Physics System of Particles and Rotational Motion

Using conservation of energy:

$mgh=\frac{1}{2}{l}_{cm}{\omega }^2+\frac{1}{2}m{v}^2=\frac{1}{2}{l}_{cm}\frac{V^2}{r^2}+\frac{1}{2}m{v}^2$

$v^2=\frac{mgh}{\frac{1}{2}\frac{l_{cm}}{r^2}+m{v}^2}$

So, sphere has the greatest velocity of COM and ring has least.

${?}^2={?}_0^2+2???{\left(\frac{2?}{60}*1800\right)}^2={\left(\frac{2?}{60}*600\right)}^2+2\left(\frac{2?}{60}*\frac{1200}{10}\right)?$

$?{\left(60?\right)}^2?{\left(20?\right)}^2=2\left(4?\right)???=\frac{80?*40?}{2*4?}=400?\text{?}\text{?}rad$

Number of rotations = $\frac{?}{2?}=\frac{400?}{2?}=200$

Case – I : When disk slides down

$t_1=\sqrt[]{\frac{2L}{gsin\theta }}..........\left(i\right)$              

Case – II : When disk rolls down

$t_2=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\beta }^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\left(\frac{k}{R}\right)}^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\left(\frac{R/\sqrt[]{2}}{R}\right)}^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+\frac{1}{2}}}}=\sqrt[]{\frac{4L}{3gsin\theta }}......\left(ii\right)$             

$\Rightarrow \frac{t_1}{t_2}=\sqrt[]{\frac{2L}{gsin\theta }}*\sqrt[]{\frac{3gsin\theta }{4L}}=\sqrt[]{\frac{3}{2}}$              

Using conservation of linear momentum, we can write

$mu=M{v}_1\Rightarrow v_1=\frac{mu}{M}........\left(1\right)$           

Using conservation of angular momentum about centre of mass of Rod, we can write

$v_1+\frac{\mathcal{l}\omega }{2}=u\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\Rightarrow \frac{m}{M}=\frac{1}{4}$     . (2)

Using definition of e, we can write

$v_1+\frac{\mathcal{l}\omega }{2}=u\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\Rightarrow \frac{m}{M}=\frac{1}{4}$  

$l=\frac{m{\mathcal{l}}^2}{3}$  

Energy conservation Low

$mg\mathcal{l}=\frac{1}{2}\frac{m{\mathcal{l}}^2}{3}{\omega }^2.....\left(i\right)$                

And speed V =    $\omega r=\omega \mathcal{l}$

then    $V=\sqrt[]{6g\mathcal{l}}=\sqrt[]{6*10*.6}$

->6 m/s

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