LOOPING THE LOOP!

Electricity and Magnetism Level 3

A particle of mass $m=1\text{ kg}$ and charge $q=3\text{ C}$ is hung from a light inextensible string of length $l=2\text{ m}$ . The entire system is placed in a uniform horizontal electric field of magnitude $8\text{ V/m}$ . What must be the minimum initial horizontal velocity with which the particle must be projected so that it completes a vertical circle?

Details and Assumptions

The gravitational acceleration is $g=-10\text{ m/s}^2$ .
The particle is initially at the lowest point.

The answer is 14.

1 solution

Pinak Wadikar
May 9, 2014

Some discussions about the problem

At first, When we release the charge at the bottommost point, it will not stay there.
As soon as we release it at bottom point, there will be two forces acting on the mass namely, the gravitational pull and Electric field.
The particle will be suspended in equilibrium. {in fact it will perform SHM!}

The particle will make an angle of $\theta$ w.r.t. horizontal.

This $\theta$ is given by the expression

$\tan \theta = \frac{mg}{qE}$

What we have to find out is the minimum velocity from this point so that it completes vertical circle. This is because as soon as we release the mass, it will spontaneously go to this position.

My equations were as follows

$\frac{mgl}{2} = mgl \times [1 + \sin \theta] + \frac{mgl}{2} + lqE\cos \theta$

This I arrived at by Work Energy Principle

This is the expression for minimum velocity with which particle should be projected in order to complete a circle.

whats wrong with this, from conservation of energy, 1/2mv^2+(E/q)*2l=mgh

Sanad Kadu - 3 years, 1 month ago

Hello Everyone, I am really sorry, I just found out that this solution does not contain the term of velocity. I would like to ask someone to take a look at this as I have lost touch of the subject. You can revert back with corrections by replying to this comment. I suspect that the LHS part of my equation is should be the change in Kinetic energy? Please tell me if I am right! Again, I have completely lost touch off the subject! Please help! Thanks

Pinak Wadikar - 3 years ago

I think he was trying to write: "m(v0)^2/2=mgl(1+sin(θ))+qEcos(θ)+mgl/2", namely "Kinetic Energy-Work=Potential Energy".

Sami Minea - 3 years ago

Sorry "m(v0)^2/2=mgl(1+sin(θ))+lqE+mgl/2", because Lorentz force acting in the x direction"

Sami Minea - 3 years ago

I'm getting the ans as 12.

Apar Agarwal - 2 years, 8 months ago

I dont understnad the equation

Vamsi Krishna Appili - 5 years, 2 months ago

your equation doesn't involve velocity. How you found that?? Work Energy Principle is W(net)=change in KE

A Former Brilliant Member - 5 years, 1 month ago

You failed to clarify without a drawing whether the sum of the forces vector is closer to further from the starting position than gravity

Jake Ok - 4 years, 7 months ago

or further than *

Jake Ok - 4 years, 7 months ago

Your ans is incorrect , it shoulf be 18 metres per second, i solved it whole

Raunak Agrawal - 4 years, 2 months ago

Yes 18.4 seems my ans also

Jyotisman Rath - 3 years, 8 months ago

I got 14...

Praveen Kumar - 3 years, 3 months ago

@Praveen Kumar – Hello @Praveen Kumar , 14 is the correct answer! However, I find some minor discrepancies in my suggested answer. Can you please help me by suggesting a correction to my equation? I have recently left a comment and you can reply to it with the suggested answer. Thanks!

Pinak Wadikar - 3 years ago

No no, sorry, i tried again and got the right ans

Raunak Agrawal - 4 years, 2 months ago

Hello @Raunak Agrawal If you did get the right answer can you please let me know what is wrong in my explanation above? I have left a comment recently, you can reply to it with the suggested correction. Thanks!

Pinak Wadikar - 3 years ago

effective g= \sqrt {g^2+(E*q/m)^2} now min velocity will be =\sqrt{5g(effective)l}

prashant singh - 5 years, 6 months ago

LOOPING THE LOOP!

The answer is 14.

1 solution

0 pending reports