Moment of inertia of a cone

[ehilge]

Homework Statement

Calculate the moment of inertia of a uniform solid cone about an axis through its center. The cone has mass M and altitude h. The radius of its circular base is R. (see attached photo)

Homework Equations

I know I need to somehow use the equation I= intergral r^2 dm
also, I have an equation from my proffessor, dm=rho dv I'm not sure if I need this though since its unifrom density so it doesn't seem like $$\rho$$ should matter.

The Attempt at a Solution

I don't have a solution right now. I know the answer is 3/10 MR² but I don't know how to get there. From class, since we did some examples, I think I need an equation that has a triple integral in it but I don't know what to integrate and to where. Do I need to get dm in terms of something like d$$\vartheta$$, dr, and dh?

Thanks for your help. Also, if you have any general suggestions on how to complete moment of inertia problems like this that would be great. I know we're going to be doing a lot of them.

Thanks again!

 

[hage567]

Yes dm = $$\rho$$dV will be used. Have you tried setting up dV in cylindrical coordinates? Then try to look for your limits of integration.

 

[baba89]

The moment of inertia of a cone can be calculated using the equation I= (3/10)MR^2, as you mentioned. However, it is important to understand the derivation of this equation in order to fully grasp the concept and be able to apply it to other problems.

To calculate the moment of inertia of a cone, we can use the equation I= intergral r^2 dm, where r is the distance from the axis of rotation to an infinitesimal piece of mass dm. In order to use this equation, we need to express dm in terms of the variables we are given, which are M, h, and R.

Since the cone has a uniform density, we can use the equation dm= \rho dv, where \rho is the density and dv is the infinitesimal volume element. In this case, the density is constant throughout the cone, so we can write \rho=M/V, where V is the total volume of the cone.

Now, we can express dv in terms of the given variables. The cone can be divided into infinitesimal circular discs, each with a radius r and thickness dr. The volume of each disc can be calculated as dV=\pi r^2 dh, where dh is the infinitesimal height of the disc. Therefore, the total volume of the cone can be expressed as V= \int_0^h \pi r^2 dh.

Substituting these expressions into the equation for dm, we get dm= \rho dv= \frac{M}{V} \pi r^2 dh= \frac{M}{\int_0^h \pi r^2 dh} \pi r^2 dh= \frac{M}{h} r^2 dh.

Now, we can substitute this expression for dm into the equation for moment of inertia, giving us I= \int r^2 dm= \int_0^R \frac{M}{h} r^4 dr. Solving this integral gives us I= \frac{1}{5} \frac{M}{h} R^5.

However, this is not the final answer, as we need to take into account the fact that the axis of rotation is through the center of the cone, not the base. This means that the moment of inertia will be different for different values of h, with the maximum value occurring when h=R.

To account for this, we need to multiply our