Physics System of Particles and Rotational Motion

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

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

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

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

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

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

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

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

 

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

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

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

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

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

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

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

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

The direction of torque and angular momentum defines how and in which orientation an object will rotate or sustain its spin. This is important to understand in machines, athletic movements, and even natural phenomena, such as planetary motion. 

Both torque and angular momentum have direction. We determine their direction based on the right-hand rule. That makes them vector quantities in rotational motion. 

Yes, an object can have angular momentum without torque. The physics behind it is that if no external force act on an object, its angular momentum is constant. That is based on the law of conservation of momentum. 

Torque is the measure of rotational force on an object (rigid or extended body). It changes the object's angular momentum. To calculate in physics, torque equals the rate of change of angular momentum. The formula for that is

$?=\frac{dL}{dt}$