Determine the product of inertia of the thin strip of area with respect to the $x$ and $y$ axes. The strip is oriented at an angle $\theta$ from the $x$ axis. Assume that $t \ll l$
Added by Janet A.
Step 1
This can be expressed as: $I_x = \int_{0}^{L} (s \sin(\theta))^2 \cdot t \, ds$ $I_x = t \int_{0}^{L} s^2 \sin^2(\theta) \, ds$ Show more…
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