4.1 Conditions for a rod in a vertical circular motion

  Circular Motion

    1.0 Introduction
    2.0 Dynamics of circular motion
    3.0 Motion in a vertical circle
    4.0 Rigid body rotating in a vertical circle
    5.0 Circular turning of roads
    6.0 Conical Pendulum
    7.0 Death well
    8.0 Rotor
    9.0 Bending of a cyclist or motorcyclist while taking turn
    10.0 Centrifugal force

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4.1 Conditions for a rod in a vertical circular motion

When the rod rotates in a vertical circle, then whether the rod 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 rod

When the rod rotates in a vertical circle, then whether the rod 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 rod

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4.1.1 Condition for rotating through the complete circle

The initial horizontal velocity $u$ decides whether a rod rotates in a complete circle or not.

Let the velocity at the lowest point $A$ and the highest point $B$ is $u $ and $v$ respectively.

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

For velocity $v$ at point $B$ to be minimum, LHS has to be minimum i.e. zero. So, $$\begin{equation} \begin{aligned} {T_B} + mg = 0 \\ {T_B} = - mg \\\end{aligned} \end{equation} $$ So, $$v=0$$

In rod the tension can act away from the rod.

As tension is a vector quantity. Therefore $T_B=-mg$ , means that the magnitude of tension is equal to $mg$ but in the opposite direction.

So, the velocity of rod at its highest point $B$ in a circular motion is zero. But the rod will complete its rotation due to the inertia gained in the previous motion.

Now,

$v=0$ (velocity of rod at point $B$)

$h=+2l=+L$ (Rise in height of the centre of mass $COM$ and it is positive as $COM$ moves upward).

$u=?$ (velocity of rod at point $A$)

Therefore, $$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ 0 = {u^2} - 2g(2L) \\ u = 2\sqrt {gL} \\\end{aligned} \end{equation} $$

Therefore for rod to rotate through the complete circle, the initial velocity $(u)$ at point $A$ should be imparted in horizontal direction such that $u \geqslant 2\sqrt {gL} $.

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4.1.2 Condition for oscillation

The rod will oscillate in a vertical circle about point $O$ through out the circle except at its highest point $B$ as shown in the figure.

So, the minimum velocity for a rod to complete the circular motion is $u = 2\sqrt {gL} $.

For rod to oscillate in a vertical circle, the initial velocity $(u)$ at point $A$ should be imparted in horizontal direction such the rod does not reaches the highest point B. Therefore,

$$u < 2\sqrt {gL} $$

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4.1.3 Condition for rod to leave the circular path

Rod or any rigid body will never leave the circular path about the hinge point $O$.

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Question 4. A stone of mass $m$ is tied to one end of an inextensible massless string of length $l$ and is whirled in a vertical circle. Find the minimum speed the stone can have at the highest point of the circle.

Solution: Let $v$ be the velocity at the highest point $A$ in the circular motion.

$h$ is the vertical distance moved by the particle upwards.

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

For velocity $v$ to be minimum, LHS should be minimum. Therefore,

$${(T + mg)_{{\text{minimum}}}} = mg$$

As weight $mg$ is constant, so minimum value of tension $T$ is zero because it cannot be negative or its direction cannot be towards the stone.

So, $$\begin{equation} \begin{aligned} mg = \frac{{m{v^2}}}{l} \\ {v_{\min }} = \sqrt {gl} \\\end{aligned} \end{equation} $$

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Question 5. A body of mass $m$ tied to an inextensible massless string of length $l$ is rotated about one of its end. A body at its lowest point is imparted a horizontal velocity of $2\sqrt {gl} $. Find the angle $\theta $ at which the body will leave the circular path as shown in the figure.

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

For body to leave the circular path tension $T$ at point $B$ should be zero.

Therefore, $$\begin{equation} \begin{aligned} T = 0 \\ v = \sqrt {gl\sin \theta } \quad ...(i) \\\end{aligned} \end{equation} $$

Now,

$u = 2\sqrt {gl} $ (Initial horizontal velocity)

$h = + l(1 + \sin \theta )$ (rise in height of the body)

$v = \sqrt {gl\sin \theta } $ (velocity at point $B$)

$$\begin{equation} \begin{aligned} {v^2} = {u^2} - 2gh \\ gl\sin \theta = 4gl - 2gl(1 + \sin \theta ) \\ \sin \theta = \frac{2}{3}\quad {\text{or}}\quad \theta = {\sin ^{ - 1}}\left( {\frac{2}{3}} \right) \\\end{aligned} \end{equation} $$