question_answer

A small object of uniform density rolls up a curved surface with an initial velocity \[\upsilon \]. It reaches upto a maximum height of \[\frac{3{{\upsilon }^{2}}}{4g}\] with respect to the initial position. Name the given object.

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.