3.1 Conditions for a particle in a circular motion

1.1 Angular Variables
1.2 Kinematic equation for circular motion
1.3 Relation between angular and linear variables
1.4 Unit vectors along the radius and tangent
1.5 Velocity and acceleration of particle in circular motion

2.1 Centripetal force
2.2 Application of circular motion

3.1 Conditions for a particle in a circular motion

4.1 Conditions for a rod in a vertical circular motion

5.1 By friction
5.2 By banking of roads
5.3 By both friction and banking of road

When the particle rotates in a vertical circle, then whether the particle will rotate through the complete circle, or leave the circle at any point or oscillate about the center depends on the initial velocity imparted to the particle.

Different conditions for a particle in a circular motion are,

1. Rotating through the complete circle

2. Oscillation

3. Leaving the circular path

For a particle to move through the complete circle, the horizontal velocity $u$ is imparted to it in the horizontal direction as shown in the figure.

We have to find the minimum velocity $v$ at point B, so that the particle just rotates through the complete circle.

So, when the particle reaches the highest point $B$, tension $(T_B)$ and weight $(mg)$ acts on the particle.

The net force in the radial direction will provide the necessary centripetal acceleration.

From FBD at point $B$ we get, $${T_B} + mg = \frac{{m{v^2}}}{l}$$

In the above equation, only $T_B$ and $v$ are variables.

So for minimum value of velocity $v$, tension $T_B$ has to be minimum.

The minimum value of tension $T_B$ is zero, as tension in a massless inextensible string cannot be negative because it always acts in a direction opposite to the particles. Therefore,

$${T_B} = 0$$ So, $$v = \sqrt {gl} $$

Now,

$h = + 2l$ (as particle moves upwards, so $h$ is positive)

$v = \sqrt {gl} $ (velocity at point $A$)

$u = ?$ (velocity at point $B$)

$$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ {u^2} = gl + 4gl \\ u = \sqrt {5gl} \\\end{aligned} \end{equation} $$

A particle is said to oscillate when it makes to and fro motion about a fixed point $O$. In a vertical circle a particle will oscillate only on the red path between $B$ and $D$ through the lowest point $A$.

$u$ is the horizontal velocity imparted to the particle in the horizontal direction such that the particle reaches only till point $B$ or $D$.

We have to find the maximum velocity $u$ at point $A$, so that the velocity $v$ at point $B$ or $D$ becomes zero.

From FBD at point $B$ we get,

$mg$ will provide the translational retardation.

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

For velocity $v$ to be zero, tension $T_B$ in the string should be zero. Therefore, $${T_B} = 0,\;v - 0$$

Now,

$h=+l$ (as particle moves upwards so, $h$ is positive).

$$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ u = \sqrt {2gl} \\\end{aligned} \end{equation} $$

The point at which the tension $T$ in the string is zero, the particle will leave the circular path.

A particle will leave the circular path between point $B$ and $C$, i.e. on the red path.

So, the particle has to cross point $B$ and should not reach point $C$. If the particle will not cross point $B$, then it will start oscillatory motion and if the particle will cross point $C$, then it will rotate through the complete circle.

So, we have to find the horizontal velocity $(u)$ at point $A$ that should be imparted to the particle, so that it crosses point $B$ but does not reach point $C$.

Let the velocities at point $A$, $B$ and $C$ be $u$, $v_B$ and $v_C$ respectively as shown in the figure.

From FBD at point $B$ we get,

$mg$ provides the translational retardation. $${T_B} = \frac{{mv_B^2}}{l}$$

Now,

$h=+l$ (As particle moves upward so $h$ is positive)

$$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ 0 = {u^2} - 2gl \\ u = \sqrt {2gl} \\\end{aligned} \end{equation} $$

From FBD at point $C$ we get, $${T_C} + mg = \frac{{mv_C^2}}{l}$$

For particle not to cross point $C$, the velocity $v_C$ should be minimum.

So, tension $T_C$ should be minimum i.e. zero. Tension $T_C$ cannot be negative as it always acts in a direction opposite to the particle.

Therefore, $${v_C} = \sqrt {gl} $$ Now,

$h=+2l$ ( As particle moves upward, so $h$ is positive)

$$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ gl = {u^2} - 2g( + 2l) \\ u = \sqrt {5gl} \\\end{aligned} \end{equation} $$