Recent content by Kubix

When $$\omega = \omega_e $$

It is frequency of external torque(forced frequency): $$ \omega_e = 20 $$

Do i need to find amplitude A and make its denominator equal 0? Or is it enough to compare$$ \omega^2 = 20, $$ because $$ \tau_o(t) = 10sin(20t)?$$

ok, so when $$(1/2)m_1 + m_2 = m$$ then $$\theta ''+ \frac {k\theta} {m} = \frac {m_2 g} {mr}$$, general solution is: $$\theta (t) = c_1 cos( \frac {\sqrt k} {\sqrt m} t) + c_2 sin( \frac {\sqrt k} {\sqrt m} t ) + \frac {gm_2} {kr}$$can i assume that $$ \theta (0)=0$$ and $$\theta ' (0) =...

ok, i think, finally, i found an answer: $$ \frac {(gm_2)} {(-w^2+(k/m))*rm}(t) = Asin(\omega_0 t)+Bcos(\omega_0 t)$$ Please give me some advice for further steps, i live in europe and I'm going to sleep right now. Thank you for your help

Ok, so my differential equation is θ'' - kθ/m = (gm_2)/(rm) and general solution is x(t) = (gm_2)/(kr)+c1*e((-sqrt(k)*t)/sqrt(m))+c2*e((sqrt(k)*t)/sqrt(m)), but I'm not sure about my t=0 conditions, is it θ(0)=0 and θ'(0)=0?

Yes, it is ω0=(k/m)(1/2)

My first step was to calculate Torques acting on system, I found 3, one given(external): a)torque produced by point mass: (m2)grcos(θ)=(m2)gr b)torque produced by spring krsin(θ)rcos(θ)=kr2θ c)external torque τ_o(t)=10sin(20t) I also calculated moments of inertia I=m1r2+(1/2)m1∗r2 then I...