Finding mass to resonate a system

[Kubix]

Homework Statement

Rotable disc with r = 0.5m and m_1 = 4kg is mounted on its center, in point O. It can freely rotate. On its side point mass is mounted(m_2), on other side spring is mounted(k=10000N/m). There is also external torque relative to point 0, M_o(t) = 10sin(20t).
Calculate at what m_2 system will resonate. Gravitation should be taken into account. Use small angle approximation - sin(θ) = θ, cos(θ)=1.

Relevant Equations

τ=Iα
τ=rF*sin(θ)
θ = Asin(20t+φ)

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 made differential equations
τ=Iατ=Iα
θ¨=(k(r²)θ+10sin(20t)−(m₂)gr)/((m₂)(r²)+(1/2)(m₁)(r²))

And from now i don't know what to do next.

I have tried also substitute θ with θ = Asin(20t+φ) but still couldn't find an answer.

 

[kuruman]

For a given mass $m_2$ the system has a natural frequency of small oscillations. Can you find an expression for that?

 

[Kubix]

Yes, it is ω₀=(k/m)^(1/2)

 

[kuruman]

That's for a linear spring-mass system. That's not what you have here. Writing the differential equation is a good start. You don't need the driving force torque to figure out the natural frequency of the system.

 

[Kubix]

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?

 

[kuruman]

Your differential equation is incorrect. The torque is restoring which means that the torque is always opposite to the angular displacement. What is the angular equivalent of $F=-kx$?

The one-dimensional harmonic oscillator equation has one form, $$\frac{d^2(\text{something)}}{dt^2}+\omega^2{\text{(something)}}=0$$and its general solution is $$\text{something}(t)=A\sin(\omega t)+ B\cos (\omega t). $$ Can you arrange your differential equation to look like the generic form above? If so, then whatever multiplies the "something" variable in it is the frequency squared.

 

[Kubix]

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

 

[kuruman]

ok, i think, finally, i found an answer:

$$ \frac {(gm_2)} {(-w^2+(k/m))*rm}*\theta(t) = Asin(\omega t)+Bcos(\omega t)$$

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

My advice is to write the appropriate differential equation. What you have above is not a differential equation. You need to fix your first attempt,

θ¨=(k(r2)θ+10sin(20t)−(m2)gr)/((m2)(r2)+(1/2)(m1)(r2))

It is the equation you get from a free body diagram but without the driving force. After you fix it, see if you can bring it into the form shown in post #6.

Good night. By the time you wake up, I will probably be asleep so take your time thinking about your response.

 

[Kubix]

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 $$\theta ' (0) = 0?$$So natural frequency for my system is $$ \omega = \frac {\sqrt k} {\sqrt{ (1/2)m_1 + m_2}} $$, right?

 

[kuruman]

Right. Note that you do not need to solve the differential equation to find the natural frequency and you do not need initial conditions. The system will oscillate at its natural frequency (assuming no driving torque) regardless of when or how you start it. Please re-read post #6. In the context of that post, "something" here is $\theta$. Whatever multiplies it in your equation is the frequency squared. Therefore $$\theta ''+ \frac {k\theta} {m} = \frac {m_2 g} {mr}~~\Rightarrow~\omega^2=\frac{k}{m}=\frac{k}{\frac{1}{2}m_1+m_2}.$$.See how easy? Now how will you find what the problem is asking?

 

[Kubix]

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

 

[kuruman]

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

You don't need the amplitude which you cannot find anyway because you don't have the damping term.

In the expression $10\sin(20t)$ what does the "20" represent? Please remind me.

 

[Kubix]

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

 

[kuruman]

Right and, when the system resonates, how is the driving frequency related to the natural frequency?

 

[Kubix]

When $$\omega = \omega_e $$

 

[kuruman]

Right again. How would you use this information to find what $m_2$ is needed?