3.0 Motion in a vertical circle

3.1 Conditions for a particle in a circular motion

Suppose a particle of mass $m$ is tied to a massless inextensible string of length $R$ and is whirled in a vertical circle about the fixed point $O$.

It is imparted initial velocity $u$ in horizontal direction at the lowest point $A$.

Let $v$ be the velocity at point $B$ of the circle as shown in the figure.

At point $A$, net force is, $${\overrightarrow F _A} = T - mg\quad {\text{(towards the center)}}$$

This net force $\left( {{{\overrightarrow F }_A}} \right)$ will provide the necessary centripetal acceleration at point $A$. Therefore,

$$\begin{equation} \begin{aligned} {\overrightarrow F _A} = T - mg \\ T - mg = \frac{{m{u^2}}}{R}\quad ...(i) \\\end{aligned} \end{equation} $$

At point $B$, net force is, $${\overrightarrow F _B} = T - mg\cos \theta \quad {\text{(towards the center)}}$$

This net force $\left( {{{\overrightarrow F }_B}} \right)$ will provide the necessary centripetal acceleration at point $B$. Therefore,

$$\begin{equation} \begin{aligned} {\overrightarrow F _B} = T - mg\cos \theta \\ T - mg\cos \theta = \frac{{m{v^2}}}{R}\quad ...(ii) \\\end{aligned} \end{equation} $$

Here $mg\sin \theta $ at point $B$ will provide the translational retardation as force and velocity are in opposite direction.

Now we have to find relation between the variables as shown in the figure.

$$h = R(1 - \cos \theta )$$

So, by applying law of conservation of energy at point $A$ and $B$ we get,

$$\begin{equation} \begin{aligned} {\text{Total energy at point }}A = {\text{Total energy at point }}B \\ {\left( {KE} \right)_A} + {\left( {PE} \right)_A} = {\left( {KE} \right)_B} + {\left( {PE} \right)_B} \\ \frac{1}{2}m{u^2} + 0 = \frac{1}{2}m{v^2} + m\overrightarrow g \overrightarrow h \\ {v^2} = {u^2} - 2\overrightarrow g \overrightarrow h \\\end{aligned} \end{equation} $$

Note: When the particle moves upward, $\overrightarrow h $ is positive and when the particle moves downward, $\overrightarrow h $ is negative.

Question 3. A bob of mass $m$ attached to a massless inextensible string of length $l$. The horizontal velocity $u = \sqrt {6gl} $ is imparted to it at the lowest point in the horizontal direction. Find the velocity $v$, tension in the string at point $A$ and $B$ as shown in the figure.

Solution: The velocity at bob at point $A$ and $B$ is $u$ and $v$ respectively.

Height it rises from point $A$ to $B$: $h = l(1 + \cos \theta )$

As the height increases, so it is positive

Therefore, $$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ {v^2} = 6gl - 2gl(1 + \cos 60^\circ ) \\ v = \sqrt {3gl} \\\end{aligned} \end{equation} $$

From FBD at point $A$ we can write, $$\begin{equation} \begin{aligned} T - mg = \frac{{m{u^2}}}{l} \\ T = mg + \frac{{m(6gl)}}{l} \\ {T_A} = 7mg \\\end{aligned} \end{equation} $$

From FBD, at point $B$ we can write, $$T + mg\sin \theta = \frac{{m{v^2}}}{l}$$

Here, $mg\cos \theta $ provides the translational retardation.

$$\begin{equation} \begin{aligned} T = \frac{{m(3gl)}}{l} - mg\sin 60^\circ \\ T = 3mg - \frac{{\sqrt 3 }}{2}mg \\ {T_B} = \sqrt 3 \left( {\sqrt 3 - \frac{1}{2}} \right)mg \\\end{aligned} \end{equation} $$