System of Particles and Rotational Motion

$V_P=\sqrt[]{V_0^2+V_0^2}=\sqrt[]{2}{V}_0$

Consider translational motion of a body

mg sinq - f_S = ma             .(i)

Consider rotational motion of body about its centre of mass

$f_{S}R=l\alpha \Rightarrow \alpha =\frac{f_{S}R}{l}$ .(ii)

Using the condition of rolling without slipping at contact, we have

$\alpha R=a\Rightarrow \frac{f_S{R}^2}{l}=\frac{mgsin\theta -f_S}{m}\Rightarrow f_S=\frac{lmgsin\theta }{l+m{R}^2}$

Using the condition of static friction, we have

$f_S\le \mu N\Rightarrow \frac{lmgsin\theta }{l+m{R}^2}\le \mu mgcos\theta \Rightarrow \mu \ge \frac{ltan\theta }{l+m{R}^2}...........\left(iii\right)$

$\alpha =\frac{mgRsin\theta }{l+m{R}^2}\Rightarrow a=\alpha R=\frac{mg{R}^{2}sin\theta }{l+m{R}^2}=\frac{g{R}^{2}sin\theta }{k^2+R^2}...........\left(iv\right)$

$t=\sqrt[]{\frac{2\mathcal{l}}{a}}\Rightarrow$ Time required to reach the ground

$\Rightarrow \frac{v_{ring}}{v_{cylinder}}=\sqrt[]{\frac{2g\mathcal{l}{R}^{2}sin\theta }{k_{ring}^2+R^2}}*\sqrt[]{\frac{k_{cylinder}^2}{2g\mathcal{l}{R}^{2}sin\theta }}=\sqrt[]{\frac{\frac{R^2}{2}+R^2}{R^2+R^2}}=\frac{\sqrt[]{3}}{2}$

$K_T=\frac{m{v}^2}{2}and{K}_R=\frac{l{\omega }^2}{2}=\frac{l{v}^2}{2R^2}$

According to question, we can write

$K_R=\frac{1}{2}{K}_T\Rightarrow \frac{l{v}^2}{2R^2}=\frac{1}{2}*\frac{m{v}^2}{2}\Rightarrow l=\frac{m{R}^2}{2}\Rightarrow Solidcylinder$

$l_z=l_x+l_y$

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

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

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

= 52 Mr²

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

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

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

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

$\overrightarrow{F}=5\widehat{i}+3\widehat{j}-7\widehat{k}$

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

$\overrightarrow{\tau }=\left|\overrightarrow{r}*\overrightarrow{F}\right|=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 3\hfill & \hfill -7\hfill \end{array}\right|\begin{array}{cc}\hfill =\widehat{i}\left(-k_1-3\right)-\widehat{j}\left(-14-5\right)+\widehat{k}\left(6-10\right)\hfill & \hfill =-17\widehat{i}+19\widehat{j}-4\widehat{k}\hfill \end{array}$

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.

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

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

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

$I_{OO\text{'}}=\left[\frac{1}{4}M{\left(\frac{a}{2}\right)}^2\right]*2+\left[\frac{5}{4}m{\left(\frac{a}{2}\right)}^2\right]*2$

$=\frac{6M{a}^2}{8}=\frac{3}{4}M{a}^2$

x=3