Physics System of Particles and Rotational Motion

Using parallel axis theorem:

$l=2*\left[l_{cm}+m{d}^2\right]$  

$l=2*\left[\frac{2}{5}\left(1.5\right){\left(0.5\right)}^2+\left(1.5\right){\left(2.5\right)}^2\right]$

l = 19.05 kgm²

$\overrightarrow{L}=\overrightarrow{r}*\overrightarrow{p}\Rightarrow \overrightarrow{L}=mvr\left(\widehat{k}\right)$

Direction & magnitude both remain same

 for particle moving with constant speed.

From Conservation of Angular Momentum

  $\Rightarrow l_1{\omega }_1+l_2{\omega }_2=\left(l_1+l_2\right)\omega$             

$\omega =\frac{l_1{\omega }_1+l_2{\omega }_2}{l_1+l_2}$

$K{E}_i=\frac{1}{2}{l}_1{\omega }_1^2+\frac{1}{2}{l}_2{\omega }_2^2$

$K{E}_f=\frac{1}{2}\left(l_1+l_2\right){\omega }^2$

$\Rightarrow \left|K{E}_f-K{E}_i\right|=\frac{1}{2}\left(\frac{l_1{l}_2}{l_1+l_2}\right){\left({\omega }_1-{\omega }_2\right)}^2$

$l_z=l_x+l_y$

$l_z=\frac{m{\mathcal{l}}^2}{3}+\frac{m{\mathcal{l}}^2}{3}=\frac{2}{3}m{\mathcal{l}}^2$

= $l_{xx\text{'}}=\left[l_{yy\text{'}}+M{\left(\frac{5r}{2}\right)}^2\right]+\left[l_{ZZ\text{'}}+M{\left(\frac{13r}{2}\right)}^2\right]$

$M\left(\frac{36r^2}{12}\right)+\frac{M*25r^2}{4}=\frac{M{r}^2}{2}+M*\frac{169M{r}^2}{4}$

= 52 Mr²

^{$\overrightarrow{r}=2\widehat{i}+\widehat{j}+2\widehat{k}$}

^{$\overrightarrow{\tau }=\overrightarrow{r}*\overrightarrow{F}$}

^{$=\left(2\widehat{i}+\widehat{j}+2\widehat{k}\right)*\left(3\widehat{i}+4\widehat{j}-2\widehat{k}\right)=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill -\widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill -2\hfill \end{array}\right|$}

^{$\overrightarrow{\tau }=-10\widehat{i}+10\widehat{j}+5\widehat{k}$}

l₁ = M. l of solid sphere about its diameter

$=\frac{2}{5}MR{\text{'}}^2=\frac{2}{5}M{\left(2R\right)}^2=\frac{8}{5}M{R}^2$               

l₂ = M. I of solid cylinder about its axis

$=\frac{M{R}^{12}}{2}=\frac{M{\left(2R\right)}^2}{2}=2M{R}^2$               

I₃ = M. I of solid circular disc about its diameter

$=\frac{M{R}^{12}}{4}=\frac{M{\left(2R\right)}^2}{2}=M{R}^2$               

I₄ = M. I of this circular ring about its diameter

  $=\frac{M{R}^{12}}{2}=\frac{M{\left(2R\right)}^2}{2}=2M{R}^2$              

$\Rightarrow 6M{R}^2+2M{R}^2=\frac{x8M{R}^2}{5}$               

x = 5

To keep,   $\Delta {\alpha }_{cm}=0$

      $M_1\Delta x_1+M_2\Delta X_2=0$          

$\Rightarrow 10*6+30\Delta X_2=0$      

$\Delta X_2=\frac{-10*6}{30}$

$\Delta X_2=-2cm$        

i.e. 2 cm towards the 10 kg block.

Moment of inertia of ring

$\mathrm{d}\mathrm{l}={\mathrm{d}\mathrm{m}\mathrm{r}}^2$

$\mathrm{I}={\int }_0^{\mathrm{a}}?(\mathrm{A}+\mathrm{B}\mathrm{r})2\pi \mathrm{r}\mathrm{d}\mathrm{r}\pi {\mathrm{r}}^2$

$=2\pi A{\int }_0^a?r^{3}dr+2\pi B{\int }_0^a?r^{4}dr$

 

$=2\pi \left. [A\frac{a^4}{4}+B\frac{a^5}{5}\right]$

$=2\pi {\mathrm{a}}^4\left. [\frac{\mathrm{A}}{4}+\frac{\mathrm{B}\mathrm{a}}{5}\right]$

$v=\sqrt[]{\frac{2gH}{1+k^2/R^2}}$

$\frac{V_{Cylinder}}{V_{Sphere}}=\sqrt[]{\frac{{\left(1+k^2/R^2\right)}_{Sphere}}{{\left(1+k^2/R^2\right)}_{Cylinder}}}$

= $\sqrt[]{\frac{1+2/5}{1+1/2}}=\sqrt[]{\frac{14}{15}}$