Free-body diagram: Spring-loaded cylinder sliding on a metal rod

[ymnoklan]

Homework Statement

Homework Statement: I have a cylinder of mass m1=5.0 kg sliding on a metal rod. Additionally it is connected to a spring with k=500 N/m. The height between the cylinder and where spring is connected is 0.30 m. I am asked to make a free-body diagram of the cylinder. The only forces I can identify immediately are gravity and the force from the spring, but I have a strong feeling there might be some more. What do you think?
Relevant Equations: G=m*g, F = k*x

Relevant Equations

G=mg, F=kx

G=mg (vertical down), F=kx (from the spring to the cylinder)

 

[Orodruin]

Which of the forces you mentioned is supposed to keep the cylinder constrained to the rod?

 

[Steve4Physics]

Homework Statement: Homework Statement: I have a cylinder of mass m1=5.0 kg sliding on a metal rod. Additionally it is connected to a spring with k=500 N/m. The height between the cylinder and where spring is connected is 0.30 m. I am asked to make a free-body diagram of the cylinder. The only forces I can identify immediately are gravity and the force from the spring, but I have a strong feeling there might be some more. What do you think?

I think you need to provide a diagram!

Too much information is missing. For example:
- is the rod vertical? or horizontal? or angled?
- if (for example) the rod is vertical, is the spring above or below the cylinder?
- you state "The height between the cylinder and where spring is connected is 0.30 m" but this could be interpreted in several ways.

 

[kuruman]

The only forces I can identify immediately are gravity and the force from the spring, but I have a strong feeling there might be some more. What do you think?

I think that if these are the only forces acting on the cylinder, then the rod makes no difference so you might as well forget about its existence. Only you know what the picture looks like. What would happen if the rod were magically vaporized? If the answer is "Nothing", then the spring and Earth are the only entities exerting a force each on the cylinder.

 

[ymnoklan]

Which of the forces you mentioned is supposed to keep the cylinder constrained to the rod?

I am thinking the x-component of the force from the spring.

 

[ymnoklan]

I think you need to provide a diagram!

Too much information is missing. For example:
- is the rod vertical? or horizontal? or angled?
- if (for example) the rod is vertical, is the spring above or below the cylinder?
- you state "The height between the cylinder and where spring is connected is 0.30 m" but this could be interpreted in several ways.

The rod is horizontal, not angled. I am attaching a photo of the situation

 

[ymnoklan]

I think that if these are the only forces acting on the cylinder, then the rod makes no difference so you might as well forget about its existence. Only you know what the picture looks like. What would happen if the rod were magically vaporized? If the answer is "Nothing", then the spring and Earth are the only entities exerting a force each on the cylinder.

I am thinking there may be some kind of normal force upwards from the rod on the cylinder, neutralizing the gravitational force keeping the cyllinder from falling down, in addtition to the gravity (downwards) and the force from the spring (diagonally).

 

[kuruman]

Thanks for the picture, it certainly clarifies what's going on. If the rod is frictionless, then yes, it can only exert a normal force. Now proceed with the FBD.

 

[kuruman]

The spring is horizontal, not angled. I am attaching a photo of the situation

In what sense is the spring not angled? The picture shows that it forms angle $\alpha$ with respect to the horizontal.

 

[ymnoklan]

In what sense is the spring not angled? The picture shows that it forms angel $\alpha$ with respect to the horizontal.

Sorry, yes, I meant the rod is not angled, but the spring is

 

[Steve4Physics]

Sorry, yes, I meant the rod is not angled, but the spring is

Is the spring in a state of compression or tension?

 

[kuruman]

Is the spring in a state of compression or tension?

I think that at the instant shown the spring is at its equilibrium length $l_0$.

 

[ymnoklan]

I think that at the instant shown the spring is at its equilibrium length $l_0$.

Correct

 

[kuruman]

Correct

So what is the force exerted by the spring when it is at its equilibrium length?

 

[ymnoklan]

Thanks for the picture, it certainly clarifies what's going on. If the rod is frictionless, then yes, it can only exert a normal force. Now proceed with the FBD.

The rod is indeed frictionless. Is the force from the spring going upwards or downwards?

 

[ymnoklan]

So what is the force exerted by the spring when it is at its equilibrium length?

Zero?

 

[kuruman]

Zero?

Yes.

 

[Steve4Physics]

@ymnoklan, you didn't post the complete question as given to you. For example, initially you omitted the diagram and you didn't state that the spring was at its equilibrium length.

Is there any other information (even if you don't think it's important!) - maybe relating to the variables shown in the diagram ($\alpha, l_0,h$ and $x$)?

 

[ymnoklan]

This is all the information I have:
In this assignment, we will look at a metal cylinder of mass m = 5.0 kg that has a hole in
center and is set on a metal rod. The cylinder slides horizontally along the rod. In addition, it isattached a spring to the cylinder with spring constant k = 500 N/m. When the spring is not stretched or compressed, it has a length l0 = 0.50 m (the natural length of the spring). The spring is again attached to a table placed under the metal cylinder and rod. The height between the table and the attachment point of the spring on the cylinder is h = 0.30 m. We start by assuming that
is no friction or air resistance.

@ymnoklan, you didn't post the complete question as given to you. For example, initially you omitted the diagram and you didn't state that the spring was at its equilibrium length.

Is there any other information (even if you don't think it's important!) - maybe relating to the variables shown in the diagram ($\alpha, l_0,h$ and $x$)?

 

[kuruman]

What is the question to be answered on the basis of this description of the system? My suspicion is to find the period of small oscillations about the equilibrium position. At least that's what I would ask.

 

[Steve4Physics]

@ymnoklan I agree with @kuruman in Post #20.

Are you asked to produce a free body diagram (with the spring at its equilibrium length)? Or is that just something you thought was necessary/useful to help with other parts of the question?

Typo' corrected.

 

[ymnoklan]

@ymnoklan I agree with @kuruman in Post #20.

Are you asked to produce a free body diagram (with the spring at its equilibrium length)? Or is that just something you thought was necessary/useful to help with other parts of the question?

Typo' corrected.

I am asked to make a FBD of the cylinder, however it is not clear to me if they mean at equilibrium or not.

 

[Steve4Physics]

I am asked to make a FBD of the cylinder, however it is not clear to me if they mean at equilibrium or not.

If this is work to be handed-in, you could provide 3 different free body diagrams of the cylinder:
1) at equilibrium position;
2) cylinder displaced some amount to the left of equilibrium position;
3) cylinder displaced some amount to the right of equilibrium position.
Then you are pretty much covered!

Re-read previous posts, make an attempt at each of the above diagrams and post them here.

 

[ymnoklan]

If this is work to be handed-in, you could provide 3 different free body diagrams of the cylinder:
1) at equilibrium position;
2) cylinder displaced some amount to the left of equilibrium position;
3) cylinder displaced some amount to the right of equilibrium position.
Then you are pretty much covered!

Re-read previous posts, make an attempt at each of the above diagrams and post them here.

Top: at equilibrium
Middle: spring pulled to the left
Bottom: spring pulled to the right
What do you think of this?

 

[Steve4Physics]

Top: at equilibrium
Middle: spring pulled to the left
Bottom: spring pulled to the right
What do you think of this?

Here are some comments. Note: I’m in the UK - there may be some variations with how FBDs are drawn in your country.
__________________

General Comments

1. Can you rotate the image so it’s easier for us to read?

2. You are asked for the FBD of the cylinder. The only object shown in each FBD should be the cylinder. If the long horizontal line on each diagram is meant to be the rod, it should be removed.

3. $G$ is not the usual symbol for weight (here in the UK, anyway). I’d use $W$ but maybe this is related to language.

4. Include a ‘key’ which explains each force. E.g.
$\vec G$ weight of cylinder
$\vec N$ <your description of what you think $\vec N$ is>
$\vec F$ <your description of what you think $\vec F$ is>
Apologies - I didn't notice that you had done this below the diagrams.

5. On a FBD, distributed forces (forces not acting at a point) are usually shown as a single arrow, So,rather than 2 arrows for $\vec N$, use a single arrow in the middle. (In the same way that a single arrow is used for weight, even though gravity acts on every atom in the cylinder.)

(Note: engineers may sometimes use a series of arrows for a distributed force $\uparrow \uparrow \uparrow \uparrow$. But I assume we’re talking physics here, not engineering.)
__________________

Digram 1

Looks like you’ve got the total length of the $N$ arrows correct. Good. But they need replacing with a single arrow (item 5, above).
__________________

Digrams 2 and 3

At equilibrium, the spring is angled, $\alpha \approx 45^o$ (see Post 6 diagram).

For a small left-displacement of the cylinder, say to make $\alpha= 50^o$, the spring will be in a state of compression; so what is the direction on the force on the spring? For a small displacement right, say to make $\alpha = 40^o$, the spring will be in a state of tension and will not be as shown in your diagram 3!

The dotted lines are not needed. In fact they show the y-component of $\vec F$ to be equal to the weight – which is is not generally true.

Using the approximate length of an arrow to indicate the magnitude of each force, it appear that there is a net downwards force-component on the cylinder. This would causes downwards acceleration – but the cylinder does not accelerate downwards!

Edited - LaTeX and other corrections.

 

[ymnoklan]

Thanks for your feedback! What do you think of this?

Here are some comments. Note: I’m in the UK - there may be some variations with how FBDs are drawn in your country.
__________________

General Comments

1. Can you rotate the image so it’s easier for us to read?

2. You are asked for the FBD of the cylinder. The only object shown in each FBD should be the cylinder. If the long horizontal line on each diagram is meant to be the rod, it should be removed.

3. $G$ is not the usual symbol for weight (here in the UK, anyway). I’d use $W$ but maybe this is related to language.

4. Include a ‘key’ which explains each force. E.g.
$\vec G$ weight of cylinder
$\vec N$ <your description of what you think $\vec N$ is>
$\vec F$ <your description of what you think $\vec F$ is>
Apologies - I didn't notice that you had done this below the diagrams.

5. On a FBD, distributed forces (forces not acting at a point) are usually shown as a single arrow, So,rather than 2 arrows for $\vec N$, use a single arrow in the middle. (In the same way that a single arrow is used for weight, even though gravity acts on every atom in the cylinder.)

(Note: engineers may sometimes use a series of arrows for a distributed force $\uparrow \uparrow \uparrow \uparrow$. But I assume we’re talking physics here, not engineering.)
__________________

Digram 1

Looks like you’ve got the total length of the $N$ arrows correct. Good. But they need replacing with a single arrow (item 5, above).
__________________

Digrams 2 and 3

At equilibrium, the spring is angled, $\alpha \approx 45^o$ (see Post 6 diagram).

For a small left-displacement of the cylinder, say to make $\alpha= 50^o$, the spring will be in a state of compression; so what is the direction on the force on the spring? For a small displacement right, say to make $\alpha = 40^o$, the spring will be in a state of tension and will not be as shown in your diagram 3!

The dotted lines are not needed. In fact they show the y-component of $\vec F$ to be equal to the weight – which is is not generally true.

Using the approximate length of an arrow to indicate the magnitude of each force, it appear that there is a net downwards force-component on the cylinder. This would causes downwards acceleration – but the cylinder does not accelerate downwards!

Edited - LaTeX and other corrections.

 

[Lnewqban]

 

[Steve4Physics]

Thanks for your feedback! What do you think of this?

I would draw the normal reaction centred (left/right) on the cylinder.

You might like to consider these (not free body) diagrams:

 

[ymnoklan]

I would draw the normal reaction centred (left/right) on the cylinder.

You might like to consider these (not free body) diagrams:
View attachment 352292

How would you describe the direction of the normal force on the cylinder? Is it always vertically upwards) oppsite of gravity or does the direction of the normal force vary with the horizontal displacement x?

 

[Orodruin]

How would you describe the direction of the normal force on the cylinder? Is it always vertically upwards) oppsite of gravity or does the direction of the normal force vary with the horizontal displacement x?

What do you think?

 

[Steve4Physics]

How would you describe the direction of the normal force on the cylinder? Is it always vertically upwards) oppsite of gravity or does the direction of the normal force vary with the horizontal displacement x?

The normal force of the rod on the cylinder ($\vec N$) is (by definition of 'normal') perpendicular to the contact surfaces. So in this problem ,$\vec N$ must act vertically up, or vertically down, or be zero.

The cylinder never has a vertical acceleration, so the net vertical (y) force on the cylinder is always zero: $\vec W + \vec {F_y} + \vec N = 0$

Can you see under what conditions:
$\vec N$ acts vertically up?
$\vec N$ acts vertically down?
$\vec N$ = 0?

Also, have you covered simple harmonic motion (SHM)? This could help you understand how your rod/cylinder/spring system will behave.

Edit. Of course I should simply have asked what @Orodruin asked in Post #30!

 

[Lnewqban]

You have represented the N forces correctly oriented in your diagrams, only that the origin of the vector could be better located at the top surface of the hole in the cylinder.
That is where the bar is excerting a normal force on the cylinder in order to counteract gravity and the normal component of the spring force.

 

[ymnoklan]

Now I am wondering about the equilibrium points of the system. I find that they must be at x = -0.4, x = 0 and x = 0.4 (by finding where there would be 0 force in the horizontal direction), but I struggle characterising which would be stable and which would be unstable. What does this even mean?

 

[Steve4Physics]

Now I am wondering about the equilibrium points of the system. I find that they must be at x = -0.4, x = 0 and x = 0.4 (by finding where there would be 0 force in the horizontal direction), but I struggle characterising which would be stable and which would be unstable. What does this even mean?

If x=-0.4m the cylinder will not be entirely on the rod (assuming the Post #6 diagram shows x=0.4m). So presumably you are meant to assume the rod is longer than shown.

Equilibrium states can be stable or unstable. What is the difference? Have you looked-up these terms?

By the way, you never answered my Post #31 questions relating to the direction of $\vec N$.

 

[Orodruin]

but I struggle characterising which would be stable and which would be unstable. What does this even mean?

For this, I suggest working with the potential of the system. A (un)stable equilibrium corresponds to a minimum (maximum) of the potential.

Alternatively, find how the force looks close to each equilibrium. If it points towards (away) from the equilibrium, it is a (un)stable equilibrium.