Question

A particle is whirled in a horizontal circle with speed v, with the help of a string of length L, the angle made by the thread with the vertical
if v=(3/2Lg)½ is π/n radian ,then find n​

Answer 1

Let the string makes an angle θ with the vertical.

Initially, before the magnetic field is applied, Tsinθ= mv²/r

If the particle experiences a force outward in the direction of the radius, tension T will increase.

If the particle experiences a force inward in the direction of the radius towards the center, tension T will decrease.

Answer 2

Given info : A particle is whirled in a horizontal circle with speed v, with the help of a string of length L.

To find : the angle made by the thread with the vertical , if v = $\sqrt{\frac{3}{2}gL}$ is $\frac{\pi}{n}$ then the value of n is...

solution : let the angle made by the thread with the vertical is θ.

at equilibrium,

      centripetal force acting on the particle = component of tension in the string in horizontal direction.

⇒ $\frac{mv^2}{r}=Tsin\theta$ , ...(1)

again,

vertical component of tension of string = weight of particle

⇒ Tcosθ = mg ...(2)

from equations (1) and (2) , we get

tanθ = $\frac{v^2}{rg}$

from figure, it is clear that r = Lsinθ

⇒ tanθ sinθ = $\frac{\frac{3}{2}gL}{gL}$ = 3/2

⇒ 2sin²θ = 3cosθ

⇒ 2(1 - cos²θ) = 3cosθ

⇒ 2 - 2cos²θ = 3cosθ

⇒ 2cos²θ + 3cosθ - 2 = 0

⇒ 2cos²θ + 4cosθ - cosθ - 2 = 0

⇒ (cosθ + 2)(2cosθ - 1) = 0

but cosθ + 2 ≠ 0 so, cosθ = 1/2 = cos60°

hence, θ = 60° = π/3

therefore the angle made by the thread with the vertical is π/3 hence, the value of n is 3.