Physics - Rotation Question with Solution | TestHub

If no external torque acts on a system of particle then angular momentum of the system remains constant, that is,

$\mathrm{\tau }=0$

$\Rightarrow$$\frac{\mathrm{dL}}{\mathrm{dt}}=0$

$\Rightarrow$$\mathrm{L}=\mathrm{I\omega }=$constant

$\Rightarrow$${\mathrm{I}}_1{\mathrm{\omega }}_1={\mathrm{I}}_2{\mathrm{\omega }}_2$

$\therefore$$\frac{1}{2}{\mathrm{Mr}}^2{\mathrm{\omega }}_1=\frac{1}{2}\left(\mathrm{M}+2\mathrm{m}\right){\mathrm{r}}^2{\mathrm{\omega }}_2$...(i)

Here,$\mathrm{M}=2 \mathrm{kg}$,$\mathrm{m}=\text{0.25} \mathrm{kg}$and$\mathrm{r}=\text{0.2} \mathrm{m}$

${\mathrm{\omega }}_1=30\mathrm{ }\mathrm{rad} {\mathrm{s}}^{-1}$

Hence, we get after putting the given values in Eq. (i)

$\frac{1}{2}\times 2\times {\left(0.2\right)}^2\times 30=\frac{1}{2}\times \left(2+2\times 0.25\right){\left(0.2\right)}^2\times {\omega }_2$

$\Rightarrow$$60=2.5{\omega }_2$

$\therefore$${\omega }_2=24 \mathrm{r}\mathrm{a}\mathrm{d} {\mathrm{s}}^{-1}$