Find the moment of inertia of a cone of base radius a
about axis.
Answers
Answer:
Moment of inertia of a cone can be expressed using different formulas depending on the structure of the cone. We have to take into account two main types – hollow and solid cones.
Solid Cone
For a solid cone the moment of inertia is found by using the given formula;
I = 3MR2 / 10
Hollow Cone
For a hollow cone, we determine the moment of inertia using;
I = MR2 / 2
Moment of Inertia of Solid Cone Derivation
Here we will look at the derivation as well as the calculation for finding the moment of inertia of a uniform right circular cone about an axis.
Moment of Inertia of Circular Cone
We will divide the cone into a small elemental disc where we consider the cone’s radius to be r at a distance x from the top and of thickness dx. We will need to find the mass of the disc and it is given as;
dm = \frac{M}{\frac{1}{3}\pi R^{2}H}.(\pi r^{2})dxdm=
3
1
πR
2
H
M
.(πr
2
)dx dm = \frac{3M}{R^{2}H}. r^{2}dxdm=
R
2
H
3M
.r
2
dx ——-(1)
[using the concept of similar triangles we get r = (R/H). x
Using the value of r in equa (1)
dm = \frac{3M}{R^{2}H}. \frac{R^{2}}{H^{2}}x^{2}dxdm=
R
2
H
3M
.
H
2
R
2
x
2
dx
dm = \frac{3M}{H^{3}}. x^{2}dxdm=
H
3
3M
.x
2
dx
Now, let us find the moment of inertia of the elemental disc
dI = (1/2) dmr2
dI = \frac{1}{2}\frac{3M}{H^{3}}.x^{2}dx.\frac{R^{2}}{H^{2}}x^{2}dI=
2
1
H
3
3M
.x
2
dx.
H
2
R
2
x
2
dI = \frac{3MR^{2}}{2H^{5}}x^{4}.dxdI=
2H
5
3MR
2
x
4
.dx
\int dI = \frac{3MR^{2}}{2H^{5}}\int_{0}^{H}x^{4}.dx∫dI=
2H
5
3MR
2
∫
0
H
x
4
.dx
= \frac{3MR^{2}}{2H^{5}}\frac{H^{5}}{5}=
2H
5
3MR
2
5
H
5
I = 3MR2 / 10
Moment of inertia of solid cone
So if we consider the z-axis then we get;
IZ = 3MR2 / 10
For x-axis;
Ix = Iy = 3m (r2 / 4 + h2) / 5
We will look at one of the simple problems below.
Solved Example To Find Moment Of Inertia Of A Solid Cone
Calculate the moment of inertia of the right circular cone with regards to the x and y-axis. Given, M = 20, R= 4, Height = 2 m.
Moment Of Inertia Of A Circular Cone
Solution:
We will solve the problem by using the right formulas.
For the z-axis;
Iz = 3 MR2/ 10
Substituting the values;
Iz = 3 x 20 x 4 x 4/ 10
Iz = 96 kg m2
For the x-axis;
Ix = Iy = 3 m [(r2 / 4) + H2]/ 5
Ix = Iy = 3 x 20 x [( 42 / 4) + 22)] / 5
Ix = 3 x 20 x [( 16 / 4) + 4)]/ 5
Ix = 3 x 20 x 8/5
Ix = 96 kg m2