Question 9.SP.13: A thin steel plate that is 4 mm thick is cut and bent to for...
A thin steel plate that is 4 mm thick is cut and bent to form the machine part shown. The density of the steel is 7850 kg/m³. Determine the moments of inertia of the machine part with respect to the coordinate axes.
STRATEGY: The machine part consists of a semicircular plate and a rectangular plate from which a circular plate has been removed (Fig. 1). After calculating the moments of inertia for each part, add those of the semicircular plate and the rectangular plate, then subtract those of the circular plate to determine the moments of inertia for the entire machine part.
MODELING and ANALYSIS:
Computation of Masses. Semicircular Plate
\begin{aligned}V_1 & =\frac{1}{2} \pi r^2 t=\frac{1}{2} \pi(0.08 \mathrm{~m})^2(0.004 \mathrm{~m})=40.21 \times 10^{-6} \mathrm{~m}^3 \\m_1 & =\rho V_1=\left(7.85 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\right)\left(40.21 \times 10^{-6} \mathrm{~m}^3\right)=0.3156 \mathrm{~kg}\end{aligned}
Rectangular Plate
\begin{aligned}V_2 & =(0.200 \mathrm{~m})(0.160 \mathrm{~m})(0.004 \mathrm{~m})=128 \times 10^{-6} \mathrm{~m}^3 \\m_2 & =\rho V_2=\left(7.85 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\right)\left(128 \times 10^{-6} \mathrm{~m}^3\right)=1.005 \mathrm{~kg}\end{aligned}
Circular Plate
\begin{aligned}V_3 & =\pi a^2 t=\pi(0.050 \mathrm{~m})^2(0.004 \mathrm{~m})=31.42 \times 10^{-6} \mathrm{~m}^3 \\m_3 & =\rho V_3=\left(7.85 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\right)\left(31.42 \times 10^{-6} \mathrm{~m}^3\right)=0.2466 \mathrm{~kg}\end{aligned}
Moments of Inertia. Compute the moments of inertia of each component, using the method presented in Sec. 9.5C.
Semicircular Plate. Observe from Fig. 9.28 that, for a circular plate of mass m and radius r,
I_x=\frac{1}{2} m r^2 \quad I_y=I_z=\frac{1}{4} m r^2
Because of symmetry, halve these values for a semicircular plate. Thus,
I_x=\frac{1}{2}\left(\frac{1}{2} m r^2\right) \quad I_y=I_z=\frac{1}{2}\left(\frac{1}{4} m r^2\right)
Since the mass of the semicircular plate is m_1=\frac{1}{2} m, you have
\begin{aligned}& I_x=\frac{1}{2} m_1 r^2=\frac{1}{2}(0.3156 \mathrm{~kg})(0.08 \mathrm{~m})^2=1.010 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\& I_y=I_z=\frac{1}{4}\left(\frac{1}{2} m r^2\right)=\frac{1}{4} m_1 r^2=\frac{1}{4}(0.3156 \mathrm{~kg})(0.08 \mathrm{~m})^2=0.505 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2\end{aligned}
Rectangular Plate
\begin{aligned}& I_x=\frac{1}{12} m_2 c^2=\frac{1}{12}(1.005 \mathrm{~kg})(0.16 \mathrm{~m})^2=2.144 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\& I_z=\frac{1}{3} m_2 b^2=\frac{1}{3}(1.005 \mathrm{~kg})(0.2 \mathrm{~m})^2=13.400 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\& I_y=I_x+I_z=(2.144+13.400)\left(10^{-3}\right)=15.544 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2\end{aligned}
Circular Plate
\begin{aligned}I_x & =\frac{1}{4} m_3 a^2=\frac{1}{4}(0.2466 \mathrm{~kg})(0.05 \mathrm{~m})^2=0.154 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\I_y & =\frac{1}{2} m_3 a^2+m_3 d^2 \\& =\frac{1}{2}(0.2466 \mathrm{~kg})(0.05 \mathrm{~m})^2+(0.2466 \mathrm{~kg})(0.1 \mathrm{~m})^2=2.774 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\I_z & =\frac{1}{4} m_3 a^2+m_3 d^2=\frac{1}{4}(0.2466 \mathrm{~kg})(0.05 \mathrm{~m})^2+(0.2466 \mathrm{~kg})(0.1 \mathrm{~m})^2 \\& =2.620 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2\end{aligned}
Entire Machine Part
\begin{aligned}& I_x=(1.010+2.144-0.154)\left(10^{-3}\right) \mathrm{kg} \cdot \mathrm{m}^2 \quad I_x=3.00 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\& I_y=(0.505+15.544-2.774)\left(10^{-3}\right) \mathrm{kg}^{-\mathrm{m}^2} \quad I_y=13.28 \times 10^{-3} \mathrm{~kg}^{-3} \mathrm{~m}^2 \\& I_z=(0.505+13.400-2.620)\left(10^{-3}\right) \mathrm{kg} \cdot \mathrm{m}^2 \quad I_z=11.29 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^2 \\&\end{aligned}
