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Step-by-Step Solution
Step 1: Understand the Physical Quantity Involved
The quantity of interest here is the moment of inertia of a uniform circular disc about its central axis. This depends on the mass distribution and the square of the radius.
Step 2: Write the General Formula for the Moment of Inertia of a Disc
For a uniform disc of radius $R$ and mass $M$, the moment of inertia about its central axis is:
$I = \frac{1}{2} M R^2 \,.$
Step 3: Relate Mass to Geometric Factors
The disc is made of the same material and has the same thickness. If $\rho$ denotes the density of the material, $A$ denotes the area of the disc, and $t$ denotes the uniform thickness, then:
$M = \rho \times A \times t \,.$
For a disc of radius $R$, the area $A = \pi R^2$. Hence:
$M = \rho \times \pi R^2 \times t \,.$
Step 4: Express the Moment of Inertia in Terms of $R$
Substituting $M$ into the moment of inertia formula:
$I = \frac{1}{2} \bigl( \rho \,\pi R^2 \,t \bigr)\, R^2 = \text{(constant)} \times R^4 \,,
where the constant includes factors like $\rho$, $\pi$, $t$, and $\tfrac{1}{2}$. We can denote this constant by $K$, so:
$I = K \, R^4 \,.$
Step 5: Set Up the Ratio of Moments of Inertia
We have two discs with radii $R_1 = R$ and $R_2 = \alpha R$. Their moments of inertia are:
$I_1 = K\, R_1^4 = K\,R^4 \,, \\[6pt]
I_2 = K\, R_2^4 = K \,(\alpha R)^4 = K\, \alpha^4 R^4 \,.
Given $I_1 : I_2 = 1 : 16$, we write:
$\frac{I_1}{I_2} = \frac{K\,R^4}{K\,\alpha^4 \, R^4} = \frac{1}{\alpha^4} \,.
Thus:
$\frac{1}{\alpha^4} = \frac{1}{16} \,.$
Step 6: Solve for $\alpha$
Equating both sides gives:
$\alpha^4 = 16 \,.
Taking the fourth root, we get:
$\alpha = \sqrt[4]{16} = 2 \,.
Final Answer
The value of $\alpha$ is 2.
