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Question
Derive expressions for the linear velocity at the lowest position, mid-way position and top-most position for a particle revolving in a vertical circle, if it has to just complete circular motion without string slackening at the top.
Solution
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:
- Its weight is mg, vertically downwards, which is constant.
- 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).
∴ TA + mg = `(m"v"_A^2)/r`
At the topmost position, A, the tension in the string is 0.
∴ TA = 0
∴ mg = `(m"v"_A^2)/r`
∴ `("v"_A)_"min" = sqrt(rg)` ....(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.
∴ TB - mg = `(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.
∴ mg(2r) = `1/2m"v"_B^2 - 1/2m"v"_A^2`
∴ `"v"_B^2 - "v"_A^2 = 4rg`
∴ `"v"_B^2 - rg = 4rg` .............[from (i)]
∴ `("v"_B)_"min" = sqrt(5rg)` ....(ii)
Mid-way Positions (C and D):
Let vC be the velocity at point C.
Total energy at point C = Kinetic energy + Potential Energy
= `1/2m"v"_C^2 + mgr`
Total energy at point B = Kinetic energy + Potential Energy
= `1/2m"v"_B^2 + 0 = 5/2 mgr` ....[From (ii)]
From the law of conservation of energy,
Total energy at point C = Total energy at point B
∴ `1/2m"v"_C^2 + mgr = 5/2mgr`
∴ `"v"_C^2 + 2rg = 5rg`
∴ `("v"_C)_"min" = sqrt(3rg)`
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