Question

plzzzz do it its urgent....​

Answer 1

Answer:-

       $\boxed{\sf \omega=\dfrac{2}{R}\;.\;\sqrt{\dfrac{gh}{3}}}$

Explanation:-

$\rule{300}{1.5}$

Here, we need to apply law of conservation of energy as conservative forces are acting here. So, Increase in Kinetic energy will equal the decrease in Potential energy.

$\longmapsto \textsf{Decrease in P.E}=\textsf{Increase in K.E}$

Here, the kinetic energy will be sum of kinetic energy of the translating motion and the kinetic energy of circular motion.

$\longmapsto\sf mgh=\dfrac{1}{2}\times mv^{2}+\dfrac{1}{2}\times I\omega^{2}\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[mv^{2}+I\omega^{2}\Bigg]\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[m\Big(R\;\omega\Big)^{2}+I\omega^{2}\Bigg]\ \ \ \ \because\Big(v=R\;\omega\Big)\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[mR^{2}\;\omega^{2}+I\omega^{2}\Bigg]$

We know for a circular disc the Moment of inertia is mR²/2

$\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[mR^{2}\;\omega^{2}+\dfrac{mR^{2}}{2}\omega^{2}\Bigg]\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\;mR^{2}\omega^{2}\Bigg[1+\dfrac{1}{2}\Bigg]\\\\\\\\\longmapsto\sf \not m gh=\dfrac{1}{2}\; \not m R^{2}\omega^{2}\Bigg[\dfrac{2+1}{2}\Bigg]\\\\\\\\\longmapsto\sf gh=\dfrac{1}{2}\times\dfrac{3}{2}\times R^{2}\omega^{2}\\\\\\\\\longmapsto\sf gh=\dfrac{3}{4}\times R^{2}\omega^{2}\\\\\\\\\longmapsto\sf \omega^{2}=\dfrac{4gh}{3\;R^{2}}$

$\longmapsto\sf \omega=\sqrt{\dfrac{4gh}{3\;R^{2}}}\\\\\\\\\longmapsto\sf \omega=\dfrac{2}{R}\;.\;\sqrt{\dfrac{gh}{3}}\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf \omega=\dfrac{2}{R}\;.\;\sqrt{\dfrac{gh}{3}}}}}}$

∴ We got Angular velocity of the circular disc.

$\rule{300}{1.5}$

Answer 2

Answer:

$\longmapsto \textsf{Decrease in P.E}=\textsf{Increase in K.E}$

$\longmapsto\sf mgh=\dfrac{1}{2}\times mv^{2}+\dfrac{1}{2}\times I\omega^{2}\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[mv^{2}+I\omega^{2}\Bigg]\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[m\Big(R\;\omega\Big)^{2}+I\omega^{2}\Bigg]\ \ \ \ \because\Big(v=R\;\omega\Big)\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[mR^{2}\;\omega^{2}+I\omega^{2}\Bigg]$

We know for a circular disc the Moment of inertia is mR²/2

$\longmapsto\sf mgh=\dfrac{1}{2}\Bigg[mR^{2}\;\omega^{2}+\dfrac{mR^{2}}{2}\omega^{2}\Bigg]\\\\\\\\\longmapsto\sf mgh=\dfrac{1}{2}\;mR^{2}\omega^{2}\Bigg[1+\dfrac{1}{2}\Bigg]\\\\\\\\\longmapsto\sf \not m gh=\dfrac{1}{2}\; \not m R^{2}\omega^{2}\Bigg[\dfrac{2+1}{2}\Bigg]\\\\\\\\\longmapsto\sf gh=\dfrac{1}{2}\times\dfrac{3}{2}\times R^{2}\omega^{2}\\\\\\\\\longmapsto\sf gh=\dfrac{3}{4}\times R^{2}\omega^{2}\\\\\\\\\longmapsto\sf \omega^{2}=\dfrac{4gh}{3\;R^{2}}$

$\longmapsto\sf \omega=\sqrt{\dfrac{4gh}{3\;R^{2}}}\\\\\\\\\longmapsto\sf \omega=\dfrac{2}{R}\;.\;\sqrt{\dfrac{gh}{3}}\\\\\\\\\longmapsto\large{\underline{\boxed{\blue{\sf \omega=\dfrac{2}{R}\;.\;\sqrt{\dfrac{gh}{3}}}}}}$

∴ We got Angular velocity of the circular disc.

$\rule{300}{1.5}$.