Recent content by TSny

Very nice explanation! I think it gets to the heart of the matter. By choosing the tangential force appropriately as a function of time, we can switch back and forth between oscillatory motion and uniform circular motion while maintaining constant acceleration magnitude. This problem reminds...

Yes, I agree. And the arc length comes out nice.

I don't understand why you say that "dEx is Rcosθ distance away". Both dE and dEx are associated with the same point, namely the origin: In what sense is dEx located a distance Rcosθ away?

Your answer is wrong because you can't get the x-component dEx by replacing R in the denominator of dE by Rcosθ. Note that Rcosθ < R when θ≠0. So, kdQ/(Rcosθ)2 is greater than kdQ/R2 when θ≠0. That is, your expression for dEx is greater than the expression for dE. But the component of a...

yes

No. Let's forget dEx for now. Just concentrate on dE. dE is the magnitude of the electric field vector dE at the origin O produced by the charge dQ in the diagram in post #2. In the general formula dE = kdQ/r2, what would you use for the distance r in this case?

What was your reason for writing the denominator of the integrand as ##(R\cos \theta)^2##? This is not correct. To find ##E_x## due to the entire charge ##Q##, you need to understand that ##E_x = \int dE_x## where ##dE_x## is the x-component of the field vector ##\overrightarrow{dE}## due to an...

For the element of charge ##dQ## shown, how would you write an expression for ##dE## at the origin? How would you write the x-component of ##dE##?

The vertical side of each triangle is the y-component of the velocity, which remains constant in time. The horizontal side is the x-component of velocity, which increases with time. The constant y-component of velocity represents the distance the object moves in the y-direction during one unit...

Yes. For any value of ##k##, you get the same expression for the minimum force necessary to cause movement of ##m_2##. [Edit: This expression is independent of ##k##.] Therefore, taking the limit as ##k## goes to infinity still gives the same expression. Yet, if we replace the spring with an...

In the 4th edition (1975), the relevant page is 257, not 277. I believe I was able to derive the equation for ##D_{\alpha}## in (2) by following the suggestions in the problem. I don't see any misprint or error here.

Your method is fine. To shortcut it, you can note that all factors in ##[y p_z, x p_z]## commute among themselves. Thus, ##x## commutes with ##y## and ##p_z## , ##y## commutes with ##x## and ##p_z## , and ##p_z## commutes with itself and ##x## and ##y##. So, the result of the commutator is...

👍 If the string is not vertical, then the string will exert a force on the ring that has a nonzero horizontal component. But there are no other horizontal forces acting on the ring. So, if the string is not vertical, there will be a nonzero net horizontal force on the ring and the ring will...

Yes. From post #8, @Electrodude realizes that the "point charge model" won't work for the original problem, but the model should work for charge distributions along the axis of the dipole. I agree. Perhaps the question is asking for just the magnitude of the force.

Suppose you want to find the force on a dipole ##p## produced by a linear charge distribution with uniform linear charge density ##\lambda## as shown. The force that the linear charge distribution exerts on the dipole is equal and opposite to the net force that the dipole exerts on the charge...