Question 9.SP.11: Determine the moment of inertia of a right circular cone wit......

MODELING and ANALYSIS: Choose the differential element of mass shown in Fig. 1. Express the radius and mass of this disk as

r=a \frac{x}{h} \quad d m=\rho \pi r^2 d x=\rho \pi \frac{a^2}{h^2} x^2 d x

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.

d I_x=\frac{1}{2} r^2 d m=\frac{1}{2}\left(a \frac{x}{h}\right)^2\left(\rho \pi \frac{a^2}{h^2} x^2 d x\right)=\frac{1}{2} \rho \pi \frac{a^4}{h^4} x^4 d x

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 h

Since 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. Applying 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 m

Substituting the expressions for r and dm into this equation yields

d I_y=\left(\frac{1}{4} \frac{a^2}{h^2} x^2+x^2\right)\left(\rho \pi \frac{a^2}{h^2} x^2 d x\right)=\rho \pi \frac{a^2}{h^2}\left(\frac{a^2}{4 h^2}+1\right) x^4 d x

I_y=\int d I_y=\int_0^h \rho \pi \frac{a^2}{h^2}\left(\frac{a^2}{4 h^2}+1\right) x^4 d x=\rho \pi \frac{a^2}{h^2}\left(\frac{a^2}{4 h^2}+1\right) \frac{h^5}{5}

Introducing the total mass of the cone m, you can rewrite I_y as

I_y=\frac{3}{5}(\frac{1}{4}a^2 + h^2) \frac{1}{3}\rho \pi a^2 h                       I_y=\frac{3}{5}m(\frac{1}{4}a^2 + h^2)

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

\bar{I}_{y^{\prime \prime}}=I_y-m \bar{x}^2=\frac{3}{5} m\left(\frac{1}{4} a^2+h^2\right)-m\left(\frac{3}{4} h\right)^2

\bar{I}_{y^{\prime \prime}}=\frac{3}{20} m\left(a^2+\frac{1}{4} h^2\right)

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.