Question

Consider a solid cone of radius R, height H, and mass M. The volume of a cone is 1/3 πHR²

a. What is the distance from the apex (the point) to the center of mass?
b. What is the moment of inertia for rotation about the axis of the cone?
Hint: The moment of inertia can be calculated as the sum of the moments of inertia of lots of small pieces.

Short answer

a) Center of mass is at 3/4 H from vertex.

b) The moment of inertia is 3Mr²/10

Step-by-step solution

Part(a) Step 1: Given

A solid cone of radius R, height H, and mass M.

The volume of a cone is 1/3 πHR²

Part(a) Step2: Explanation

Lets consider a differential disc with radius r at a distance of x from the vertex as shown in figure below.

The mass of the differential disc is dm.

density of the disk is can be calculated as

$\rho =\frac{M}{V}=\frac{M}{{\displaystyle \frac{1}{3}}\pi H{R}^2}=\frac{3M}{{\displaystyle \pi H{R}^2}}........................\left(1\right)$

From the figure we can find

$$\begin{gathered} \frac{r}{R}=\frac{x}{H} \\ r=\frac{R}{H}x \end{gathered}$$

Now we can find the value of dm

$dm=\pi r^2\times dx\times s$

Substitute the value we get

localid="1649152801449" $=\frac{3M}{H^3}{x}^{2}dx...........................\left(2\right)$

Now find the center of mass

$$\begin{gathered} x_{\text{com}}=\frac{1}{M}\int xdm \\ =\frac{1}{M}\times \frac{3M}{H^3}{\int }_0^H{x}^{3}dx \\ =\frac{3}{4}H \end{gathered}$$

Part(b) Step 1: Given information

A solid cone of radius R, height H, and mass M.
The volume of a cone is 1/3 πHR²

Part(b) Step 2: Explanation

Find the moment of inertia using dm from equation (2)

$dm=\frac{3M}{H^3}{x}^{2}dx$

So moment of inertia is

$dI=\frac{(dm\times r^2)}{2}=\frac{\left(3M{R}^2\right)}{\left(2H^5\right)}{x}^{4}dx$

Now integrate to find moment of inertia

$$\begin{gathered} I=\int dI \\ =\frac{3M{R}^2}{2H^5}{\int }_0^H{x}^{4}dx \\ =\frac{3M{R}^2}{10} \end{gathered}$$