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nataly862011
1 month ago
8

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m

agnitude 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.
Physics
1 answer:
Maru [3.3K]1 month ago
3 0

Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by \frac { m{ v }^{ 2 } }{ r }.

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =\frac { m{ v }^{ 2 } }{ r }

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

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

From this equation, it's clear that the initial tension has quadrupled.

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

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