Answer:
The
kinetic energy of the rolling object is converted into potential energy at
height h
\[=3{{\upsilon
}^{2}}/4g\], Fig. 5(HT).2.
\[\therefore
\] \[\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}=mgh\]
\[\frac{1}{2}m{{\upsilon
}^{2}}+\frac{1}{2}I{{\frac{\upsilon }{{{r}^{2}}}}^{2}}=mg\left(
\frac{3{{\upsilon }^{2}}}{4g} \right)\] \[\left(
\because \,\,\,\omega =\frac{\upsilon }{r} \right)\]
or \[\frac{1}{2}I\frac{{{\upsilon
}^{2}}}{{{r}^{2}}}=\frac{3}{4}m{{\upsilon }^{2}}-\frac{1}{2}m{{\upsilon
}^{2}}=\frac{1}{4}m{{\upsilon }^{2}}\] or \[I=\frac{1}{2}m{{r}^{2}}\]
As \[\left(
\frac{1}{2}m{{r}^{2}} \right)\] is moment of inertia of a circular disc of mass
m and radius r about an axis passing through its centre and perpendicular to
its plane, the object must be a circular disc.
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