Motion In A Vertical Circle | Circular Motion

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 $m g$ 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 $m g \cos θ$ and the centripetal force of the body. $T = F + m g \cos θ$ $T = \frac{m v^{2}}{r} + m g \cos θ$

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. $At point A, θ = 0 °,$ $T_{max} = \frac{m v^{2}}{r} + m g \cos 0$ $T_{max} = \frac{m v^{2}}{r} + m g$

Minimum Tension in the String

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

Tension in the String at B and D

$At point B or D, θ = 90 °,$ $T = \frac{m v^{2}}{r} + m g \cos 90$ $T = \frac{m v^{2}}{r}$

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. $0 + m g = \frac{m v_{min}^{2}}{r}$ $m g = \frac{m v_{min}^{2}}{r}$ $v_{min}^{2} = g r$ $v_{min} = \sqrt{g r}$ 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, $K.E. of body at A = (K.E.+P.E.) of the body at C$ $\frac{1}{2} m v^{2} = \frac{1}{2} m v_{min}^{2} + m g (2 r)$ $\frac{1}{2} v^{2} = \frac{1}{2} v_{min}^{2} + 2 g r$ $v^{2} = v_{min}^{2} + 4 g r$ $Here, v_{min}^{2} = g r$ $v^{2} = g r + 4 g r$ $v = \sqrt{5 g r}$ Hence, the velocity must be equal to or greater than $\sqrt{5 g r}$ 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 $\sqrt{5 g r}$ , 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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