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  4. An object of radius R and mass M is rolling horizontally without slipping with speed v. It then rolls up the hill to a maximum height is h=(3 v2/4 g). The moment of inertia of the object is ( g= acceleration due to gravity):

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Q. An object of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v.$ It then rolls up the hill to a maximum height is $h=\frac{3 v^{2}}{4 g}.$ The moment of inertia of the object is
( $g=$ acceleration due to gravity):

NTA AbhyasNTA Abhyas 2020

Solution:

Loss in $KE=$ Gain in $PE$
$\frac{1}{2}mv^{2}\left(1 + \frac{K^{2}}{R^{2}}\right)=mgH$
$\frac{v^{2}}{2}\left(1 + \frac{K^{2}}{R^{2}}\right)=g\left(\frac{3 v^{2}}{4 g}\right)$
$\frac{K^{2}}{R^{2}}=\frac{1}{2}$
$I=MK^{2}=\frac{M R^{2}}{2}$