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Q5.26:
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative] $T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

Solution :

Step 1: Analyzing Forces at the Lowest Point

At the lowest point of the vertical circle, the stone is subjected to two primary forces acting along the vertical axis:

  • Tension ($T_1$) acting vertically upwards towards the center of the circle.
  • Weight ($mg$) acting vertically downwards.

The net force provides the necessary centripetal force directed towards the center. Therefore, the equation of motion is:

$T_1 - mg = \frac{mv_1^2}{R}$ [Per Newton's Second Law of Motion]

The question asks for the net force directed downwards. Since the centripetal force is directed upwards (towards the center at the bottom), the downward force is the negative of the centripetal force:

$F_{\text{net, down}} = -(T_1 - mg) = mg - T_1$

Step 2: Analyzing Forces at the Highest Point

At the highest point of the vertical circle, both forces act in the same direction:

  • Tension ($T_2$) acting vertically downwards towards the center.
  • Weight ($mg$) acting vertically downwards.

The sum of these forces provides the centripetal force required for circular motion:

$T_2 + mg = \frac{mv_2^2}{R}$ [Per Newton's Second Law of Motion]

Since both forces are acting vertically downwards, the net force directed downwards is simply the sum:

$F_{\text{net, down}} = T_2 + mg$

Step 3: Visual Representation of Forces

T₁ mg T₂ mg Vertical Circle Geometry

Step 4: Comparative Summary

Point Net Force (Downwards)
Lowest Point $mg - T_1$
Highest Point $T_2 + mg$

Final Solution: The net force directed vertically downwards at the lowest point is $(mg - T_1)$ and at the highest point is $(T_2 + mg)$.


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