We assume there are only 4 spokes per wheel (minimum number for wheel stability) and that we want to compute the moments of inertia with respect to all axes x, y, z. See the figure below.

When computing the moment of inertia of the whole figure we an take advantage of the fact that it is composed of three different objects (a wheel and two by two perpendicular spokes) which add together algebraically their moments of inertia.

From the symmetry of the figure we have

`I =I_x =I_y `

Only one rod formed by two spokes each rotating about its end, and the wheel rotating about its diameter will contribute to the moment of inertia along x and y axis. The other rod is along the axis.

` I(=I_x=I_y)= I_("wheel") +I_("rod") =(M_("wheel")*R^2)/2 + 2*(M_("spoke")*R^2)/3`

` ` `I = 5*0.2^2/2 +2*(0.5*0.2^2)/3 =0.113 kg*m^2`

For the moment of inertia about the z axis (perpendicular in the center of wheel to the figure), we have two rods rotating about their common center (or 4 spokes rotating about their ends) and one wheel rotating about an axis in its center perpendicular to it.

`I' = I_z = M_("wheel")*R^2 + 4*(M_("spoke")*R^2)/3`

`I' =I_z = 5*0.2^2 +4*0.5*0.2^2/3 =0.226 kg*m^2`

Final observation: It can be shown (as can be observed from above numerical values) that for any plane figure and any Cartesian axis system `Oxyz` one has

`I_x =I_y = 2*I_z` .

Thus it is convenient to compute first `I_z` and only to divide this value by 2 to find `I_x` and `I_y` .

**Answer: The moment of inertia about an axis in the plane of figure is `0.113 kg*m^2` and the moment of inertia about a central axis perpendicular to the plane of figure is `0.226 kg*m^2` **

**Further Reading**

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