Question 9.SP.11: Determine the moment of inertia of a right circular cone wit...
Determine the moment of inertia of a right circular cone with respect to (a) its longitudinal axis, (b) an axis through the apex of the cone and perpendicular to its longitudinal axis, (c) an axis through the centroid of the cone and perpendicular to its longitudinal axis.
STRATEGY: For parts (a) and (b), choose a differential element of mass in the form of a thin circular disk perpendicular to the longitudinal axis of the cone. You can solve part (c) by an application of the parallel-axis theorem.
MODELING and ANALYSIS: Choose the differential element of mass shown in Fig. 1. Express the radius and mass of this disk as
a. Moment of Inertia I_{x}. Using the expression derived in Sec. 9.5C
for a thin disk, compute the mass moment of inertia of the differential element with respect to the x axis.
Integrating from x = 0 to x = h gives you
I_{x}=\int d I_{x}=\int_{0}^{h} \frac{1}{2} \rho \pi \frac{a^{4}}{h^{4}} x^{4} d x=\frac{1}{2} \rho \pi \frac{a^{4}}{h^{4}} \frac{h^{5}}{5}=\frac{1}{10} \rho \pi a^{4} hSince the total mass of the cone is m=\frac{1}{3} \rho \pi a^{2} h, you can write this as
I_{x}=\frac{1}{10} \rho \pi a^{4} h=\frac{3}{10} a^{2}\left(\frac{1}{3} \rho \pi a^{2} h\right)=\frac{3}{10} m a^{2} \quad I_{x}=\frac{3}{10} m a^{2}b. Moment of Inertia I_{y}. Use the same differential element. Apply-ing the parallel-axis theorem and using the expression derived in Sec. 9.5C for a thin disk, you have
d I_{y}=d I_{y^{\prime}}+x^{2} d m=\frac{1}{4} r^{2} d m+x^{2} d m=\left(\frac{1}{4} r^{2}+x^{2}\right) d mSubstituting the expressions for r and dm into this equation yields
Introducing the total mass of the cone m, you can rewrite I_{y} as
I_{y}=\frac{3}{5}\left(\frac{1}{4} a^{2}+h^{2}\right) \frac{1}{3} \rho \pi a^{2} h \quad I_{y}=\frac{3}{5} m\left(\frac{1}{4} a^{2}+h^{2}\right)c. Moment of Inertia \bar{I}_{y^{\prime \prime}} Apply the parallel-axis theorem to obtain
I_{y}=\bar{I}_{y^{\prime \prime}}+m \bar{x}^{2}Solve for \bar{I}_{y^{\prime \prime}} and recall from Fig. 5.21 that \bar{x}=\frac{3}{4} h (Fig. 2). The result is
REFLECT and THINK: The parallel-axis theorem for masses can be just as useful as the version for areas. Don’t forget to use the reference figures for centroids of volumes when needed.
