Find Hinge Force of Rod Undergoing Circular Motion

[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