Thank you for reporting, we will resolve it shortly
Q.
In figure $E$ and $V _{ cm }$ represent the total energy
and speed of centre of mass of an object of mass $1 \,kg$ in pure rolling. The object is:
KCETKCET 2021System of Particles and Rotational Motion
Solution:
$ E =\frac{1}{2} mv _{ cm }^{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right] $
$\frac{ E }{ v _{ cm }^{2}}=\frac{1}{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right]$
From graph, $\frac{ E }{ v _{ cm }^{2}}=\frac{3}{4}$
$\frac{3}{4}=\frac{1}{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right] $
$\frac{3}{2}-1=\frac{ K ^{2}}{ R ^{2}} $
$\frac{ K ^{2}}{ R ^{2}}=\frac{1}{2}$
For disc, $I =\frac{ mR ^{2}}{2}$
$\frac{m R^{2}}{2}=m K^{2} $
$\frac{K^{2}}{R^{2}}=\frac{1}{2}$
