Question

An object of mass 0.5 kg is whirled at the end of a string 0.8 long.If the string makes 3 revolutions in 1.2 second, find the tension in the string​

Answer 1

Answer:

The tension of the string will be 24.65N.

Answer 2

Tension in the string​ is 65.84 N .

Given :

Mass of object , m = 0.5 kg .

Length of string , l = 0.8 m .

It is also given that string makes 3 revolutions in 1.2 second .

So , its angular frequency is , $v=\dfrac{2\pi r(3)}{1.2}=12.57\ m/s$ .

We know , tension in rotating string is given by :

$T=m\dfrac{v^2}{R}$

Putting given values in above equation .

We get ,

$T=0.5\times\dfrac{(12.57)^2}{1.2}\\\\T=65.84\ N$

Hence , this is the required solution .

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