Moment of Inertia of Uniform Cone: 3/10*M*R^2

AI Thread Summary
The moment of inertia of a uniform cone about the axis joining the center of its base and its apex is confirmed to be 3/10*(M*R^2), where M is the mass and R is the radius of the base. The calculation involves using calculus to integrate over the cone's dimensions. The formula is derived from the integral I_z = 2πρ ∫_0^h ∫_0^{R(h-z)/h} r^2 * r * dr * dz. Understanding this concept requires a grasp of calculus principles. The discussion emphasizes the correctness of the derived formula for the moment of inertia of a cone.
saksham
Messages
16
Reaction score
0
What is the Moment of inertia of a uniform cone about the axis joining the center of its base and its apex.
Iis it 3/10*(M*R^2) where M is the mass of the cone and R the radius of the base of the cone?
 
Physics news on Phys.org
That's correct.

It's not hard to figure out, if you know some calculus :

I_z = 2\pi\rho \int _0 ^h \int _0 ^{R(h-z)/h} r^2*r*dr*dz = \frac {3}{10}MR^2
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top