of the entire assembly

Assume there are \(n = 4\) vanes (since it is a typical wind - mill as shown in the figure). The total moment of inertia of the vanes about axis \(B\) is \(I_{vanes - total}=4I_{B1}\)

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The moment of inertia of the entire assembly \(I_{total}=I_{vanes - total}\) (since we ignore the hub)

Answer:

\(28045.6\space kg\cdot m^{2}\)