A disk of radius \(R\) and thickness \(w\) has a mass density that increases from the center outward, given by \(\rho=\rho_{0} r / R,\) where \(r\) is the distance from the disk axis. Calculate (a) the disk's total mass \(M\) and (b) its rotational inertia about its axis in terms of \(M\) and \(R\). Compare with the results for a solid disk of uniform density and for a ring.

Short Answer

Expert verified
For the disk with radially increasing mass density \(\rho = \rho_{0} r / R\), the total mass \(M\) is \(\frac{4}{3} \pi \rho_{0} w R^{2}\) and the rotational inertia \(I\) is \(\frac{8}{15} MR^{2}\). Compared to a solid disk of uniform density, we see that the rotational inertia is more due to the radially outward increase in mass. Contrarily, for a ring of uniform density, the rotational inertia is less due to mass concentrated at the circumference.

Step by step solution

01

Calculating disk's total mass

The total mass for a disk of variable density will be achieved by integrating the mass elements \(dm = \rho dv\), where the volume element \(dv = r dr d\phi dz\). By substituting \(\rho = \rho_{0} r / R\) the mass element changes to \(dm = \rho_{0} r^2 dr d\phi dz / R\). Lastly, setting the integration bounds; \(r\) ranges from 0 to \(R\), \(z\) from \(-w/2\) to \(w/2\), and \(\phi\) from 0 to \(2\pi\), we can integrate to find \(M\).
02

Calculating rotational inertia

Rotational inertia for a thin disk is given by \(I = \int r^2 dm\). Substituting \(dm\) from the first step, we can integrate to find \(I\). The integration bounds remain the same as in step 1.
03

Compare with solid disk and a ring.

For a solid disk of uniform density, the distribution is \(\rho = \rho_{0}\) and the rotational inertia is \(I = \frac{1}{2}MR^2\). For a ring, the density remains constant but it is distributed on the circumference and thus, the rotational inertia is \(I = MR^2\). Comparing these with the previously derived expression for \(I\) will shed light on the impact of variable mass distribution.

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