A circular plate of mass M and radius R has i | vedclass

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(A) The mass $M$ of the plate is given by integrating the density over the area: $M = \int_0^R \rho(r) (2\pi r dr) = \int_0^R \rho_0 r (2\pi r dr) = 2

Vedclass · Accepted Answer

$8/5$

Vedclass · Answer

(A) The mass $M$ of the plate is given by integrating the density over the area: $M = \int_0^R \rho(r) (2\pi r dr) = \int_0^R \rho_0 r (2\pi r dr) = 2\pi \rho_0 \int_0^R r^2 dr = \frac{2\pi \rho_0 R^3}{3}$.The moment of inertia about the center of mass $(I_{CM})$ is: $I_{CM} = \int_0^R r^2 dm = \int_0^R r^2 (\rho_0 r \cdot 2\pi r dr) = 2\pi \rho_0 \int_0^R r^4 dr = \frac{2\pi \rho_0 R^5}{5}$.Expressing $I_{CM}$ in terms of $M$: $I_{CM} = (\frac{2\pi \rho_0 R^3}{3}) \cdot \frac{3}{5} R^2 = \frac{3}{5} MR^2$.Using the parallel axis theorem,the moment of inertia about an axis passing through the edge and perpendicular to the plate is: $I = I_{CM} + MR^2 = \frac{3}{5} MR^2 + MR^2 = \frac{8}{5} MR^2$.Comparing this with $I = aMR^2$,we get $a = 8/5$.

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