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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]

                Lowest Point                              Highest Point

  (a)     $mg - T_1$                                          $mg + T_2$

(b)     $mg + T_1$                                          $mg - T_2$

(c)     $mg + T_1 - (mv_1^2))/R$           $mg - T_2 + (mv_1^2))/R$

(d)     $mg - T_1 - (mv_1^2))/R$            $mg + T_2 + (mv_1^2))/R$

$T_1$ and $v_1$ denote the tension and speed at the lowest point,

$T_2$ and $v_2$ denote corresponding values at the highest point.

Answers (1)

best_answer

The FBD of the stone at the lowest point is shown below :

Laws of motion, 20275

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,

$Fnet = T\ -\ mg\ =\ \frac{mv1^2}{r}$

Where, v = Velocity at the lowest point

The FBD of the stone at the highest point is given below :

Laws of motion, 20275 (2)

Using Newton's law of motion we have :

$T\ +\ mg\ =\ \frac{mv^2}{r}$

Where v = Velocity at the highest point

It is clear from the above equations that the net force acting at the lowest and the highest points are respectively (T – mg) and (T + mg).

Posted by

Devendra Khairwa

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