Equation of motion for a rotational mass and spring system

[Isow]

Homework Statement

A small ball with mass M attached to a uniform rod of mass m and length l pivoted at point o and attached to two springs with spring constant k₁ and k₂ at distance d₁ and d₂ from point o as shown in Fig 2. The system oscillates around the horizontal line. Assume the system is in a fluid and is damped by a viscous drag force proportional to the speed. Find the equation of motion for the small angle oscillation and its solution if it starts from rest with initial angular position θ₀. Find the quality factor. Ignore the buoyancy force.

Homework Equations

[/B]
Moment of inertia for rod: I_rod = (1/3)*m*l²
Moment of inertia for ball: I_ball = M*l^2
Equation for torque: τ = I*θ''
Drag force: F_D = -b*cosθ
Spring force: F_S = -k*sinθ
Small angle approx for sin: sinθ = θ
and for cos: cosθ = 1

The Attempt at a Solution

My attempt involves trying to convert all forces to torques and to get an equation of motion in terms of theta. I'm having trouble in a few places. For one, I'm not sure how to handle finding the center of mass symbolically without knowing the mass of the rod. It seems like the CoM could be almost anywhere on the rod, depending on the rod's mass... and the placement of CoM with respect to d₁ and d₂ seems important.

If I can figure out how to set the differential equation up, I don't think solving it will be much trouble.

 

[OldEngr63]

Don't worry about the CM location; take moments about the left end. It will all work out quite easily.

 

[Isow]

Hi, thank you for the reply. The examples in my book all deal with a uniform rod and no mass attached, or the deal with a "massless" rod and mass. The former explicitly uses the CoM (l/2) to calculate torque and insert in my third equation. The latter are simple pendulums.

So I have a free body diagram drawn, but I'm not confident about how to get these forces in terms of theta, or sum them. I'm pouring over the book, but everything in the relevant chapters involves very simple systems compared to this problem.

 

[OldEngr63]

You are making this waaaay to hard!

First, from the integral definition of MMOI, it is evident that these terms simply add. Thus the total MMOI with respect to the left end is simply the sum of the MMOI for the rod + MMOI for the lumped mass. You have both of these terms.

When the system is displaced downward by θ as shown in your diagram, the compression in spring 1 is (approx) d1*θ, so the force in this spring is F1 = K1*d1*θ and the moment of this force about the left end is M1 = d1*F1 = K1*d1²*θ. A similar analysis applies for the second spring.

I leave it to you to figure out how to put it all together and take the damping into account. (Be brave, go ahead!)

 

[Isow]

So I think I have the torque of the springs: T = k*d^2*θ

When it comes to the mass, I'm not sure if I should be taking into account gravity, since equilibrium is at the horizontal. Would I just take into account the momentum of the mass working against the system, adding to drag?

p = mv

So would the momentum term for the mass become -M*l*θ' ?

And then -(m*l*θ')/2 for the rod?

And finally, the drag force, -b*θ' ?

Once you have the sum of torques, can you plug it into T = I*θ'' and solve for the moment?

Normally, I think, we divide through by mass to get terms for ω (√k/m) and γ (b/m), but if the above is correct there are two masses which I'm not sure how to handle when it comes to getting factors of ω and γ.

 

[Isow]

Hey, thanks again. I wrote my last post right as you were posting.

What you're calling the moment I'm calling the torque... but it seems like we're talking about the same thing. My understanding is that Torque = F*d and F_spring = kdθ so T = kd²θ. Why are you calling that the moment?

Plus, the questions above. If I'm making this too hard, it's because it's been a LONG time since I've dealt with classical mechanics and this class is vibrations and waves so it's not really touching on this stuff I should probably already know, just asking me to use it...

 

[Isow]

Okay, so it seems like you're saying I_total = l²(m/3 + M)

And we can sum the torques... so T_total = k₁d₁²θ + k₂d₂²θ + b*l*θ' ? I'm not sure about the drag part. I know the drag force is bθ', but converting that to torque by multiplying by l seems weird since it's not a point force, and would be distributed unevenly across the rod and mass.

Is there any chance somebody can just tell me what the equation of motion will be, and why? I'll still have to solve the ODE and find the other quantities asked for. I feel utterly lost trying to set this equation up.

 

[haruspex]

Okay, so it seems like you're saying I_total = l²(m/3 + M)

And we can sum the torques... so T_total = k₁d₁²θ + k₂d₂²θ + b*l*θ' ? I'm not sure about the drag part. I know the drag force is bθ', but converting that to torque by multiplying by l seems weird since it's not a point force, and would be distributed unevenly across the rod and mass.

Is there any chance somebody can just tell me what the equation of motion will be, and why? I'll still have to solve the ODE and find the other quantities asked for. I feel utterly lost trying to set this equation up.

Since you're not given a drag coefficient, you can just write the drag torque as ##-b\dot\theta##. The 'b' encompasses the distribution of the drag along the rod and ball.

 

[Isow]

I think I've got it now, but for the drag I ended up integrating over the length of the rod:

[itex]\int_{0}^{l}-br\theta'dr = -\frac{1}{2}bl^{2}\theta'[/itex]

Which seems right since it gives units of torque (the [itex]l^{2}[/itex] term).

 

[haruspex]

I think I've got it now, but for the drag I ended up integrating over the length of the rod:

[itex]\int_{0}^{l}-br\theta'dr = \frac{1}{2}bl^{2}\theta'[/itex]

Which seems right since it gives units of torque (the [itex]l^{2}[/itex] term).

OK, but since you are inventing b and l is constant you might just as well have invented ##c = \frac{1}{2}bl^{2}##. And why the sign flip?

[This is weird. On my screen, your post says 1/3. It also says 1/3 if I right click and 'show math as'. But when I clicked reply it came up as 1/2.]

 

[Isow]

The sign flip was a typo, and I did have 1/3 initially but fixed it so that probably explains the bug.

The [itex]l^{2}[/itex] term here actually ends up cancelling out when you divide through by the moment, [itex]l^{2}(m/3+M)[/itex].

I did absorb the 1/2 term into the b, which I hope is an okay liberty to take.

 

[haruspex]

The sign flip was a typo, and I did have 1/3 initially but fixed it so that probably explains the bug.

The [itex]l^{2}[/itex] term here actually ends up cancelling out when you divide through by the moment, [itex]l^{2}(m/3+M)[/itex].

I did absorb the 1/2 term into the b, which I hope is an okay liberty to take.

That all sounds fair.