0
$\begingroup$

Imagine that you have a cylinder of radius $a$, with total charge $Q$ uniformly distributed through the surface, infinite height and coaxial with $z$ axis. Let be $s$, $\phi$, z (cylindrical) our system of coordinates.

Imagine also that we have a certain circumferential electric field $\vec{E(s)}=E(s)\vec{u_{\phi}}$, whose value depends only upon $s$ coordinate (the distance from axis $z$), and we´re asked for the torque exerted to the cylinder by the field. In the problem where I´ve found this situation, it´s due to the fact that the cylinder is inside a solenoid whose current is changing, and therefore, according to Faraday´s law, a circumferential electric field is created by a changing magnetic field, but I don´t find it relevant for the purpose of my question.

According to my sources, the torque is simply $\tau=\vec{r} \times Q\vec{E}= a\vec{u_{s}} \times Q\vec{E}$

This is equivalent to suppose that the cylinder is a point charge $Q$ located at a distance $a$ from axis $z$, what can seem "logical", but I´m unable to prove it.

My questions are:

How can this fact be proven from the point of view of rigid body mechanics?

Is there a more general theorem for rigid body that says that, given certain symmetries in the problem, such as uniformity of the force on the surface or the shape of the body, torque or angular momentum can be written this way?

$\endgroup$
2
$\begingroup$

Let us break the cylinder down into elementary rings, and then break the ring into elementary particles. It is easy to see that the torque on an elementary particle will be $a\vec{u_s} \times \mathrm (dq)\vec{E} $, where $\mathrm dq$ is the charge on the elementary particle. Summing over the ring, we get the torque on the elementary ring equal to $\left(aq\cdot E(s)\right)\vec{u_\text{axis}}$, where $q$ is the charge on the elementary ring. Finally summing this on the entire cylinder, we get the required expression: $$\tau_\text{net} = \left( aQ \cdot E(s) \right)\vec{u_\text{axis}}$$ enter image description here

$\endgroup$
1
$\begingroup$

A point at a distance $\vec r$ from the $Z$-axis and rotating in a plane parallel to the $XY$ plane is similar to a cylinder of radius $r$ having symmetry axis $Z$ and rotating about $Z$, in the fact that they have the same moment of inertia, i.e: for both M.I (about $Z$-axis) = $mr^2$. So, a blind person who has some means to measure the angular acceleration of the system, given that the forces on the particle and cylinder are equivalent, cannot tell whether its a cylinder or a particle rotating. This may be incorrect, but the only rotational mechanics element that is same for both is the moment of inertia of the two.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.