Charged Ring Rolling!

A non-conducting ring of mass m m and radius R R , the charge per unit length λ \lambda is shown in figure. It is then placed on a rough non-conducting horizontal plane. At time t = 0 t=0 , a uniform electric field E = E 0 i ^ \vec{E} = E_0 \hat{i} is switched on and the ring starts rolling without sliding. Find the frictional force ( f ) (f) acting on the ring.

Given - E 0 = 3 , R = 4 m , λ = 6 |E_0| = 3 , R = 4m , |\lambda| = 6

Find the value of f 2 R \sqrt{\frac{f}{2R}} .


The answer is 3.000.

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2 solutions

Nathanael Case
Jul 18, 2015

Net electric force is zero, so the linear acceleration is the force of friction divided by the mass. Since it rolls without slipping, the angular acceleration is the linear acceleration divided by the radius. Therefore we have:

α = f m R \alpha = \frac{f}{mR}

The differential charge is λ R d θ \lambda Rd\theta and so if θ \theta is measured with respect to the horizontal, then the torque from the electric force will be 2 λ E 0 R 2 0 π / 2 s i n θ d θ = 2 λ E 0 R 2 2\lambda E_0 R^2\int \limits_0^{\pi/2} sin\theta d\theta = 2\lambda E_0 R^2

(The factor of 2 comes from symmetry: the torque from the negative charge is the same as the torque from the positive charge.)

Friction will oppose the rotation caused by the electric force, so it will supply an opposite torque, therefore we have:

τ = 2 λ E 0 R 2 f R = I α = m R 2 α = f R \tau = 2\lambda E_0 R^2-fR=I\alpha = mR^2\alpha = fR

Solving yields f = λ E 0 R f=\lambda E_0 R

What exactly has been asked? Friction or the quantity sqrt(f/2r) ?

Niranjan Rao - 5 years, 10 months ago

Why integrate from 0 0 to π / 2 \pi/2 ? When in the question only 1 / 4 t h 1/4^{th} of the ring is charged?

Akshay Yadav - 4 years ago

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Exactly...

Md Zuhair - 3 years, 2 months ago

I have seen some of your questions of electrostats and many of them including this one belong to a specific book

Its from cengage

Karan Doshi - 4 years, 1 month ago

1 pending report

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