# Question f483d

Jan 13, 2017

Let us consider that a particle of mass $m$ is executing a vertical circular motion,centering the point $O$ as shown in figure and having been attached to an in-extensible string of length $R$.

Let $P$ represents an arbitrary position of the particle , where the string makes an angle $\theta$ with the vertical. If $v$ be the velocity of the particle at this position then considering forces on the particle we can write

For position P

$\frac{m {v}^{2}}{R} = T - m g \cos \theta \ldots \ldots . . \left[1\right]$

For position L (lowest bottom position where $\theta = 0$)

(mv_"bot"^2)/R=T_"bot"-mgcos0^@,

where ${v}_{\text{bot}}$ is the velocity of the particle at L

=>(mv_"bot"^2)/R=T_"bot"-mg...........[2]

For position H (highest top position where $\theta = {180}^{\circ}$)

(mv_"top"^2)/R=T_"top"-mgcos180^@,

where ${v}_{\text{top}}$ is the velocity of the particle at H

=>(mv_"top"^2)/R =T_"top"+mg...........[3]

Considering conservation of energy at $L \mathmr{and} H$ we can write

$K E \text{ at H}$

$= K E \text{ at L" -"gain of PE due to lift of height 2R (L to H) }$

$\frac{1}{2} m {v}_{\text{top"^2=1/2mv_"bot}}^{2} - 2 m g R$

$\implies {v}_{\text{top"^2=v_"bot}}^{2} - 4 g R \ldots \ldots . \left[4\right]$

The particle will complete the circle , if the string doesn't slack at the highest point when $\theta = {180}^{\circ}$ There must be centripetal force to make this happen. The minimum centripetal force required can be had by putting ${T}_{\text{top}} \approx 0$ in equation [3].

(m(v_"top"^2)_"min")/R =0+mg

$\implies {\left({v}_{\text{top}}\right)}_{\min} = \sqrt{g R} \ldots \ldots . \left[5\right]$

Inserting this value in equation [4] we can get the minimum required velocity of the particle at position $L$

$\implies {v}_{\text{top"^2=v_"bot}}^{2} - 4 g R$

=>(v_"top"^2)_ min=(v_"bot"^2)_"min"-4gR

=>(v_"bot"^2)_"min"=(sqrt(gR))^2+4gR

=>(v_"bot")_"min"=sqrt(5gR)#