Find Hinge Force of Rod Undergoing Circular Motion

[utkarshakash]

I may be wrong but I think the equation in post#15 is incorrect .You cannot write torque equation about the hinge .

In this problem it has to be written about the COM.

You can write torque equation about either of them. Chestermiller took it about the hinge to make the reaction forces disappear in the torque equation and thus save a lot of work which was not possible otherwise. I don't find anything incorrect in his solution as it gives me the correct answer.

 

[paisiello2]

Correct...So forget energy conservation :)

Except we know what that work is as it is explicitly given in the problem. In this case the work done is only in one direction and if we are only interested in the rotational energy then we can solve directly for ω without doing any integration.

In addition, gravity also does work on the system. Therefore by your logic we could never do a conservation of energy which is not correct.

 

[utkarshakash]

Congrats for getting the correct answer .Quite a difficult problem .

Could you share exactly what equations you used and what is the correct answer ?

OK here's what I did:

Since Chet already figured out ω I used that to put it in this equation

$R_x - mg = m \omega ^2 l$

which gives $R_x = 4mg$

Since the rod rotates about the hinge,
$a_y = \dfrac{\alpha l}{2}$

Using the equation $\tau = I \alpha$
I get

$\dfrac{mgl}{2} = \dfrac{ml^2}{3} \alpha$

Now using Newton's second law

$mg - R_y = ma_y$

Solve the above equations and you'll get the answer.

 

[Saitama]

OK here's what I did:

Since Chet already figured out ω I used that to put it in this equation

$R_x - mg = m \omega ^2 l$

Why do you use $l$ instead of $l/2$?

 

[Saitama]

Okay, I have tried something which gives the correct answer of $\sqrt{257}mg/4$.

Move to the frame accelerating with an acceleration a=g. This results in a pseudo force of mg acting on the rod towards left.

Doing moment balance when the rod is at an angle $\theta$, gives:
$$ mg\frac{l}{2}\sin\theta+mg\frac{l}{2}\cos\theta=\frac{ml^2}{3}\alpha $$
Rewriting alpha as $\dfrac{1}{2}\dfrac{d\omega^2}{d\theta}$ and simplifying gives:
$$\frac{d\omega^2}{d\theta}=\frac{3g}{l}(\sin\theta+\cos\theta)$$
Solving the D.E gives:
$$\omega^2=\frac{3g}{l}(1-\cos\theta+\sin\theta)$$
At $\theta=\pi/2$, $\omega^2=6g/l$. From Newton's second law:
$$R_x-mg=m\omega^2\frac{l}{2} \Rightarrow R_x=4mg$$
This is the same value which utkarshakash posted and I am inclined to think that what I have done above makes more sense because I wrote $R_x-mg=m\omega^2\frac{l}{2}$ instead of $R_x-mg=m\omega^2l$. But then Chet said that we don't take into account the pseudo forces when taking moments.

EDIT: If I work in this accelerating frame, I can get the value of $\omega^2$ directly from energy conservation.

$$\frac{mgl}{2}+\frac{mal}{2}=mgl=\frac{1}{2}\frac{ml^2}{3}\omega^2$$
$$\Rightarrow \omega^2=\frac{6g}{l}$$

 

[paisiello2]

I may be wrong but I think the equation in post#15 is incorrect .You cannot write torque equation about the hinge .

In this problem it has to be written about the COM.

Not true. You can take any frame of reference you want. Taking moments about the hinge simplifies the problem.

 

[Chestermiller]

OK. I'm going to do this a another way, by taking moments about the center of mass. This should focus on Tanya's concern about taking moments about the hinge, and should also address utkarshakash and Pranav-Arora's contention that the pseudo force needs to be included if we take moments about the hinge. Taking moments about the center of mass gives:

$$R_x\frac{l}{2}\cosθ+R_y\frac{l}{2}\sinθ=\left(m\frac{l^2}{12}\right) \frac{d^2θ}{dt^2}$$

Note that, on the right hand side of this equation, I've used the moment of inertia about the center of mass. Now, from the force balance and the kinematics,
$$R_x=m\left(g-\frac{l}{2}\cosθ\frac{d^2θ}{dt^2}+ \frac{l}{2}\sinθ\left(\frac{dθ}{dt}\right)^2\right)$$

$$R_y=mg-m\left(\frac{l}{2}\sinθ\frac{d^2θ}{dt^2}+\frac{l}{2}\cosθ\left(\frac{dθ}{dt}\right)^2\right)$$
Now if we substitute these equations for the reaction forces into the moment balance equation and combine terms, we obtain:
$$mg\frac{l}{2}\cosθ+mg\frac{l}{2}\sinθ=m\left(\frac{l^2}{3}\right)\frac{d^2θ}{dt^2}$$
This is identical to Pranav-Arora's equation in the previous post. So it looks like Tanya was right about the need to take moments about the center of mass, and utkarshakash and Pranav-Arora were right about the need to include the pseudo reaction force if you take moments about the hinge.

Chet

 

[Saitama]

Very nicely done, thanks a lot Chestermiller. Its always great to know about different ways to solve a problem.

I got to learn a lot from this thread, thanks to utkarshakash for putting up such an excellent problem. :)

 

[paisiello2]

So it looks like Tanya was right about the need to take moments about the center of mass, and utkarshakash and Pranav-Arora were right about the need to include the pseudo reaction force if you take moments about the hinge.

The pseudo forces or inertial forces are there regardless of what reference frame you pick. So it's wrong to say you have to take it only at the center of mass. I think using conservation of energy is another way to do it but doesn't involve solving a differential equation.

 

[utkarshakash]

Okay, I have tried something which gives the correct answer of $\sqrt{257}mg/4$.

Move to the frame accelerating with an acceleration a=g. This results in a pseudo force of mg acting on the rod towards left.

Doing moment balance when the rod is at an angle $\theta$, gives:
$$ mg\frac{l}{2}\sin\theta+mg\frac{l}{2}\cos\theta=\frac{ml^2}{3}\alpha $$
Rewriting alpha as $\dfrac{1}{2}\dfrac{d\omega^2}{d\theta}$ and simplifying gives:
$$\frac{d\omega^2}{d\theta}=\frac{3g}{l}(\sin\theta+\cos\theta)$$
Solving the D.E gives:
$$\omega^2=\frac{3g}{l}(1-\cos\theta+\sin\theta)$$
At $\theta=\pi/2$, $\omega^2=6g/l$. From Newton's second law:
$$R_x-mg=m\omega^2\frac{l}{2} \Rightarrow R_x=4mg$$
This is the same value which utkarshakash posted and I am inclined to think that what I have done above makes more sense because I wrote $R_x-mg=m\omega^2\frac{l}{2}$ instead of $R_x-mg=m\omega^2l$. But then Chet said that we don't take into account the pseudo forces when taking moments.

EDIT: If I work in this accelerating frame, I can get the value of $\omega^2$ directly from energy conservation.

$$\frac{mgl}{2}+\frac{mal}{2}=mgl=\frac{1}{2}\frac{ml^2}{3}\omega^2$$
$$\Rightarrow \omega^2=\frac{6g}{l}$$

But why should we use l/2 here? Since the rod rotates about the hinge, shouldn't it be l? It'd be nice if you could explain me.

 

[Chestermiller]

The pseudo forces or inertial forces are there regardless of what reference frame you pick. So it's wrong to say you have to take it only at the center of mass.

If you don't want to have to include a pseudo force, (and you are using an inertial frame of reference) then you have to take moments about the center of mass. Tanya already pointed out the error of not taking moments around the center of mass, and I confirmed Tanya's assertion in my most recent post.

I think using conservation of energy is another way to do it but doesn't involve solving a differential equation.

So far you've said that several times, but we still haven't seen your demonstration of how it can be done.

Of course, in the accelerated frame, doing it using energy conservation is no problem because there the pseudo force becomes an actual body force with a horizontal "graviational" potential (as reckoned by observers in the accelerated frame.)

Chet

 

[paisiello2]

If you don't want to have to include a pseudo force, (and you are using an inertial frame of reference) then you have to take moments about the center of mass. Tanya already pointed out the error of not taking moments around the center of mass, and I confirmed Tanya's assertion in my most recent post.

That's not what I said. What I said was the inertia forces are there regardless of what reference frame you choose. So Tanya made an erroneous statement.

 

[Saitama]

But why should we use l/2 here? Since the rod rotates about the hinge, shouldn't it be l? It'd be nice if you could explain me.

Lets consider a rod of mass m rotating with a constant angular velocity $\omega$. In the rotating frame, a centrifugal force acts on it. Consider a small length ($dx$) of rod at a distance x from the hinge. The centrifugal force on it is given by $(m/l)\omega^2x\,dx$. Integrating to calculate the net centrifugal force gives $m\omega^2l/2$ which is the result I used in the problem.

I hope that helps.

 

[Saitama]

That's not what I said. What I said was the inertia forces are there regardless of what reference frame you choose. So Tanya made an erroneius statement.

Hi paisiello2!

How about writing down a few equations and see for yourself? ;)

 

[Tanya Sharma]

OK. I'm going to do this a another way, by taking moments about the center of mass. This should focus on Tanya's concern about taking moments about the hinge, and should also address utkarshakash and Pranav-Arora's contention that the pseudo force needs to be included if we take moments about the hinge. Taking moments about the center of mass gives:

$$R_x\frac{l}{2}\cosθ+R_y\frac{l}{2}\sinθ=\left(m\frac{l^2}{12}\right) \frac{d^2θ}{dt^2}$$

Note that, on the right hand side of this equation, I've used the moment of inertia about the center of mass. Now, from the force balance and the kinematics,
$$R_x=m\left(g-\frac{l}{2}\cosθ\frac{d^2θ}{dt^2}+ \frac{l}{2}\sinθ\left(\frac{dθ}{dt}\right)^2\right)$$

$$R_y=mg-m\left(\frac{l}{2}\sinθ\frac{d^2θ}{dt^2}+\frac{l}{2}\cosθ\left(\frac{dθ}{dt}\right)^2\right)$$
Now if we substitute these equations for the reaction forces into the moment balance equation and combine terms, we obtain:
$$mg\frac{l}{2}\cosθ+mg\frac{l}{2}\sinθ=m\left(\frac{l^2}{3}\right)\frac{d^2θ}{dt^2}$$
This is identical to Pranav-Arora's equation in the previous post. So it looks like Tanya was right about the need to take moments about the center of mass, and utkarshakash and Pranav-Arora were right about the need to include the pseudo reaction force if you take moments about the hinge.

Chet

Brilliant...Simply brilliant !

Thank you so much Chet

 

[utkarshakash]

Lets consider a rod of mass m rotating with a constant angular velocity $\omega$. In the rotating frame, a centrifugal force acts on it. Consider a small length ($dx$) of rod at a distance x from the hinge. The centrifugal force on it is given by $(m/l)\omega^2x\,dx$. Integrating to calculate the net centrifugal force gives $m\omega^2l/2$ which is the result I used in the problem.

I hope that helps.

Thanks!

I have one more doubt. Do we associate potential energy corresponding to a pseudo force just like you did in energy conservation?

 

[paisiello2]

Hi paisiello2!

How about writing down a few equations and see for yourself? ;)

See what for myself? That I can't take moments about any point I want?

 

[Saitama]

Thanks!

I have one more doubt. Do we associate potential energy corresponding to a pseudo force just like you did in energy conservation?

I did not associate any potential energy to the pseduo force. I simply wrote the work done by the pseudo force. Remember, the work done is force times the displacement in the direction of force. The pseudo force is mg and the displacement in its direction is l/2.

 

[utkarshakash]

I did not associate any potential energy to the pseduo force. I simply wrote the work done by the pseudo force. Remember, the work done is force times the displacement in the direction of force. The pseudo force is mg and the displacement in its direction is l/2.

I got it. Energy conservation was a much more simpler method.

 

[Chestermiller]

That's not what I said. What I said was the inertia forces are there regardless of what reference frame you choose. So Tanya made an erroneous statement.

Hi paisiello.

I can see that I still haven't done a good job of explaining what I'm trying to say. So I'm going to try again.

When most of us learned to do moment balances, we learned that we are supposed to include only moments resulting from body forces and contact forces, not ma (through the center of mass) and not pseudo forces. We also learned that, if the center of mass of the rigid body is accelerating relative to our frame of reference, then, in order to get the correct answer, we needed to do the moment balance about the center of mass. Otherwise, we would get the wrong answer (as I did in my post #15, when I failed to honor this rule). We also learned that, if the center of mass of our rigid body is not accelerating relative to our inertial frame of reference, then we can take moments about any convenient axis (and get the right answer).

Moment balances can also, of course, be done relative to a non-rotating frame of reference that is at rest relative to the center of mass of the body, even if the center of mass is accelerating relative to inertial frames of reference. In this type of situation, the pseudo force reckoned from an inertial frame of reference becomes a real body force as reckoned from the accelerating frame of reference. So, accordingly, this real body force must be included in the moment balance for the accelerating frame of reference. Since, as reckoned from the accelerating frame of reference, the center of mass is at rest, the moment balance can be taken about any convenient axis (again, provided the body force is included).

I hope I did a better job of explaining this time. I am soliciting comments from others on how this articulation can be improved.

Chet

 

[paisiello2]

So far you've said that several times, but we still haven't seen your demonstration of how it can be done.

Of course, in the accelerated frame, doing it using energy conservation is no problem because there the pseudo force becomes an actual body force with a horizontal "graviational" potential (as reckoned by observers in the accelerated frame.)

I believe Pranav-Arora showed this already in a post previous to yours. This contradicts what Tanya said about "forgetting conservation of energy" and it is a much simpler way to solve the problem.

 

[Saitama]

I believe Pranav-Arora showed this already in a post previous to yours. This contradicts what Tanya said about "forgetting conservation of energy" and it is a much simpler way to solve the problem.

Hey paisiello! :)

Tanya is correct. Conservation of energy won't work in the lab or inertial frame as we don't know the work done by external force in that frame. But when we move to the accelerated frame, as Chet said, we have got a actual body force so there is no problem in using the conservation of energy.

I hope that addresses your doubt. ;)

 

[paisiello2]

Hi Chet:

Thanks very much for your civility in your post. I will do the same in my response and maybe we can both learn something in the discussion of our different view points. I certainly admit I might be wrong here for some reason that I cannot see but maybe you might admit the same.

...we are supposed to include only moments resulting from body forces and contact forces, not ma (through the center of mass) and not pseudo forces.

The inertial forces are still there, however:
∑F = ma
→ ∑F - ma = 0

You can interpret the ma term as the inertial force acting opposite to the other body forces. If you take moments about the center of mass then of course the lever arm is zero and so the moments are zero but the inertial forces are still there.

...if the center of mass of the rigid body is accelerating relative to our frame of reference, then, in order to get the correct answer, we needed to do the moment balance about the center of mass. Otherwise, we would get the wrong answer (as I did in my post #15, when I failed to honor this rule).

I disagree in the statement that you have to do the moment balance about the center of mass. You got the wrong answer because you didn't include the inertial forces which I claim are there regardless of where you take the moment axis.

...if the center of mass of our rigid body is not accelerating relative to our inertial frame of reference, then we can take moments about any convenient axis (and get the right answer).

I think this is just trivial statics but I agree of course.

You introduce the term inertial frame of reference here but I am not sure that it means anything in the context of this problem. It might mean something in some broader context of Einstein relativity or something but technically we are assuming that the problem is occurring in a rotating reference frame that matches that of the earth's. And as a result we are neglecting the centrifugal, Coriolis, and Euler forces assuming they are negligible. But we do need to still include the inertial forces (which would of course be zero in the case of statics).

Moment balances can also, of course, be done relative to a non-rotating frame of reference that is at rest relative to the center of mass of the body, even if the center of mass is accelerating relative to inertial frames of reference. In this type of situation, the pseudo force reckoned from an inertial frame of reference becomes a real body force as reckoned from the accelerating frame of reference. So, accordingly, this real body force must be included in the moment balance for the accelerating frame of reference. Since, as reckoned from the accelerating frame of reference, the center of mass is at rest, the moment balance can be taken about any convenient axis (again, provided the body force is included).

Here is where you sort of confuse me because I think you are contradicting what you said in the second quote above:

1) In the second quote you said you have to take moments about the center of mass otherwise you get the wrong answer. You didn't qualify this statement with what frame of reference you were taking. You just made a blanket statement.

2) Now you are saying you can take moments about any axis after all, but you qualify it by saying it has to be a non-rotating inertial frame of reference.

Can you see there the contradiction that I see if you don't bother to qualify 1)? Maybe that is just trivial but still important when you make a blanket statement like Tanya did about only being able to take moments about the center of mass.

Even it you were to qualify 1), which I assume you agree that you should do, I also think you might have it backwards anyway. If you take an inertial frame that is non-rotating then by definition there are supposed to be no inertial forces. Therefore in order to get the same answer as a non-inertial frame you have to take moments about the center of mass. This is just my understanding about what is meant by an inertial frame of reference but maybe I have it backwards.

Now, more importantly in this kind of specific problem, why wouldn't you always just take a non-inertial frame of reference by default? Technically I think you have to assume you are anyway for the planet earth.

Maybe you can enlighten me as to where my reasoning is faulty here. I'm actually interested in learning something that maybe I didn't know before.

Thanks again for your time.

 

[paisiello2]

Hey paisiello! :)

Tanya is correct. Conservation of energy won't work in the lab or inertial frame as we don't know the work done by external force in that frame. But when we move to the accelerated frame, as Chet said, we have got a actual body force so there is no problem in using the conservation of energy.

I hope that addresses your doubt. ;)

But she didn't say "in the lab or inertial frame". She said just forget about it. So her blanket statement is wrong.

Also, I claim that because no other information is given then we should assume by default that the problem is occurring on the planet Earth which is technically a rotating non-inertial frame.

Additionally, we do know the work done by the external force in that frame: mgl/2

 

[Chestermiller]

Hi Chet:

Thanks very much for your civility in your post. I will do the same in my response and maybe we can both learn something in the discussion of our different view points. I certainly admit I might be wrong here for some reason that I cannot see but maybe you might admit the same.

The inertial forces are still there, however:
∑F = ma
→ ∑F - ma = 0

You can interpret the ma term as the inertial force acting opposite to the other body forces. If you take moments about the center of mass then of course the lever arm is zero and so the moments are zero but the inertial forces are still there.

Hi paisiello2,

Most of us have learned to do moment balances without including the pseudo force. Whether or not the rigid body is rotating, if you have an accelerating body and you do the moment balance without including the pseudo force, you will get the wrong answer unless you take moments about the center of mass. This is pretty easy to show mathematically. However, if you include the pseudo force in the moment balance, you can take moments about any convenient axis and get the right answer.

I think that this pretty much confirms what both of us have been saying (in our own ways).

I disagree in the statement that you have to do the moment balance about the center of mass. You got the wrong answer because you didn't include the inertial forces which I claim are there regardless of where you take the moment axis.

Apparently, there is no one right way of doing the moment balance. If you include the pseudo force, then you can take moments about any convenient axis, and, if you don't include the pseudo force, you must take moments about the center of mass (unless ma = 0). Most of us learned to do it without the pseudo force, and are unaccustomed to (and uncomfortable with) using the pseudo force approach.

2) Now you are saying you can take moments about any axis after all, but you qualify it by saying it has to be a non-rotating inertial frame of reference.

In a non-rotating frame of reference that is accelerating at the same rate as the center of mass (I didn't say "inertial"), the pseudo force becomes a real force, and the pseudo force reckoned from this accelerated frame of reference becomes zero. This means that you are justified in taking moments about any convenient axis.

Apparently, the entire source of our inability to communicate on this issue stems from our different orientations on how we are accustomed to doing a moment balance. The two methods are entirely equivalent mathematically. But this difference in orientation is an interesting phenomenon (depending on training an background) that I have encountered in other technical areas on Physics Forums.

 

[paisiello2]

Chet:

Do you agree with this post made by Tanya:

You cannot write torque equation about the hinge .

In this problem it has to be written about the COM.

Simply, yes or no?

And do you agree with this post made by Tanya:

Energy conservation can not be applied in this problem.

Again simply, yes or no?

If you are consistent with your reasoning then you have to agree the above statements made by Tanya are false irrespective of what anybody learned in school.

Apparently, there is no one right way of doing the moment balance. If you include the pseudo force, then you can take moments about any convenient axis, and, if you don't include the pseudo force, you must take moments about the center of mass (unless ma = 0). Most of us learned to do it without the pseudo force, and are unaccustomed to (and uncomfortable with) using the pseudo force approach.

What I am saying is that the default assumption should be to assume a non-inertial reference frame (which is what the Earth is) and to always include the inertial forces. If you were taught to ignore them then it can make problems more difficult if not impossible to solve in my opinion.

For example, what if the hinge were to have a mass M? or the CM was difficult to determine or was not constant? Would you be uncomfortable using inertial forces then?

I assume your answer would be no and that including the inertial forces is generally the best approach to solving dynamics.

 

[Chestermiller]

Hi Paisiello2. Please see my private message to you.

Chet

 

[haruspex]

I may be wrong but I think the equation in post#15 is incorrect .You cannot write torque equation about the hinge .

No, it's fine. You can take torque and angular momentum about the CoM or about any non-accelerating reference point. We can make that reference point wherever the hinge happens to be at some fixed instant. For torque, we only care about that instant, so the fixity of the reference point becomes irrelevant. We would have to be more careful if assessing angular momentum.

 

[Tanya Sharma]

Hi haruspex...

No, it's fine.

I beg to differ .It's not fine.

It's fine only when you take into account the torque due to pseudo force.In post#15 torque equation about the hinge was written without considering the pseudo force which is incorrect.

You can take torque and angular momentum about the CoM or about any non-accelerating reference point.

Exactly .But since the hinge is accelerating,it would have been a good idea to consider writing torque equation about the COM .

The torque equation should be written about either a stationary point or about COM even if it is accelerating.Please have a look at http://en.wikipedia.org/wiki/Torque

From Wiki

"This equation has the limitation that the torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motion - either motion is pure translation, pure rotation or mixed motion. I = Moment of inertia about point about which torque is written (either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then the torque equation is same about all points in the plane of motion."

We can make that reference point wherever the hinge happens to be at some fixed instant. For torque, we only care about that instant, so the fixity of the reference point becomes irrelevant.

Again, I would respectfully disagree..but the choice of reference is quite relevant.

Either we work with COM or we need to take into account pseudo force if working with accelerating reference point.

 

[haruspex]

I beg to differ .It's not fine.

You're right, I withdraw that statement. I was misled because I got the same final answer taking a fixed point where the hinge is at some instant. But on closer inspection, the equations are not the same. With the rod length 2L and x, y as coordinates of rod's CoM, I get
$I\alpha = mgL \sin(\theta) + mL\ddot y\sin(\theta)+mL\ddot x \cos(\theta)$
From the constraints on the hinge motion:
$\ddot y = - L\sin(\theta) \ddot \theta$
$\ddot x = a - L\cos(\theta) \ddot \theta$
Whence
$\ddot \theta = \frac{3}{4L}(a\cos(\theta)+g\sin(\theta))$
With a = g this gives
${\dot \theta}^2 = \frac{3g}{2L}(1-\sqrt(2)\cos(\theta+\pi/4))$
At θ=pi/2, ${\dot \theta}^2 = \frac{3g}{2L}(1+1) = \frac{3g}{L}$
Force at joint $= ma + mL{\dot \theta}^2 = 4mg$

 

[Tanya Sharma]

You're right, I withdraw that statement. I was misled because I got the same final answer taking a fixed point where the hinge is at some instant. But on closer inspection, the equations are not the same. With the rod length 2L and x, y as coordinates of rod's CoM, I get
$I\alpha = mgL \sin(\theta) + mL\ddot y\sin(\theta)+mL\ddot x \cos(\theta)$
From the constraints on the hinge motion:
$\ddot y = - L\sin(\theta) \ddot \theta$
$\ddot x = a - L\cos(\theta) \ddot \theta$
Whence
$\ddot \theta = \frac{3}{4L}(a\cos(\theta)+g\sin(\theta))$
With a = g this gives
${\dot \theta}^2 = \frac{3g}{2L}(1-\sqrt(2)\cos(\theta+\pi/4))$
At θ=pi/2, ${\dot \theta}^2 = \frac{3g}{2L}(1+1) = \frac{3g}{L}$
Force at joint $= ma + mL{\dot \theta}^2 = 4mg$

Thank goodness .It requires a lot of courage to disagree with you :tongue2: