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- https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.03%3A_Dynamics_of_Rotational_Motion_-_Rotational_InertiaA hoop’s moment of inertia around its axis is therefore \(MR^2\), where \(M\) is its total mass and \(R\) its radius. (We use \(M\) and \(R\) for an entire object to distinguish them from \(m\) and \(...A hoop’s moment of inertia around its axis is therefore \(MR^2\), where \(M\) is its total mass and \(R\) its radius. (We use \(M\) and \(R\) for an entire object to distinguish them from \(m\) and \(r\) for point masses.) In all other cases, we must consult Figure 10.4.3 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for \(I\) that have been derived from integration over the continuous body.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/02%3A_Moments_of_Inertia/2.01%3A_Definition_of_Moment_of_InertiaThe quantity \( m_{i} r_{i}^{2} \) is the second moment of the \( i \) th mass with respect to (or "about") the axis, and the sum \( \sum m_{i} r_{i}^{2} \) is the second moment of mass of all the mas...The quantity \( m_{i} r_{i}^{2} \) is the second moment of the \( i \) th mass with respect to (or "about") the axis, and the sum \( \sum m_{i} r_{i}^{2} \) is the second moment of mass of all the masses with respect to the axis. But the question should be asked: "What is the purpose of calculating the squares of the distances of lots of particles from an axis, multiplying these squares by the mass of each, and adding them all together?
- https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_I_(2211)/01%3A_Introduction_to_Physics_Measurements_and_Mathematics_Tools/1.09%3A_Math_Review_of_Other_Topics/1.9.24%3A_Calculating_Centers_of_Mass_and_Moments_of_Inertiathe centroid of a region is the geometric center of the region; laminas are often represented by regions in the plane; if the lamina has a constant density, the center of mass of the lamina depends on...the centroid of a region is the geometric center of the region; laminas are often represented by regions in the plane; if the lamina has a constant density, the center of mass of the lamina depends only on the shape of the corresponding planar region; in this case, the center of mass of the lamina corresponds to the centroid of the representative region
- https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_II_(2212)/01%3A_Introduction_to_Physics_Measurements_and_Mathematics_Tools/1.09%3A_Math_Review_of_Other_Topics/1.9.24%3A_Calculating_Centers_of_Mass_and_Moments_of_Inertiathe centroid of a region is the geometric center of the region; laminas are often represented by regions in the plane; if the lamina has a constant density, the center of mass of the lamina depends on...the centroid of a region is the geometric center of the region; laminas are often represented by regions in the plane; if the lamina has a constant density, the center of mass of the lamina depends only on the shape of the corresponding planar region; in this case, the center of mass of the lamina corresponds to the centroid of the representative region
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/02%3A_Moments_of_Inertia/2.03%3A_Moments_of_Inertia_of_Some_Simple_Shapes"For how many different shapes of body must I commit to memory the formulas for their moments of inertia?" I would be tempted to say: "None". However, if any are to be committed to memory, I would sug..."For how many different shapes of body must I commit to memory the formulas for their moments of inertia?" I would be tempted to say: "None". However, if any are to be committed to memory, I would suggest that the list to be memorized should be limited to those few bodies that are likely to be encountered very often (particularly if they can be used to determine quickly the moments of inertia of other bodies) and for which it is easier to remember the formulas than to derive them.
- https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Celestial_Mechanics_(Tatum)/05%3A_Gravitational_Field_and_Potential/5.12%3A_Gravitational_Potential_of_any_Massive_Bodyso \[ψ = -G \left[ \frac{1}{R} \int ρ dτ + \frac{1}{R^2} \int ρ r \cos θ d τ + \frac{1}{R^3} \int ρ r^2 P_2 (\cos θ) dτ + \frac{1}{R^4} \int ρ r^3 P_3 (\cos θ) d τ ... \right]. \label{5.12.2} \tag{5.1...so \[ψ = -G \left[ \frac{1}{R} \int ρ dτ + \frac{1}{R^2} \int ρ r \cos θ d τ + \frac{1}{R^3} \int ρ r^2 P_2 (\cos θ) dτ + \frac{1}{R^4} \int ρ r^3 P_3 (\cos θ) d τ ... \right]. \label{5.12.2} \tag{5.12.2}\] The integral is to be taken over the entire body, so that \(∫ ρdτ = M\), where \(M\) is the mass of the body. But \(A + B + C\) is invariant with respect to rotation of axes, so it is also equal to \(A_0 + B_0 + C_0\), where \(A_0, \ B_0, \ C_0\) are the principal moments of inertia.
