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(A) The moment of inertia $(I)$ of a uniform circular disc is given by $I = \frac{1}{2} M R^2$.Since the mass $(M)$ and thickness $(t)$ are constant,

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the one made from the more dense material

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(A) The moment of inertia $(I)$ of a uniform circular disc is given by $I = \frac{1}{2} M R^2$.Since the mass $(M)$ and thickness $(t)$ are constant, we express the radius $(R)$ in terms of density $(\rho)$:$M = \pi R^2 t \rho \Rightarrow R^2 = \frac{M}{\pi t \rho}$.Substituting this into the formula for $I$:$I = \frac{1}{2} M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.Since $M$, $t$, and $\pi$ are constants, we have $I \propto \frac{1}{\rho}$.Therefore, the disc made from the more dense material (higher $\rho$) will have a smaller rotational inertia.

Two uniform circular discs having the same mass and the same thickness but different radii are made from different materials. The disc with the smaller rotational inertia is

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