Rotation of a rod fastened to a wire

[roadrunner1994]

Homework Statement

A thin homogeneous rigid rod has length L and mass M.
One endpoint of the rod is fastened to a light, rigid steel wire, also of length L.
The other end of the wire is fastened to a horizontal axis O in the laboratory.
The system can turn in a vertical plane. The steel wire cannot bend, and its
mass can be ignored relative to M.
Initially the system is held at rest in a horizontal position. It is then set
free to move under the influence of gravity. The acceleration of-gravity is g.
Ignore air resistance.
(1) At the moment the system begins to move, i.e., while the rod is still
horizontal, the wire acts on the rod with a force F1 at the endpoint A.
Determine F1.
(2) At the moment when the rod is vertical, the wire acts on the rod with a
force F2. Determine F2.
(3) Assume now that the steel wire breaks at the moment when the rod is
vertical. The lower point B of the rod is then at some height h above the
laboratory floor. When the rod hits the floor is has turned so that it is
exactly parallel to the floor. Determine h.

Relevant Equations

-

Hi,
I started with calculating the moment of inetria of the rod:
I = ⅓ML^2 + M(3/2 * L)^2 = 31/12 ML^2
and I thought that the reaction force in the first case will be equal to centrifugal force:
F1 = Mω^2*(3/2)L
Angular velocity is calculated from the conservation of energy:
Mg3/2*L=1/2 * Iω^2

But this solution seems to be incorrect.

 

[BvU]

But this solution seems to be incorrect.

How do you know that ?

And: would that be for part (1), (2) or (3) ?

You write

in the first case

but at that point ##\omega## is still zero !

 

[BvU]

I started with calculating the moment of inertia of the rod:

Not needed for part (1) !

And you make a mistake that I can't point out because you don't say how you are doing it.

 

[haruspex]

I = ⅓ML^2 + M(3/2 * L)^2

You appear to be applying the parallel axis theorem, but one of those two terms on the right is incorrect for that,

 

[roadrunner1994]

How do you know that ?

And: would that be for part (1), (2) or (3) ?

I got the result: F1=(54/31)Mg, and the correct answer is (1/28)Mg.
I'm trying to solve the 1)st part

You appear to be applying the parallel axis theorem, but one of those two terms on the right is incorrect for that,

Yes, I wanted to apply parallel axis theorem. It seems that I should rather use (1/12)ML^2 instead of (1/3)ML^2. In part 2) it gives the result: F2 = Mg + F(3/2)Lω^2 = (41/14)Mg, which is correct .

But I'm still struggling with the 1) part.

 

[BvU]

Initially the system is held at rest in a horizontal position. It is then set
free to move under the influence of gravity. The acceleration of-gravity is g.
Ignore air resistance.
(1) At the moment the system begins to move, i.e., while the rod is still
horizontal, the wire acts on the rod with a force F1 at the endpoint A.
Determine F1.

Relevant Equations:: --------

This does not mean that you think you can solve this without any equations, I hope ?

But I'm still struggling with the 1) part.

Ok, that helps. You follow @haruspex and my advice to come up with a better moment of inertia [edit: I see you already did that! Kudos ! ], and try to set up the proper equations for what we shall call ##t=0##.

A free-body diagram might be extremely helpful .

 

[roadrunner1994]

I guess the center of mass will be affected by the force F=Ma_tangential
a_tangential can be calculated from Iα=Mg(3/2)L
Therefore α=(9/14)g/L, and a_tangential=(27/28)g.

But both forces: Mg and Ma_tangential should be pointed downward, so why shoud they be subtracted to obtain correct result?

 

[BvU]

Free body diagram of rod alone is within yellow marker line.
Free body diagram of rod+stiff wire is without all the vertical crud.

[edit] and without ##F_1## but with another force elsewhere ... ​

(I should have been more specific ?)

What kinds of motion for the rod are going to happen with these forces acting ?
How are they interconnected ?
Equations, equations

why should they be subtracted to obtain correct result

Because they point in opposite directions, perhaps ?

##a_{\text{tangential}}\ ## can be calculated from ##I\alpha = {3\over 2} MgL##

Is that so ? All of a sudden ##F_1## is out of the game ?

Equations, equations

 

[wrobel]

generally speaking the wire acts on the rod by means of a force and a torque

 

[BvU]

The general is correct, generally speaking

 

[wrobel]

Yes and in particular one should prove that the torque is equal to zero. If it is so indeed. The force F1 is not obliged a priori to be directed vertically as well

 

[BvU]

The force F1 is not obliged a priori to be directed vertically as well

Not a priori, no. Generally speaking
But yes at ##t=0##

 

[wrobel]

But yes at

this must be verified

 

[BvU]

Done.
Note that this is not my exercise, but roadrunner's ...

 

[wrobel]

Perhaps these remarks will be of some use for the topic starter

 

[etotheipi]

luckily here we easily verify this by seeing that the centre of mass ##S## is constrained to move on a circle of radius ##r = 3L/2## with ##\dot{r} = 0##, so it's acceleration ##\mathbf{a}_S = r\ddot{\theta} \boldsymbol{e}_{\theta}(t) -mr {\dot{\theta}}^2 \boldsymbol{e}_r(t)##, where ##\theta## is the angle from the horizontal. so with ##\dot{\theta}(0) = 0##, the ##\mathbf{a}_S(0) = r\ddot{\theta} \boldsymbol{e}_{\theta}(0) = g\boldsymbol{e}_{\theta}(0) + \frac{1}{m} \mathbf{F}(0)## implies ##\mathbf{F}(0) \parallel \boldsymbol{e}_{\theta}(0)##

 

[haruspex]

generally speaking the wire acts on the rod by means of a force and a torque

It is unclear. We are told the wire cannot bend, but not whether its joint with the rod can flex.

 

[BvU]

Post #5 tells us

 

[haruspex]

Post #5 tells us

Does it? It implies that when the rod is vertical the wire is also vertical. Can that only be the case if the joint is rigid?

 

[etotheipi]

@haruspex I presumed that the rod and the wire form a single rigid body; or in other words, that the wire is welded to the rod and can exert both an axial and shear component of force on the rod, and the shear component has to be non-zero whenever the rod's angular momentum about its centre of mass is changing.

 

[haruspex]

@haruspex I presumed that the rod and the wire form a single rigid body; or in other words, that the wire is welded to the rod and can exert both an axial and shear component of force on the rod, and the shear component has to be non-zero whenever the rod's angular momentum about its centre of mass is changing.

Yes, I agree that is probably what is intended. But.. why does the question only ask about forces, not torques?

 

[BvU]

Does it?

the correct answer is (1/28)Mg.

 

[haruspex]

OK, I thought you were referencing the second part of post #5,

 

[wrobel]

It is unclear. We are told the wire cannot bend,

Exactly! The wire is rigid and it is rigidly connected with the rod. . It is the cause why the wire can rotate the rod and thus the wire can apply a torque to the rod.
You can take the wire in your hand and rotate the rod by means the wire

 

[haruspex]

Exactly! The wire is rigid and it is rigidly connected with the rod. . It is the cause why the wire can rotate the rod and thus the wire can apply a torque to the rod.
You can take the wire in your hand and rotate the rod by means the wire

"The wire cannot bend" is different from "the connection between the wire and the rod cannot bend". But the intent is made clear by the textbook answer to q1, as @BvU notes.