Question

wind energy is gaining increased attention, generating an increased interest in windmill technology. because windmill blades (vanes) rotate about a central axis, one of the most important physical properties of a windmill is its moment of inertia.
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 ) kg·m² and about axis b is ( 5860 ) kg·m², 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: Apply 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, and \(d\) is the distance between the two axes.

For axis \(A\) and the center of mass of the vane: \(I_{A}=I_{cm}+m(k_{1})^{2}\). For axis \(B\) and the center of mass of the vane: \(I_{B}=I_{cm}+m(k_{2})^{2}\).

Subtract the first equation from the second: \(I_{B}-I_{A}=m(k_{2}^{2}-k_{1}^{2})\).

Step2: Calculate the moment of inertia of one vane about the hub axis

Let the moment of inertia of one vane about the hub axis be \(I_{vane}\). Using the parallel - axis theorem again, if we know \(I_{cm}\) (from \(I_{A}=I_{cm}+m(k_{1})^{2}\), so \(I_{cm}=I_{A}-m(k_{1})^{2}\)), then \(I_{vane}=I_{cm}+m(k_{2})^{2}\).

Since \(I_{B}-I_{A}=m(k_{2}^{2}-k_{1}^{2})\), and \(I_{vane}\) can also be found as \(I_{vane}=I_{A}-m(k_{1})^{2}+m(k_{2})^{2}=I_{A}+m(k_{2}^{2}-k_{1}^{2})\)

We know \(m = 105\space kg\), \(k_{1}=3.15\space m\), \(k_{2}=4.88\space m\), \(I_{A}=241\space kg\cdot m^{2}\), \(I_{B}=5860\space kg\cdot m^{2}\)

First, check \(m(k_{2}^{2}-k_{1}^{2})=105\times(4.88^{2}-3.15^{2})=105\times(23.8144 - 9.9225)=105\times13.8919 = 1458.6495\space kg\cdot m^{2}\)

\(I_{vane}=I_{A}+m(k_{2}^{2}-k_{1}^{2})=241+1458.6495 = 1699.6495\space kg\cdot m^{2}\)

Step3: Calculate the total moment of inertia of the assembly

Assume there are \(n = 3\) vanes (a typical wind - mill has 3 vanes). Then \(I_{total}=n\times I_{vane}\)

\(I_{total}=3\times1699.6495=5098.9485\approx5100\space kg\cdot m^{2}\)

Answer:

\(5100\space kg\cdot m^{2}\)