Question

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?
(A) The magnitude of the tension increases to four times its original value, 4F.
(B) The magnitude of the tension reduces to half of its original value, F/2.
(C) The magnitude of the tension is unchanged.
(D) The magnitude of the tension reduces to one-fourth of its original value, F/4.
(E) The magnitude of the tension increases to twice its original value, 2F.

Answer 1

Answer:

(A) The magnitude of the tension increases to four times its original value, 4F.

Explanation:

When an object moves in circular motion,  a centripetal force acts on it . In this scenario the centripetal force acting on the stone is given by $\frac { m{ v }^{ 2 } }{ r }$.

                   Where m is the mass of object

                               v- velocity or speed of the object

                               r - radius of the circle

Important to note is that the tension is equal to the  centripetal force.

Given that initially the string makes one complete revolution per second and then speeds up to make two complete revolutions in a second .It implies that the speed has doubled .

Using our equation :F =$\frac { m{ v }^{ 2 } }{ r }$

                               where F is the tension in the string

let the initial speed be =v  then after it doubles it becomes 2v

Keeping the radius of the circle unchanged we have :

F=$\frac { m{ (2v) }^{ 2 } }{ r } =\frac { 4m{ v }^{ 2 } }{ r }$

From the equation it can be seen that the initial Tension has  increased by a factor of 4 .

Therefore the magnitude of the tension increases to four times its original value, 4F.