Non-uniform Inertia of a Cylinder: I = (3MR^3)/5

[LASmith]

Homework Statement

A cylinder with radius R and mass M has density that increases linearly with radial distance r from the cylinder axis, ie. $\rho$=$\rho$$_{0}$(r/R), where $\rho$$_{0}$ is a positive constant. Show that the moment of inertia of this cylinder about a longitudinal axis through the centre is given by I=(3MR$^{3}$)/5

Homework Equations

I=$\int$r$^{2}$.dm
volume = 2$\pi$rL.dr

The Attempt at a Solution

I=$\int$r$^{2}$$\rho$.dv
=$\int$(r$^{3}$$\rho$$_{0}$/R.)dv
=$\int$(r$^{3}$$\rho$$_{0}$/R.)(2$\pi$rL).dr

integrate between 0 and R to obtain
2$\rho_{0}$$\pi$R$^{4}$L/5

However, I do not understand how to express this without using the term $\rho_{0}$

 

[Doc Al]

However, I do not understand how to express this without using the term $\rho_{0}$

Find an expression for M in terms of ρ₀.

 

[LASmith]

Find an expression for M in terms of ρ₀.

I realize this, however as the density is not constant, I am unsure of how to do this.

 

[Doc Al]

I realize this, however as the density is not constant, I am unsure of how to do this.

Set up an integral to solve for the total mass, just like you set one up for the rotational inertia.

Once you get M in terms of ρ₀, you can rewrite your answer in terms of M instead of ρ₀.

 

[rwisz]

.I would approach this problem by first understanding the concept of inertia. Inertia is the resistance of an object to change its state of motion. It is directly related to the mass and distribution of the object. In this case, we are dealing with the inertia of a cylinder, which is a measure of how difficult it is to change its rotational motion.

To calculate the moment of inertia of a cylinder, we use the formula I = (MR^2)/2, where M is the mass of the cylinder and R is the radius. However, in this scenario, the density of the cylinder is not constant, but increases as we move away from the axis of rotation. This means that the mass of the cylinder is not evenly distributed, and we need to take that into account when calculating the moment of inertia.

To do this, we can divide the cylinder into infinitesimally thin concentric rings, each with a thickness of dr. The mass of each ring can be calculated using the formula dm = \rho(r)2\pirL.dr, where \rho(r) is the density at a distance r from the axis of rotation. We can then integrate this expression from 0 to R to obtain the total mass of the cylinder.

I = \int_{0}^{R} \rho(r)2\pirL.dr

Now, to simplify the expression, we can substitute \rho(r) with \rho_{0}(r/R), as given in the problem statement. This allows us to express the moment of inertia as:

I = \int_{0}^{R} \rho_{0}(r/R)2\pirL.dr

= \rho_{0}2\piL \int_{0}^{R} rdr

= \rho_{0}2\piL \frac{r^2}{2} \Big|_{0}^{R}

= \rho_{0}\piLR^2

= \frac{2}{5}M \piR^2

= \frac{2}{5} \cdot \frac{M}{\piR^3} \piR^4

= \frac{2}{5} \cdot \frac{M}{V} \cdot V \piR^4

= \frac{2}{5} \cdot \frac{M}{V} \cdot \frac{V}{\piR^3} \pi