Falling Dumbbell: Determine Force vs Angle

[sergiokapone]

Homework Statement

Determine force vs angle, that acts on the vertical wall from the falling dumbbell. Dumbbell falls without initial velocity. The mass of each bead of dumbbell is m.

Homework Equations

The problem need to be solved using only Newton's equations by writing them for each of the material points.

The Attempt at a Solution

I do not know how I find the acceleration of the points.

 

[tiny-tim]

hi sergiokapone!

start by drawing two free-body diagrams, one for each ball

(call the positions of the two balls x and y, and find how x'' and y'' depend on θ and on each other)

show us what you get ​

 

[haruspex]

Determine force vs angle, that acts on the vertical wall

What wall? Is there a diagram missing?

 

[tiny-tim]

What wall? Is there a diagram missing?

wot? no wall? ​

 

[sergiokapone]

What wall? Is there a diagram missing?

 

[sergiokapone]

start by drawing two free-body diagrams, one for each ball

(call the positions of the two balls x and y, and find how x'' and y'' depend on θ and on each other)

show us what you get

x=lcosα
y=lsinα
l - the length of the conjunction
x²+y²=l²

 

[tiny-tim]

good morning!

The problem need to be solved using only Newton's equations by writing them for each of the material points.

x=lcosα
y=lsinα
l - the length of the conjunction
x²+y²=l²

(ahh, it's that way round!)

ok, that's the full diagram

now draw the two free-body diagrams, one for each ball, marking the forces clearly on each

 

[sergiokapone]

I solved the problem. But in addition to Newton's laws, I had to use the law of conservation of energy.

In the projections on the direction along the dumbbell:
mgsinα-R=mv²/l
R - reaction from the rod.
x=lcosα
y=lsinα
Differentiating, we find velocity
v=l dα/dt

The law of conservation of energy: mgl=ml²dα/dt²/2+mglsinα

Using Newton's third law, force, that acts on the vertical wall F=Rcosα
or, finally
F=mg(3sinα-2)cosα

How can you do without the law of conservation of energy?

 

[haruspex]

x=lcosα
y=lsinα
l - the length of the conjunction
x²+y²=l²

So far so good. What about the forces on the falling end?

 

[haruspex]

wot? no wall? ​

Love it! Would he be Chad, Kilroy or Foo to you?

 

[sergiokapone]

So far so good. What about the forces on the falling end?

See my previous post, please, where I have already written, it is the reaction force from the connecting rod and the force of gravity.

 

[tiny-tim]

How can you do without the law of conservation of energy?

divide the reaction at the wall into R_x and R_y, and call the interaction between the two balls T

do ∑F = 0 for the wall ball, and ∑F = ma for the other ball

show us what you get ​

Love it! Would he be Chad, Kilroy or Foo to you?

he was (i understand) prolific in and after the second world war here in britain, announcing shortages, but i don't think he usually had a name, or was known by one (though apparently technically he was mr chad)

 

[sergiokapone]

divide the reaction at the wall into R_x and R_y, and call the interaction between the two balls T

do ∑F = 0 for the wall ball, and ∑F = ma for the other ball

show us what you get ​

he was (i understand) prolific in and after the second world war here in britain, announcing shortages, but i don't think he usually had a name, or was known by one (though apparently technically he was mr chad)

OK
1st ball:
R_x=Tcosα
R_y=Tsinα

2nd (falling) ball:
Tcosα=ma_x
Tsinα-mg=ma_y

 

[tiny-tim]

Tcosα=ma_x
Tsinα-mg=ma_y

ok, now write a_x and a_y in terms of α (or write cosα and sinα in terms of x and y), and solve

 

[sergiokapone]

ok, now write a_x and a_y in terms of α (or write cosα and sinα in terms of x and y), and solve

?

cosα = x/l
sinα = y/l

Concerning what to solve?
I do not understand.
Moreover, a(α) depend on second derivative of α, which is unknown.
a_x=-..α lsinα-l.α²cosα

(.. - second time deriv, . - first time deriv )

 

[tiny-tim]

might be easier to do it the other way …

mx'' = Tcosα = Tx/l which is a lot easier to solve

(write ' and '' for deriviatives)

 

[sergiokapone]

Real exponent?

 

[tiny-tim]

uhh?

 

[sergiokapone]

x=aexp(sqrt T/ml t)

 

[tiny-tim]

oh, hold on, T isn't constant, so you'll have to eliminate T first, before trying to integrate …

that should give you the same equation you'd get if you treated it as a single rigid-body and took moments of forces about the corner

 

[sergiokapone]

I went back to the problem.
For the center of mass of the dumbell I have:
$2m \vec a_{cm} = 2m\vec g + \vec R$ (1)
$a_{\tau}=gcos\alpha$
$v_{cm}=-\frac{d\alpha}{dt}l$
$l$ - halflength of the dumbell.
$\frac{dv_{cm}}{dt}=gcos\alpha$
or
$v_{cm}{dv_{cm}}=-gcos\alpha d\alpha$
Integrating, I find:
$a_{n}=2g(1-sin\alpha)$
Substituting in equation (1) of $a_{n}$

$2ma_{n}=2mgsin\alpha-R$
Finally
$R=2mg(3sin\alpha-2)$

Is the factor $2$ near $mg$ is correct?

 

[haruspex]

I went back to the problem.
For the center of mass of the dumbell I have:
$2m \vec a_{cm} = 2m\vec g + \vec R$ (1)
$a_{\tau}=gcos\alpha$
$v_{cm}=-\frac{d\alpha}{dt}l$
$l$ - halflength of the dumbell.
$\frac{dv_{cm}}{dt}=gcos\alpha$
or
$v_{cm}{dv_{cm}}=-gcos\alpha d\alpha$
Integrating, I find:
$a_{n}=2g(1-sin\alpha)$
Substituting in equation (1) of $a_{n}$

$2ma_{n}=2mgsin\alpha-R$
Finally
$R=2mg(3sin\alpha-2)$

Is the factor $2$ near $mg$ is correct?

I'm not sure what all your unknowns represent, so I can't follow your logic. The lower mass doesn't play any role, so you can ignore it. It has no horizontal acceleration, and since no radius is given for it you have to ignore any KE for it.
I don't get the factor of 2.

 

[sergiokapone]

The lower mass doesn't play any role, so you can ignore it.

But what about the law of motion cernter of mass? In this law we need use total mass of the dumbell. Looks like paradox.

 

[haruspex]

But what about the law of motion cernter of mass? In this law we need use total mass of the dumbell. Looks like paradox.

You don't have to treat the dumbbell as a single object. You can treat each end mass and the bar joining them as separate objects.

 

[sergiokapone]

You can treat each end mass and the bar joining them as separate objects.

Then I need to include a unknown force from the joinig bar in equation for the falling mass.
$m\frac{d\vec v}{dt} = m\vec g + \vec F_{bar\,reaction}$

 

[sergiokapone]

And I basically want to understand how to apply the equation of motion of the center of mass. Therefore I want
to consider dumbell as a single object .

 

[haruspex]

And I basically want to understand how to apply the equation of motion of the center of mass. Therefore I want
to consider dumbell as a single object .

Ok, but as I said you need to define all your variables. If $a_\tau$ is the tangential acceleration of the mass centre, how do you calculate that it equals $g\cos(\alpha)$?

 

[sergiokapone]

Ok, but as I said you need to define all your variables. If aτa_\tau is the tangential acceleration of the mass centre, how do you calculate that it equals ##gcos(α)g\cos\alpha## ?

May be, it was my mistake. I think the correct answer is:
$a_{\tau_{CM}}=g/2 \cos\alpha$
Using the law of motion of the center of mass: $2m\vec a_{CM}=2m\vec g + \vec R$ in projection to the tangential direction ($\vec R$ - acts along the bar)
$2ma_{\tau_{CM}}=2mg \cos\alpha$
I get $a_{\tau_{CM}}=g \cos\alpha$

 

[sergiokapone]

$\vec R$ - acts along the bar
I think, this is my primary mistake.

 

[haruspex]

May be, it was my mistake. I think the correct answer is:
$a_{\tau_{CM}}=g/2 \cos\alpha$
Using the law of motion of the center of mass: $2m\vec a_{CM}=2m\vec g + \vec R$

In that equation, R is the total reaction from the ground. This is not the same as the force along the bar. The difference is the weight of the lower mass.