Physics - W.E.P. Question with Solution | TestHub

Let$m$be the mass of the disc. Then translational kinetic energy of the disc is:

$K_T=\frac{1}{2}m{v}^2$....(1)

When it ascends on a smooth track its rotational kinetic energy will remain same while translational kinetic energy will go on decreasing. At highest point Translational kinetic energy is transferred to Potential Energy.

$K_T=mgh$

or$\frac{1}{2}m{v}^2=mgh$

or$h=\frac{v^2}{2g}=\frac{{\left(6\right)}^2}{2\times 10}=1.8\mathrm{ }\mathrm{m}$