Circular Motion

# Motion In A Vertical Circle

Here, we will discuss about the different cases and analyse the motion of a body in a vertical circle.

​Suppose a body of mass $m$ is tied at the end of a string. Let the body be moving in a vertical circle with constant speed $v$. Let the radius of the circle be $r$ which is the length of the string. Let $P$ be any point on the circle at angle $θ$ from mean position. Consider four positions $A$, $B$, $C$ and $D$ on the circle. $A$ is the lowest point and $C$ is the highest point. Different components of force at $P$ are shown in figure given below.

​Let $F$ be the centripetal force, $T$ be the tension in the string which always acts towards the point $O$ and $mg$ is the weight of the body which always acts downward. From the figure, it is clear that the tension in the string is balanced by $mg\cos θ$ and the centripetal force of the body. $T = F + mg\cos θ$ $T = \frac{mv^2}{r} + mg\cos θ$

## Maximum Tension in the String

​At the lowest point $A$, tension $T$ acts upward while the weight of the body acts downward. So, tension is maximum in the string at this point. $\text{At point A, }θ=0°,$ $T_{\text{max}} = \frac{mv^2}{r} + mg\cos 0$ $T_{\text{max}} = \frac{mv^2}{r} + mg$

## Minimum Tension in the String

​At highest point $C$, tension $T$ and weight $mg$ both acts downward. So, tension in the string at this point is minimum. $\text{At point C, }θ=180°,$ $T_{\text{min}} = \frac{mv^2}{r} + mg\cos 180$ $T_{\text{min}} = \frac{mv^2}{r} – mg$

## Tension in the String at B and D

​$\text{At point B or D, }θ=90°,$ $T = \frac{mv^2}{r} + mg\cos 90$ $T = \frac{mv^2}{r}$

## Minimum Velocity required to Loop in the Vertical Circle

​At highest point $C$, tension is minimum. If the tension is zero at this point, then the weight of the body provides necessary centripetal force to loop in a vertical circle. $0+mg=\frac{mv_{\text{min}}^2}{r}$ $mg=\frac{mv_{\text{min}}^2}{r}$ $v_{\text{min}}^2=gr$ $v_{\text{min}}=\sqrt{gr}$ This is the minimum velocity also called as critical velocity required at the top so that the string does not slack.

​At lowest point $A$, tension is maximum. According to the principle of conservation of energy, $\text{K.E. of body at A}=\text{(K.E.+P.E.) of the body at C}$ $\frac{1}{2}mv^2=\frac{1}{2}mv_{\text{min}}^2+mg(2r)$ $\frac{1}{2}v^2=\frac{1}{2}v_{\text{min}}^2+2gr$ $v^2=v_{\text{min}}^2+4gr$ $\text{Here, }v_{\text{min}}^2=gr$ $v^2=gr+4gr$ $v=\sqrt{5gr}$ Hence, the velocity must be equal to or greater than $\sqrt{5gr}$ for the body to move in a vertical circle.

• If a bucket containing water is rotated along a vertical circle such that its velocity at the lowest point is equal to or greater than $\sqrt{5gr}$, the water will not spill, even at the highest point.
• The pilot of a jet plane who is not tied to his seat will not fall down while looping the vertical circle.
• For the same reason, in circus, the motorcyclists are able to move in a vertical circle inside a cage.

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