Question

Each of the objects above is composed of identical thin sticks of uniformly distributed mass 5.51 kg and length 0.737 m. What is the moment of inertia of each object about a rotation axis perpendicular to the screen and passing through the black dot?

Answer 1

In case \(A\), the four rods are rotating about an axis passing through the center.

The moment of inertia in case \(A\) is given by,

$$ \begin{aligned} I_{A} &=4 I \\ &=4\left(\frac{1}{12} M L^{2}\right) \\ &=4\left(\frac{1}{12}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2}\right) \\ &=0.560 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{aligned} $$

In case \(\mathrm{B}\), one rod is rotating about an axis passing through the center and another rod is rotating about an axis passing through end.

The moment of inertia in case \(\mathrm{B}\) is given by,

$$ \begin{aligned} I_{B} &=I_{1}+I_{2} \\ &=\frac{1}{12} M L^{2}+\frac{1}{3} M L^{2} \\ &=\frac{1}{12}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2}+\frac{1}{3}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2} \\ &=0.70 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{aligned} $$

The moment of inertia in case \(\mathrm{C}\) is given by, \(I_{C}=I_{1}+I_{2}\)

$$ \begin{array}{l} =\frac{1}{3} M L^{2}+\left(\frac{1}{3} M(2 L)^{2}+M L^{2}\right) \\ =\frac{1}{3}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2}+\frac{7}{3}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2} \mid \\ =\frac{8}{3}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2} \\ =4.48 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{array} $$

The moment of inertia in case \(\mathrm{D}\) is given by, \(I_{D}=\frac{1}{12} M L^{2}+\frac{1}{3} M L^{2}+\frac{1}{12} M L^{2}+M L^{2}\)

$$ \begin{array}{l} =\frac{18}{12}(4.47 \mathrm{~kg})(0.613 \mathrm{~m})^{2} \\ =2.52 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{array} $$