Uniform circular motion with a pendulum

[Femme_physics]

Homework Statement

http://img685.imageshack.us/img685/9250/drawingmo.jpg

In the drawing is depicted a pendulum hammer of an impact device (in both projections). The hammer is made out of shaft AB which upon it is held pendulum OM. At point M of the pendulum is a hammer. According to experiment they settle the pendulum in a condition close to that in the drawing, release it and it drops freely, as depicted in the dashed line in the drawing. At the lowest point of the fall (point N), the hammer reaches the model and breaks it.

Given:

Hammer Weight = 200 [N]
Length of the pendulum = 800mm

A) Calculate the max tension in pendulum OM.

[there is also a clause B, but I'm getting stuck at A]

The Attempt at a Solution

I've written my question in the attempts

http://img824.imageshack.us/img824/3124/fy1q.jpg

http://img405.imageshack.us/img405/5020/fy2p.jpg

 

[tiny-tim]

Hi Femme_physics!

Sorry, I don't understand what you've done.

Try conservation of energy ​

 

[I like Serena]

Hi Fp :)

Short hint this time: it's not a uniform circular motion, so those formulas do not apply.

 

[Femme_physics]

It's NOT uniform circular motion?

Hmmm. We haven't studied about conservation of energy. This exercise might be off my league for now, then.

 

[Femme_physics]

It's NOT uniform circular motion?

Hmmm. We haven't studied about conservation of energy. This exercise might be off my league for now, then.

Thanks for letting me know!

 

[I like Serena]

It's NOT uniform circular motion?

Hmmm. We haven't studied about conservation of energy. This exercise might be off my league for now, then.

Thanks for letting me know!

Well, the hammer falls and accelerates while falling. That's why.

But you do have a formula that you've hardly used yet.
The one that says:

g h = v² / 2

That would apply...

 

[Femme_physics]

Well, the hammer falls and accelerates while falling. That's why.

But you do have a formula that you've hardly used yet.

How do you know I hardly used it? What else do you know about my usages?!?

g h = v² / 2

Hmm...how is this formula called?

We were given a list of such formulas but I don't see it. Or maybe I don't have yours and tiny-tim's super math vision feat.

http://img9.imageshack.us/img9/4261/listofsuchformulas.jpg

 

[tiny-tim]

We were given a list of such formulas but I don't see it.

It's the formula in the bottom-right box.

It's the conservation of energy equation for something falling vertically.

Since energy depends only on the endpoints, and not on how you get there (direct or devious),
the same conservation of energy equation applies to this circular motion.

 

[Femme_physics]

Thanks, tiny-tim! I wish he's given us NAMES to all the formulas, and not just the formulas themselves. I guess next class a decent question to the lecturer would be "can you name all those formulas?"

It's the conservation of energy equation for something falling vertically.

Duly noted and written.

Since energy depends only on the endpoints, and not on how you get there (direct or devious),

Oh yea, we were even told that in biology.

the same conservation of energy equation applies to this circular motion.

*nods*

Then I shall spur myself to action and reply back sometimes later! :) Thanks, tiny-tim.

 

[tiny-tim]

Thanks, tiny-tim! I wish he's given us NAMES to all the formulas, and not just the formulas themselves. I guess next class a decent question to the lecturer would be "can you name all those formulas?"

They don't have names.

This one is only conservation of energy if a happens to be gravity.

Oh yea, we were even told that in biology.

erm … it doesn't work in biology …

you use energy just standing still! ​

 

[I like Serena]

How do you know I hardly used it? What else do you know about my usages?!?

There!
https://www.physicsforums.com/showpost.php?p=3297647&postcount=24

My bad, I have to admit it looks a little different, and I guess it takes a bit of a super math/physics vision feat to see it.
You will, just give it another level, and then you will have this feat as well!

Hmm...how is this formula called?

Yeah, conservation of energy (for gravity and kinetic energy).

More specifically it is:

E = m g h + m v² / 2

This is the energy of the system that never changes (assuming no friction).
So after any change, the energy E before, must be equal to the energy after.

Perhaps it's premature for you to see this, but I think you'll get it soon enough and it probably doesn't hurt if you'd recognize it.

 

[Femme_physics]

You will, just give it another level, and then you will have this feat as well!

*looks at character sheet*

This says I need 2948xp to ding! You have any ideas how many tough physics bosses problems I'll have to solve to get there?!? :(
I need an item that imbues me with that power ;)

They don't have names.

This one is only conservation of energy if a happens to be gravity.

Really? Nameless formulas? Well, someone must've come up with them and named them sometimes in the past. It can't be that he just came up with it and left it nameless?

erm … it doesn't work in biology …

I think we may have studied the physical properties that are also important in biology, but I clearly and distinctly recall this quote

"Since energy depends only on the endpoints, and not on how you get there (direct or devious),"

Perhaps it's premature for you to see this, but I think you'll get it soon enough and it probably doesn't hurt if you'd recognize it.

I tend to only get stuff after lots of problem solving, it appears. That's also how I like it :)

I used the diameter for "h" -- is that right? it's the change in position, delta x. So, the change in position is 1.6m!

http://img546.imageshack.us/img546/9615/vsolutiona.jpg

 

[tiny-tim]

I used the diameter for "h" -- is that right? it's the change in position, delta x. So, the change in position is 1.6m!

you're assuming that α = 0

 

[Femme_physics]

you're assuming that α = 0

Oh yes, sorry I didn't mention that. A rather important fact!

We're told to:

"ignore the weight of the shaft and the pendulum and assume the angle at the beginning of the motion is 0."

 

[tiny-tim]

 

[Femme_physics]

Ah...I missed that didn't you? ;)

Now that I got V I'm a little bit hesitant how to continue.

Hi Fp :)

Short hint this time: it's not a uniform circular motion, so those formulas do not apply.

Even now? Because, to find what they ask of me (tension at pendulum) don't I need to find the normal force? And doesn't that mean sum of all forces?...

[edit: heading out to college to help a student in statics :) , might not reply for a while]

 

[I like Serena]

Ah...I missed that didn't you? ;)

Now that I got V I'm a little bit hesitant how to continue.



Even now? Because, to find what they ask of me (tension at pendulum) don't I need to find the normal force? And doesn't that mean sum of all forces?...

[edit: heading out to college to help a student in statics :) , might not reply for a while]

All right!

I'll give you a couple more hints...

What forces act (or are required to act) on the hammer?
And at what point of the trajectory will these forces generate the greatest tension in the pendulum?
Finally, how large will these forces be at that point?


EDIT: As for your previous calculations ...
I'm missing ...
A little more of you! :P

 

[tiny-tim]

Now that I got V I'm a little bit hesitant how to continue.

Find the centripetal acceleration …

then use net force = mass times acceleration. ​

 

[Femme_physics]

And at what point of the trajectory will these forces generate the greatest tension in the pendulum?

Clearly when the hammer drops and just hits the sample

Finally, how large will these forces be at that point?

Well, I'll need to find out T before I can tell you. So I could tell how long does it take to make half a circle.

Hmm...how do I find it T? Do I use this formula? (I hope I used it correctly)

http://img39.imageshack.us/img39/783/triedtodecent.jpg

EDIT: As for your previous calculations ...
I'm missing ...
A little more of you! :P

;) I would but I'd hate for people to start to think I'm modeling and not trying to do my homework lol

Find the centripetal acceleration …
then use net force = mass times acceleration.

Hmm...I'm still missing T to find a, (since "a" has "v" in its formula, and "v" has "t" in its formula. It all comes down to T, it appears!)

Of course, if my above calculation is correct, then I got T. Is that right?

 

[I like Serena]

T is the time for a complete revolution of a uniform circular motion.

It's a pity that the hammer just falls and breaks something and is not neatly swinging around at an even speed to break something else.

The formula you need is the one that relates the (required) centripetal force to speed and radius.

That is:

F = m v² / R

 

[Femme_physics]

It's a pity that the hammer just falls and breaks something and is not neatly swinging around at an even speed to break something else.

More destruction! More destruction! :D

The formula you need is the one that relates the (required) centripetal force to speed and radius.

Brilliant.


And we were given this list of formulas.

http://img24.imageshack.us/img24/6849/frfrh.jpg

I can see that formula there :)
The first 1 is just general
F = ma
The second is the one that "relates the (required) centripetal force to speed and radius."
The 3rd one is angular velocity.
The 4th one is angular velocity squared? I'm not sure what is that.
And the last one...wow. Not even sure what's that either.

But I need a formula I'm sure of :) Thanks to you. I'll use it.

 

[Femme_physics]

Did I get it?

http://img11.imageshack.us/img11/4030/theretherethere2.jpg

Clause B is easy, same as the other exercise getting me back to statics :) That's my thing!

 

[I like Serena]

Did I get it?

Clause B is easy, same as the other exercise getting me back to statics :) That's my thing!

Almost, but there are 2 forces: the tensile force and the force of gravity.
The resultant force must be equal to the centripetal force.

However, this means the tensile force is a little bigger!

 

[I like Serena]

More destruction! More destruction! :D

:D

And we were given this list of formulas.

I can see that formula there :)
The first 1 is just general
F = ma
The second is the one that "relates the (required) centripetal force to speed and radius."
The 3rd one is angular velocity.
The 4th one is angular velocity squared? I'm not sure what is that.
And the last one...wow. Not even sure what's that either.

But I need a formula I'm sure of :) Thanks to you. I'll use it.

Actually, all these formulas identify the centripetal force.

The omegas in between are intermediate steps to find the last formula, which is also the centripetal force, just with different variables.

Note the little r behind a in the first formula F = m a_r
This identifies the centripetal acceleration.


Note in particular that once you introduce T or f, you're assuming it's a uniform circular motion.
Before T and f are introduced the formula is also applicable to non-uniform circular motion (such as a hammer smashing stuff ).

 

[Femme_physics]

Almost, but there are 2 forces: the tensile force and the force of gravity.
The resultant force must be equal to the centripetal force.

However, this means the tensile force is a little bigger!

OH! That's brilliant, I got to say :) Because if the centripetal force was smaller the thing would break. And, if the centripetal force was bigger...well, that's impossible, but the thing would be pulled in closer to the rotation axis.

:) Must... learn... more... (!)

I will probably review this exercise for a long time now (esp. since everybody in class going to ask me about it).

So, ahem...right, the solution! Add 200 [N] you get 1000 [N]! or 999.18

Awesome :)

Thank you! Soooooooo much.

 

[Femme_physics]

Actually, all these formulas identify the centripetal force.

As I'm still not imbued with super level 58 math vision, I'll have to ask: Are you saying all these formulas are essentially the same? Do I really need so many of them then? And by the way, I took a crack at drawing the free body diagrams. Did I get it?

http://img854.imageshack.us/img854/3959/fbd1.jpg

http://img848.imageshack.us/img848/2611/fbd2.jpg

 

[Femme_physics]

Actually, I was just thinking of something when it comes to the free body diagrams:

I think "Vt" (tangential velocity) in drawing is wrong, since at this point tangential velocity turns into a force that hits the sample, so there are two forces at play to add for max tension--> the force of the tangential velocity and the reaction force of the sample. Is that right?

 

[tiny-tim]

Hi Femme_physics!

No, your free body diagram (a) is wrong …

i] the velocity V_t should not be on it at all, it's neither a force nor an acceleration

ii] you must decide which body the diagram refers to: since you're using the acceleration of the hammer (M, at the end of the shaft), the body must be the hammer, so the reaction force N at O is irrelevant: instead you need the tension T, which does act directly on the hammer

iii] you have shown the acceleration the same way as a force: you should show it differently (I refer a double arrow, off to the side of the diagram)

(alternatively, you could use the non-inertial rotating frame of the hammer, in which case the acceleration is zero, and you must instead show the extra, centrifugal force )

… at this point tangential velocity turns into a force that hits the sample, so there are two forces at play to add for max tension--> the force of the tangential velocity and the reaction force of the sample. Is that right?

The reaction force between the hammer and the sample will be horizontal, so it will not affect the tension (which at that point is vertical).

 

[I like Serena]

As I'm still not imbued with super level 58 math vision, I'll have to ask: Are you saying all these formulas are essentially the same? Do I really need so many of them then?

I think that what you need is to understand in how far they are the same and what use they are.

The only thing you need to know of the centripetal force is that:

$$F_{centripetal} = \frac {m v^2} r$$

But you also need to know that:

$$v = \omega \times r$$

And if the circular motion is uniform:
$$\omega = \frac {2 \pi} T$$
$$f = \frac 1 T$$

The other formulas can be derived from these.
It may be useful to keep them at hand for easier calculation, but it would help if you see that they are the same.

For instance, if you substitute v from the second formula in the first one, you'll find the other form of the centripetal force: F = m ω² r

 

[Femme_physics]

ii] you must decide which body the diagram refers to: since you're using the acceleration of the hammer (M, at the end of the shaft), the body must be the hammer, so the reaction force N at O is irrelevant: instead you need the tension T, which does act directly on the hammer

Ah! Settings me straight as always :) Of course! If we're looking at the "hammer head" only it emerges "from" the hammer head, not from the support. But wait, the hammer head hasn't been defined to me in terms of length and width. Does it matter?

iii] you have shown the acceleration the same way as a force: you should show it differently (I refer a double arrow, off to the side of the diagram)

Acceleration is always towards the axis of rotation. Of course N would have to point the same way because it's opposite of W. It's normal force... hmm..what am I missing here?

The reaction force between the hammer and the sample will be horizontal, so it will not affect the tension (which at that point is vertical).

Interesting! A great fact. And duly noted :)

But you also need to know that:$$v = \omega \times r$$

That I know :)

And if the circular motion is uniform:
$$\omega = \frac {2 \pi} T$$
$$f = \frac 1 T$$

Also know :)

The only thing you need to know of the centripetal force is that:

$$F_{centripetal} = \frac {m v^2} r$$

Aha! Then this is the one that lost me. A very important one too! Thanks, formula master :)

(alternatively, you could use the non-inertial rotating frame of the hammer, in which case the acceleration is zero, and you must instead show the extra, centrifugal force )

You wrote it in small text- any reason? I'll look into it later tomorrow. Heading to sleep now-- thanks for all your great awesome help, tiny-t, ILS! :) You rrrrrrrrrrrrrrrrrock.

 

[I like Serena]

Ah! Settings me straight as always :) Of course! If we're looking at the "hammer head" only it emerges "from" the hammer head, not from the support. But wait, the hammer head hasn't been defined to me in terms of length and width. Does it matter?

No, you don't need length and width of the hammerhead.
We'll treat it as a "point mass"!

Acceleration is always towards the axis of rotation. Of course N would have to point the same way because it's opposite of W. It's normal force... hmm..what am I missing here?

You're right.
I think tiny-tim only means you should make a distinction between force vectors, acceleration vectors, and speed vectors.
A different color would do as well. ;)
Furthermore, there's no real need to draw the acceleration, and since this is a FBD...

You wrote it in small text- any reason? I'll look into it later tomorrow. Heading to sleep now-- thanks for all your great awesome help, tiny-t, ILS! :) You rrrrrrrrrrrrrrrrrock.

This is about the centrifugal pseudo-force, which operates in a so called non-inertial frame, meaning there are special rules to it...

 

[Femme_physics]

Had time to sneak in one more reply :)

We'll treat it as a "point mass"!

Good! They should've mentioned it in the question, come to think of it! Why should it just occur to me naturally that we're treating it as point mass?

You're right.
I think tiny-tim only means you should make a distinction between force vectors, acceleration vectors, and speed vectors.
A different color would do as well. ;)

Sold. I'm convinced. :)

(plus, I like colors!)

Furthermore, there's no real need to draw the acceleration, and since this is a FBD...

There's no need to draw acceleration? Really?

This is about the centrifugal pseudo-force, which operates in a so called non-inertial frame, meaning there are special rules to it...

Oh, this is the psedu-force you were talking about? Hmm...special rules... well, this may be a heavy topic and I'll need a fresh mind. Talk to you two tomorrow! :)

 

[I like Serena]

Had time to sneak in one more reply :)

:D

Good! They should've mentioned it in the question, come to think of it! Why should it just occur to me naturally that we're treating it as point mass?

No reason. :)
It's just that its size does not matter.
You can give it a size if you want, but the answers will come out the same.

Do note that the radius R is, and would be, given from the axis to the center of mass of the hammerhead.

Furthermore, there's no real need to draw the acceleration, and since this is a FBD...

There's no need to draw acceleration? Really?

Here's from wikipedia about FBD's:

"A free body diagram, also called a force diagram, is a pictorial representation often used by physicists and engineers to analyze the forces acting on a body of interest. A free body diagram shows all forces of all types acting on this body. "

That is, in a pure form it should contain only force vectors. ;)
Oh, and note that if you add all the force vectors together, you should get a resultant force that points in the same direction as the acceleration.
Their relationship is:
$$\vec F_{resultant} = m \cdot \vec a$$

 

[Femme_physics]

Okay, a fresh new day (and once again fresh new coffee spilled on my papers), I'm ready to see if I truly have this problem figured out, starting with my new free body diagram based on tiny-tim's feedback.

This diagram refers to the forces acting on the hammer.

http://img715.imageshack.us/img715/6567/newfbd.jpg

No reason. :)
It's just that its size does not matter.

Who says "size doesn't matter"? ;)

Do note that the radius R is, and would be, given from the axis to the center of mass of the hammerhead.

I did note that :)

Here's from wikipedia about FBD's:

"A free body diagram, also called a force diagram, is a pictorial representation often used by physicists and engineers to analyze the forces acting on a body of interest. A free body diagram shows all forces of all types acting on this body. "

Ah...thanks for this jewel. That's proper engineering right there.

 

[I like Serena]

Okay, a fresh new day (and once again fresh new coffee spilled on my papers), I'm ready to see if I truly have this problem figured out, starting with my new free body diagram based on tiny-tim's feedback.

This diagram refers to the forces acting on the hammer.

Looking nice! :)
(Again, where's the coffee stain?)

That is, you've drawn N smaller than W.
That doesn't leave much for any net centripetal force does it?
Perhaps the hammerhead will be flying away...
Hope it doesn't impact something else!

And you're right, the speed turns into a force.
Right there you have an entirely new beast called "impulse", which is force times time.
I don't think your studying material covered that yet! :)
Note that we have no information about how great this force would be, although we can say a little on how large the change in "momentum" will be (another, or rather the same, new beast) .

Who says "size doesn't matter"? ;)

Eep! It does?

Ah...thanks for this jewel. That's proper engineering right there.

:)