Moment of Inertia Calculation for Discs to Find Alpha

The moment of inertia ($I$) for a solid circular disc of mass $m$, radius $r$, and thickness $t$, about an axis through its center and perpendicular to its plane is given by:

$I = \frac{1}{2} m r^2$

The mass $m$ is density ($\rho$) times volume ($V$). For a disc, $V = \pi r^2 t$. So, $m = \rho \pi r^2 t$. Substituting mass into the moment of inertia formula:

$I = \frac{1}{2} (\rho \pi r^2 t) r^2 = \frac{1}{2} \rho \pi t r^4$

Disc 1 Inertia ($I_1$)

For the first disc:

• Radius = $R_1$
• Thickness = $T_1$
• Moment of Inertia ($I_1$) = $\frac{1}{2} \rho \pi T_1 R_1^4$

Disc 2 Inertia ($I_2$)

For the second disc (made of the same material, so same $\rho$):

• Radius = $R_2$
• Thickness = $T_2$
• Moment of Inertia ($I_2$) = $\frac{1}{2} \rho \pi T_2 R_2^4$

Equating Inertias and Finding Alpha

We are given that the moment of inertia of both discs is the same:

$I_1 = I_2$

$\frac{1}{2} \rho \pi T_1 R_1^4 = \frac{1}{2} \rho \pi T_2 R_2^4$

Cancelling common terms ($\frac{1}{2} \rho \pi$):

$T_1 R_1^4 = T_2 R_2^4$

Rearranging to find the ratio of thicknesses:

$\frac{T_1}{T_2} = \frac{R_2^4}{R_1^4} = \left(\frac{R_2}{R_1}\right)^4$

We are given the ratio of radii:

$\frac{R_1}{R_2} = 2 \implies \frac{R_2}{R_1} = \frac{1}{2}$

Substitute this into the thickness ratio equation:

$\frac{T_1}{T_2} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$

The problem states $\frac{T_1}{T_2} = \frac{1}{\alpha}$.

Comparing the two expressions for $\frac{T_1}{T_2}$:

$\frac{1}{\alpha} = \frac{1}{16}$

Therefore, the value of $\alpha$ is 16.