I'm solving problem number 5 from https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/resources/mit8_05f13_ps2/.
(a) Here I got:
$$
\beta = \frac{\hbar^{\frac{1}{3}}}{(\alpha m)^\frac{1}{6}}
$$
and:
$$
E = \left ( \frac{\alpha \hbar^4}{m^2} \right )^\frac{1}{3}e
$$
(b) Using Scilab I...
Are we talking only about the case (c) or any case listed above?
I think I understand what you are trying to say, but I don't understand why system has to start sliding.
I understand that when system is rolling, if the resulting velocity of the point of contact with the ground isn't zero...
I think I understand the source of my confusion. I assumed that the condition ##F_{f} \le \mu N## was closely related with the (c) case. But it is not the case.
I first considered the case of system dynamics in the middle of the movie. I found parameters of system dynamics (translational and...
If you pull harder, I assume that either string will brake or the spool will start to move with constant speed because of zero acceleratios. But I can't prove that.
When I look once more through equations, ##F_{max}=T_{max}= \frac{\mu}{cos \phi + \mu sin \phi} mg## is maximum straining force of...
First, for the case (c), I would say that the zero values of translational and rotational acceleration mean one of the following three things:
1. If the spool was accelerating (to the left or right) before meeting requirement (c), it will keep moving with constant linear and rotational speed...
I solved this problem using second Newton law for translational motion and the same law for rotational motion, and got $$a= \frac {F} {m+ \frac {I} {R^{2}}} (cosϕ−rR)$$ where m is spool mass.
Now, we have three cases:
(a) ##cos\phi>\frac{r}{R}##, when spool is accelerating to the right,
(b)...
Hi everyone!
I'm a physics anthusiast.
I fell in love with physics as soon as I got it in the middle school.
Love to read books of the famous physicists (Feynman, Susskind, B. Greene, Penrose, etc.).
If my English is sometimes clumsy, I hope you won't mind me - English is not my native...