Recent content by alalalash_kachok

Assume the part of wheel with angle ##\phi## not is cut out, but is located next to cut out part. Turn this from right to left. Then wheel's center of mass will be higher by $$R\phi \sin(\alpha/2)$$. It allows to express potential energy. But I don't know how I can express kinetic energy in this...

Okay, assume that t=0 is moment when the rotating rod is stopped. After that the load starts spring fluctuations, don't you think? Can I consider Newton's law: $$m (\omega)^2 r_1= k(r_1-r_0)$$ for the moment just before stopping and take from this starting value of r(t)? Then put it in equation...

Sorry for my misleading, I agree with you. But I have remembered about that I can consider Newton's law for moment after stopping rotating: m $$\frac{d^2r}{dt^2} = k(r-r_0)$$. And try solve this equation, taking $$z = c*e^{i\omega t}$$. After that I get r as superposition of a general solution...

I understand that after stopping of rotating I should consider second Newton's law: m d^2r/dt^2 = k(r-r_0) And using the law of energy conservation I can propose that energy of circular motion I (\omega)^2/2, where I = mr^2 - moment of intertia will be converted into spring's oscillation. But...