Calculate the moment of inertia of a rectangular prism that is 150 mm high, 100 mm wide, and 250 mm long, with respect to its longest centroidal axis. Use a density of 640 kg/m3.
Find the moment of inertia of an aluminum sphere 200 mm in diameter with respect to a centroidal axis. Aluminum has a density of 2560 kg/m3.
The Correct Answer and Explanation is:
Let’s solve each problem step by step.
1. Moment of inertia of the rectangular prism
Given:
- Height h=150 mm=0.150 mh = 150 \text{ mm} = 0.150 \text{ m}
- Width w=100 mm=0.100 mw = 100 \text{ mm} = 0.100 \text{ m}
- Length l=250 mm=0.250 ml = 250 \text{ mm} = 0.250 \text{ m} — this is the longest dimension
- Density ρ=640 kg/m3\rho = 640 \text{ kg/m}^3
Goal: Find the moment of inertia II of the prism about its longest centroidal axis, which is along the length ll.
Step 1: Find the mass mm of the prism.
Volume V=l×w×h=0.250×0.100×0.150=0.00375 m3V = l \times w \times h = 0.250 \times 0.100 \times 0.150 = 0.00375 \text{ m}^3
Mass m=ρ×V=640×0.00375=2.4 kgm = \rho \times V = 640 \times 0.00375 = 2.4 \text{ kg}
Step 2: Moment of inertia for rectangular prism about its longest centroidal axis
The longest axis is along the length ll, so the axis is along the length direction.
For a rectangular prism, the moments of inertia about its centroidal axes are:
- About the x-axis (length axis):
Ix=112m(h2+w2)I_x = \frac{1}{12} m (h^2 + w^2)
- About the y-axis (width axis):
Iy=112m(l2+h2)I_y = \frac{1}{12} m (l^2 + h^2)
- About the z-axis (height axis):
Iz=112m(l2+w2)I_z = \frac{1}{12} m (l^2 + w^2)
Since the longest axis is length ll, the moment of inertia about the length axis is: I=Ix=112×2.4×(0.1502+0.1002)I = I_x = \frac{1}{12} \times 2.4 \times (0.150^2 + 0.100^2)
Calculate the terms inside parentheses: 0.1502=0.0225,0.1002=0.0100,⇒0.0225+0.0100=0.03250.150^2 = 0.0225, \quad 0.100^2 = 0.0100, \quad \Rightarrow 0.0225 + 0.0100 = 0.0325
Calculate moment of inertia: Ix=112×2.4×0.0325=0.2×0.0325=0.0065 kg\cdotpm2I_x = \frac{1}{12} \times 2.4 \times 0.0325 = 0.2 \times 0.0325 = 0.0065 \text{ kg·m}^2
2. Moment of inertia of an aluminum sphere
Given:
- Diameter d=200 mm=0.200 md = 200 \text{ mm} = 0.200 \text{ m}
- Radius r=0.100 mr = 0.100 \text{ m}
- Density ρ=2560 kg/m3\rho = 2560 \text{ kg/m}^3
Step 1: Calculate the volume VV of the sphere
V=43πr3=43π(0.1)3=43π×0.001=0.00419 m3V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.1)^3 = \frac{4}{3} \pi \times 0.001 = 0.00419 \text{ m}^3
Step 2: Calculate the mass mm
m=ρ×V=2560×0.00419=10.726 kgm = \rho \times V = 2560 \times 0.00419 = 10.726 \text{ kg}
Step 3: Moment of inertia for solid sphere about centroidal axis
For a solid sphere rotating about any diameter: I=25mr2=25×10.726×(0.1)2=25×10.726×0.01=0.0429 kg\cdotpm2I = \frac{2}{5} m r^2 = \frac{2}{5} \times 10.726 \times (0.1)^2 = \frac{2}{5} \times 10.726 \times 0.01 = 0.0429 \text{ kg·m}^2
Summary:
- Rectangular prism moment of inertia about longest centroidal axis (length):
0.0065 kg\cdotpm2\boxed{0.0065 \text{ kg·m}^2}
- Aluminum sphere moment of inertia about centroidal axis:
0.0429 kg\cdotpm2\boxed{0.0429 \text{ kg·m}^2}
Explanation:
The moment of inertia is a physical quantity that measures how difficult it is to change the rotational motion of an object about a given axis. It depends on the mass distribution relative to the axis of rotation. For a rectangular prism, the mass is spread out unevenly depending on the axis: rotating about the longest axis means the mass is closer to the axis, leading to a smaller moment of inertia. The formula Ix=112m(h2+w2)I_x = \frac{1}{12} m (h^2 + w^2) shows that only the perpendicular dimensions to the axis contribute, not the length itself.
In contrast, the solid sphere has a symmetric mass distribution in all directions. Its moment of inertia about any diameter is given by I=25mr2I = \frac{2}{5} m r^2, which reflects how mass is uniformly distributed at varying distances from the center. The higher density of aluminum (2560 kg/m³) compared to the prism’s material (640 kg/m³) results in a significantly higher mass for the sphere, despite its smaller radius relative to the prism’s longest dimension, resulting in a larger moment of inertia.
These calculations are essential in mechanical engineering and physics, especially when analyzing rotational dynamics, structural stability, or designing components subject to rotational motion. Knowing the moment of inertia helps predict how an object will behave when subjected to torques, aiding in design optimization for safety and performance.