What are the Tensions and Accelerations in a Rotating Triangle Mass System?

[Maybe_Memorie]

Homework Statement

Two point particles of mass m are attached to two vertices of an
equilateral triangle made of three massless rigid rods of length l. The third
vertex is pivoted to a fxed point, so that the triangle can swing in a
vertical plane under the effect of gravity.
The system is released from rest with one side of the triangle in the
vertical position.
Compute the tensions on the three rods and the accelerations of the
masses right after the system is released.

Note: Since the triangle is made of rigid rods, the forces along the rods
can be either tensions or compressions.

Homework Equations

Question 3 of this
http://www.maths.tcd.ie/~kovacs/Teaching/Mechanics/PS141-5.pdf

The Attempt at a Solution

I can't use angular momentum to solve this as we haven't covered it in class yet.

Do both masses have the same acceleration?
Is the acceleration only in one direction or two?

 

[Mindscrape]

What about torque, and moment of inertia? What kind of acceleration does the problem want (angular? linear?)?

 

[Maybe_Memorie]

What about torque, and moment of inertia? What kind of acceleration does the problem want (angular? linear?)?

Haven't covered those yet.
Linear.

 

[Maybe_Memorie]

Can someone verify if the acceleration is g/root(3) ?

 

[Maybe_Memorie]

Anyone?

 

[Delzac]

I computed the angular acceleration alpha to be (sqrt(5)/8 )* g rad/s^2.

I do not think linear acceleration will be appropriate here, and i do not know how it can linearly accelerate.

And even if i use a primitive way to calculate, since you haven't covered angular momentum, the acceleration on the centre of mass i computed is g sin(30 degree)

 

[tiny-tim]

Hi Maybe_Memorie!

When the speed is still zero, the accleration must be purely "tangential", ie perpendicular to the direction to the pivot.

So call the tension (or compression) in the lower rod T, the mass m, and the acceleration a (a must be the same for both masses), and do "tangential" components of F = ma for each of the two masses (separately).

 

[Maybe_Memorie]

What should the acceleration work out as?

 

[tiny-tim]

uhh? you tell us!

 

[Maybe_Memorie]

I'm getting g(root3)/4

 

[tiny-tim]

(have a square-root: √ )

I'm getting g(root3)/4

Yup, that's what I got!

(is it wrong? )

 

[Delzac]

Hi,

Just my extensions on this question.

Does it make sense, or in fact even possible, to compute angular acceleration. Come to think of it, the angular acceleration is always changing.

My initial calculation on Angular acceleration is (sqrt(5)/8 )* g rad/s^2.
I computed by: torque = mgdsin(θ) = - I α, where d = sqrt ( l^2 + 1/4 l^2). And i simply took centre of mass as between the 2 masses.

Any ideas?

Delzac

 

[Mindscrape]

I also get g√(3)/4. Though something does seem strange in that I didn't even use the other mass, and I'm assuming you guys didn't either. I guess the implicit assumption that the acceleration must be tangential must account for it.

For angular acceleration the moment of inertia would be
I33=2ml^2

and one way to get the second order ODE would be
$$2ml^2 \ddot{\theta}=mgl(cos(\theta)+cos(\theta-\frac{\pi}{2}))$$

Then taylor expand theta somewhat or use numerics I guess.

 

[Delzac]

Hmmmm, can't we just take reference from the centre of mass, so torque at centre of mass = mg sqrt(5) / 2 sin 30degree?

Delzac

 

[Mindscrape]

Oh yeah, sure, that works too (don't know why I overcomplicated it). :)

 

[Delzac]

But if you actually think about it, the 1st mass doesn't contribute anything to the torque, so is it legal to use centre of mass to calculate the angular acceleration?

 

[tiny-tim]

Hi Mindscrape!

I also get g√(3)/4. Though something does seem strange in that I didn't even use the other mass, and I'm assuming you guys didn't either.

I did, I used it in my F_total = 0 for that mass.

 

[Mindscrape]

But if you actually think about it, the 1st mass doesn't contribute anything to the torque, so is it legal to use centre of mass to calculate the angular acceleration?

Oh yes, most definitely, after all, the center of mass still has torque on it. There will be equal torque when the CM is at 0 angle, which let's us recover mg. In fact, CM is really the way to go because the problem is in the semi-okay realm for small angle approximation (pi/3 is totally out of the question).

tiny-tim, I don't think there should be a net zero force anywhere, other than in the y direction (using y as straight up) of the perpendicular mass. I'll rework this later today because I don't remember what I did originally.

 

[tiny-tim]

tiny-tim, I don't think there should be a net zero force anywhere …

Sorry, I forgot what the question was

I meant F_total = ma.

 

[Delzac]

Alright, let's try this.

Should the torque_cm be equal to torque of its individual parts?

Meaning, should the torque of m_1 and m_2 be equal to torque_cm? It doesn't seem to be the case, since there is no torque on m_1.

 

[tiny-tim]

Hi Delzac!

Meaning, should the torque of m_1 and m_2 be equal to torque_cm? It doesn't seem to be the case, since there is no torque on m_1.

uhh? … that doesn't make sense …

the torque of A about a point plus the torque of B about the same point = the torque of the c.o.m. about the same point

 

[Delzac]

Hmmmm.

Here's my calculations, taking individual parts:
m1 has no torque, theta is 180.

Torque of m2 = Lmg sin 60 = sqrt(3) /2 Lmg

If we take Centre of mass,

torque = [ sqrt(5)/4 L ] [ 2mg sin 30]

They aren't equal, at least in my calculation. Is there anything not accounted for?

And of course, everything is taken with the same point of rotation.

 

[tiny-tim]

Hi Delzac!

If we take Centre of mass …

Are you saying that a body can be replaced by a body of the same mass at its centre of mass?

That obviously isn't true … for a start, it obviously doesn't apply to a disc.

 

[Delzac]

But isn't this manner of calculation used when we try to derive the equation for the period of a physical pendulum?

We equate torque = mgd sin(theta) = - I alpha.

 

[tiny-tim]

Yes, but that's for a "point mass".

The less a body is like a point, the less accurately it works.

For a small sphere, with the centre of mass at roughly the middle value of r, it's good enough.

For two point masses at the same distance r, the centre of mass is at distance less than r, and it doesn't work at all.

 

[Delzac]

So you are saying that i can't just calculate torque of centre of mass even if i account for the change in r?

 

[tiny-tim]

So you are saying that i can't just calculate torque of centre of mass even if i account for the change in r?

Depends how you account for it.

I suppose if you pretend the centre of mass is somewhere else, it should work!

With creative accountancy, you can do anything! ​

 

[Delzac]

Then in this case it is relatively simple.

The centre of mass is in between the 2 mass. At a distance of 1/2 L from each of them.

To calculate the distance, from axis of rotation to centre of mass, which i had already done in the previous post is:

$$Dist = \sqrt((\frac{L}{2})^2 + L^2) = \frac{\sqrt5}{2}L$$

And if i take this does torque= 2mg (Dist) Sin(30).

If so, this obviously isn't the addition of the torque of m1 and m2.

So something must be wrong.

 

[tiny-tim]

Then in this case it is relatively simple.

The centre of mass is in between the 2 mass. At a distance of 1/2 L from each of them.

To calculate the distance, from axis of rotation to centre of mass, which i had already done in the previous post is:

$$Dist = \sqrt((\frac{L}{2})^2 + L^2) = \frac{\sqrt5}{2}L$$

And if i take this does torque= 2mg (Dist) Sin(30).

If so, this obviously isn't the addition of the torque of m1 and m2.

So something must be wrong.

sorry, you've completely lost me

you haven't accounted for the centre of mass being "in the wrong place", so of course it doesn't work

 

[Delzac]

Hold on. Isn't the centre of mass in between the 2 mass. The rods have no mass.

 

[tiny-tim]

Isn't the centre of mass in between the 2 mass.

That's right … it's closer than they are (to the pivot), so it doesn't work.

 

[Delzac]

[STRIKE]Are they? Isn't the centre of mass $$\frac{\sqrt5}{2} L$$ away from pivot? Which is more that 1L.[/STRIKE]

My goodness blooper. You are right, the distance is shorter. But still, is there any proof that the formula breaks down when r becomes shorter?

 

[tiny-tim]

My goodness blooper. You are right, the distance is shorter. But still, is there any proof that the formula breaks down when r becomes shorter?

It must do … mr + mr is different from 2m times something less than r.

 

[Delzac]

Here is a question from mastering physics.

If you would take a look, they did what you said cannot work.

The distance of centre of mass is smaller than L. But they still equate it nonetheless.

 

[tiny-tim]

Yes, but they're using I as well as d …

the I of the original object.

In our case, a = Lα = Lτ/I = LmgLsin60°/2mL² = g√3/4.