Answer:
\[{{I}_{1}}=\frac{1}{2}mr_{1}^{2}\] and \[{{I}_{2}}=\frac{1}{2}mr_{2}^{2}\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] Also, \[m=\pi r_{1}^{2}t{{\rho }_{1}}=\pi r_{2}^{2}t{{\rho }_{2}}\] \[\therefore \] \[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\] Hence \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\] i.e., \[I\propto \frac{1}{\rho }\] Thus the disc with greater density will have less moment of inertia.
You need to login to perform this action.
You will be redirected in
3 sec