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A uniform ring of mass $m$ and radius $R$ can rotate freely about an axis passing through the center $C$ perpendicular to the plane of paper. Half of the ring is positively charged and the other half of the ring is negatively charged. Uniform electric field ${E_a}$ is switched along –ve x-axis (axis are shown in figure).Find the angular velocity of the ring after rotation of 180 degrees. (Magnitude charge density $\lambda $ )

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Last updated date: 20th Jun 2024
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Answer
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Hint: Use the formula of the shear stress given below and substitute the value of the parameters in it. Use the formula of the energy given below and substitute the calculated value of the shear stress and other parameters to find the value of the angular velocity.

Formula used:
(1) The formula of the shear stress is given by
$\tau = F \times 2d$
Where $\tau $ is the shear stress, $F$ is the force acting on the mass and $d$ is the distance of the force from the mass.
(2) The kinetic energy is given by
$E = \dfrac{1}{2}m{v^2}$
Where $E$ is the energy, $m$ is the mass of the object and $v$ is the velocity of it.

Complete step by step solution:
Let us interpret the diagram.
By using the formula of the shear stress,
$\tau = F \times 2d$
From the diagram it is clear that the distance is $\dfrac{{2R}}{\pi }$ , and the formula of the force is given by $F = \pi R\lambda $ . Substituting both of these values in the above formula, we get
$\Rightarrow$ $\tau = \pi R\lambda \times 2 \times \dfrac{{2R}}{\pi }$
By cancelling the similar terms and simplification of the above equation, we get
$\Rightarrow$ \[\tau = 4{R^2}\lambda {E_0}\]
Then using the formula of the energy,
$\Rightarrow$ $E = \dfrac{1}{2}m{\omega ^2}$
The shear stress is the kinetic energy produced and also the mass of the body is $R$ and hence substituting these in the above formula, we get
$\Rightarrow$ $4\pi \lambda {E_0} = \dfrac{1}{2}m{R^2}{\omega ^2}$
By cancelling the similar terms on both sides of the equation and further simplification of the above equation,
$\Rightarrow$ $\omega = \sqrt {\dfrac{{8\lambda {E_0}}}{m}} $

Note: Here the ring of the certain mass is rotated due to the presence of the half positive charge and half negatively charged. And hence the rotation motion takes place. That is the reason the kinetic energy formula is substituted with the angular velocity in place of the normal velocity of the mass.