Physics - Rotation Question with Solution | TestHub

If$I$be the moment of inertia of the cylinder about its axis, then

$I=\frac{1}{2} M{R}^2=\frac{1}{2}\times 20\times {\left(0.25\right)}^2=0.625\mathrm{ }\mathrm{kg} {\mathrm{m}}^2$

$\therefore KE$associated with the rotating cylinder is given by,

$KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.625\times {\left(100\right)}^2=3125\mathrm{ }\mathrm{J}$

Also using the relation,

$L=\sqrt{2I\left(KE\right)}$

We have$L=\sqrt{2\times 0.625\times 3125}$

$L=62.5 \mathrm{J} \mathrm{s}$