Question

A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is
(a) √glgl√
(b) √2gl2gl√
(c) √3gl3gl√
(d) √5gl5gl√

Answer 1

The minimum speed of the particle is $\sqrt{3 g l}$.

Explanation:

Let the Particulate speed be v m / s when the strings are vertical.

Now, by its own weight, it's balanced  i.e., mg.  

Therefore,  

                $\frac{m v^{2}}{l}=m g$

               $m v^{2}=m g l$

                 $v^{2}=\frac{m g l}{m}$

                 $v^{2}=g l$

Let the Particles speed be u m / s, when the strings are horizontal.

Now,  

Total Energy $=\frac{1}{2} m u^{2}$

Also, Total Energy   $=\frac{1}{2} m v^{2}+\mathrm{mgl}$

Thus,  

We're getting from this,

                          $\frac{1}{2} m u^{2}=\frac{1}{2} m v^{2}+\mathrm{mgl}$

            $\frac{1}{2} m u^{2}-\frac{1}{2} m v^{2}=\mathrm{mgl}$

              $\frac{1}{2} m\left(u^{2}-v^{2}\right)=\mathrm{mgl}$

                    $\left(u^{2}-v^{2}\right)=\frac{2 m g l}{m}$

                     $\left(u^{2}-v^{2}\right)=2 m g l$

                                $v^{2}=g l$

                         $u^{2}-g l=2 g l$

                                 $\mathrm{u}^{2}=2 \mathrm{g} +\mathrm{gl}$

                                 $u^{2}=3 g l$

                                  $u=\sqrt{3 g l}$

Thus, the speed is $u=\sqrt{3 g l}$.

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

$\sqrt{3g} l$