Question
Derive an expression for linear velocity at lowest position, midway position and the top-most position for a particle revolving in a vertical circle, if it has to just complete circular motion without string slackening at top.
✓
Answer
Vertical circular motion
Consider a bob (treated as a point mass) tied to a (practically) massless, inextensible and flexible string. It is whirled along a vertical circle so that the bob performs a vertical circular motion and the string rotates in a vertical plane. At any position of the bob, there are only two forces acting on the bob:
1. Its weight is mg , vertically downwards, which is constant.
2. The force due to the tension along the string is directed along the string and towards the centre. Its magnitude changes periodically with time and location.
Top-most position (A):
Force due to tension in the string and mg are both in the same direction (downward).
$\therefore T_A+m g=\frac{m v_A^2}{r}$
At the topmost position, A , the tension in the string is 0 .
$\therefore T_A=0$
$\therefore m g=\frac{m v_A^2}{r}$
$\therefore\left(v_A\right)_{\min }=\sqrt{T g} \ldots \text { (i) }$
Lower-most position (B):
Force due to tension in the string is in an upward direction towards the center and opposite to the direction of mg .
$\therefore T_B-m g=\frac{m v_B^2}{r}$
While moving down from the uppermost to the lowermost point, the gravitational potential energy is converted into kinetic energy because the motion is governed only by gravity.
$\therefore m g(2 r)=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2$
$\therefore v_B^2-v_A^2=4 r g$
$\therefore v_B^2-r g=4 r g \ldots \ldots \ldots(\text { from (i) })$
$\therefore\left(v_B\right)_{\min }=\sqrt{5 r g \ldots} \ldots \text { (ii) }$
Mid-way Positions (C and D):
Let $v_C$ be the velocity at point $C$.
Total energy at point $C=$ Kinetic energy + Potential Energy
$=\frac{1}{2} m v_C^2+m g r$
Total energy at point $B=$ Kinetic energy + Potential Energy
$=\frac{1}{2} m v_B^2+0=\frac{5}{2} m g r \ldots \text { (From (ii)) }$
From the law of conservation of energy,
Total energy at point $C=$ Total energy at point $B$
$\therefore \frac{1}{2} m v_C^2+m g r=\frac{5}{2} m g r$
$\therefore v_C^2+2 r g=5 rg$
$\therefore\left(v_C\right)_{\min }=\sqrt{3 rg}$
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