# Q. A particle of mass `m` tied by a string is released from rest from the horizontal position.It can rotate in a complete vertical circle.What is the tension in the string, when it swings by an...

Q. A particle of mass `m` tied by a string is released from rest from the horizontal position.It can rotate in a complete vertical circle.What is the tension in the string, when it swings by an angle of 90° ?

A) `mg`

B) `1.5mg`

C) `2mg`

D) `3mg`

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For a diagram of forces when the string swings by an angle of 90 degree (string is vertical) please see the figure below (R is the length of the string).

The motion of particle is circular it means that the forces that act on the string are the centrifugal force `Fcf = m*a_(cf) = m*v^2/R` and the weight of the particle `G= m*g` .

We can find the speed of particle at the bottom of the trajectory (when the string is vertical) from energy considerations. When the string is horizontal (just before the particle is released) there is only potential energy. `Ep = m*g*R`

When the string is vertical all the above potential energy is transformed into kinetic energy. `Ek = m*V^2/2`

Therefore `m*g*R =m*V^2/2` or `V= sqrt(2*g*R)` .

At the bottom of the trajectory the direction of this speed is horizontal (tangent to the trajectory).

Now we can compute the centrifugal force when the string is vertical

`Fcf = m*V^2/R = m*((2*g*R)/R) =2*m*g`

Therefore the total tension in the string is

`T = Fcf +G = 2*m*g +m*g =3*m*g`

**The correct answer is D)**