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प्रश्न
State an expression for the moment of intertia of a solid uniform disc, rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration:
i. about a tangent in the plane of the disc, and
ii. about a tangent perpendicular to the plane of the disc.
उत्तर
The M. I of a thin uniform disc about an axis passing through its centre and perpendicular to its plane is given by,
IC = (1/2)MR2
i. According to theorem of parallel axis,
IT = Id + Mh2 = Id + MR2 [∵ h = R]
But Id = MR2/4
`therefore I_T=(MR^2)/4+MR^2`
`therefore I_T=5/4 MR^2`
Now , radius of gyration is given by,
`K=sqrt(I/M)`
`therefore K=sqrt((5MR^2)/(4M))`
`thereforeK=sqrt5/2R`
ii. Applying theorem of parallel axis,
IT = IO + Mh2 = IO + MR2 [∵ h = R]
But IO = MR2/2
`I_T=(MR^2)/2+MR^2=3/2MR^2`
Now, radius of gyration is given by,
`K=sqrt(I/M)`
`thereforeK=sqrt((3MR^2)/(2M))`
`thereforeK=sqrt(3/2)R`
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