Example 21.1 Determine the moment of inertia of the bent rod about the Aa axis. The mass of each of the three segments are shown.
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Step 1: The moment of inertia of the bent rod about the Aa axis is the sum of the moments of inertia of the three segments. Show more…
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Find the moment of inertia of the four masses shown in Eig. $10-7$ relative to an axis perpendicular to the page and extending ( $a$ ) through point- $A$ and $(b)$ through point- $B$. (a) From the definition of moment of inertia, $$I_{A}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+\cdots+m_{N} r_{N}^{2}=(2.0 \mathrm{~kg}+3.0 \mathrm{~kg}+4.0 \mathrm{~kg}+5.0 \mathrm{~kg})\left(r^{2}\right)$$ where $r$ is half the length of the diagonal: $$r=\frac{1}{2} \sqrt{(1.20 \mathrm{~m})^{2}+(2.50 \mathrm{~m})^{2}}=1.39 \mathrm{~m}$$ Thus, $I_{A}=27 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (b) We cannot use the parallel-axis theorem here because neither $A$ nor $B$ is at the center of mass. Hence, we proceed as before. Because $r=1.25 \mathrm{~m}$ for the $2.0$ - and $3.0$ -kg masses, while $r=\sqrt{(1.20)^{2}+(1.25)^{2}}=1.733$ for the other two masses, $I_{B}=(2.0 \mathrm{~kg}+3.0 \mathrm{~kg})(1.25 \mathrm{~m})^{2}+(5.0 \mathrm{~kg}+4.0 \mathrm{~kg})(1.733 \mathrm{~m})^{2}=33 \mathrm{~kg} \cdot \mathrm{m}^{2}$.
(a) Consider a strip of length $d x$ at a perpendicular distance $x$ from the axis about which we have to find the moment of inertia of the rod. The elemental mass of the rod equals $d m=\frac{m}{l} d x$ Moment of inertia of this element about the axis $d I=d m x^{2}=\frac{m}{l} d x \cdot x^{2}$ Thus, moment of inertia of the rod, as a whole about the given axis (b) Let us imagine the plane plane taking the origin at the of the sides of the plate (Fig.). Obviously $=\frac{m a^{2}}{3}$ Similarly $I_{y}=\frac{m b^{2}}{3}$ Hence from perpendicular axis theorem
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They are connected by massless, rigid rods. Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.
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