1.

When you have learned to integrate, derive the formula for the moment of inertia of a disk.

Answer»


Solution :The differential of MASS is equal to the mass of a RING of thickness DR. We have
`(dm)/M=(2pirdr)/(piR^2)` , form which `dm= (2Mrdr)/(R^2)`
The differential of the moment of inertia is equal to the moment of inertia of this ring:
`dI=dm.r^2=(2Mr^3dr)/(R^2)`
Hence
`I = int_0^Rdm.r^2=(2M)/(R^2)int_0^Rr^3dr=(2M)/R^2=(2M)/(R^2)[r^4/4]_0^R=(2MR^4)/(R^2xx4)=(MR^2)/2`


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