Why does rotation happens about hinge point in a bar when net force acting on it are zero?

Question

I have asked this question previously but not got any proper answer relevant to its free body diagram.Few are saying the body will not be in translational equilibrium as a horizontal force will appear at hinge but how that force will appear because that point is stationary.Please anyone explain i am confused?

Yes this is not the fbd nor i am ignoring weight,but weight will be in vertical direction balanced by vertical hinge force and how the horizontal hinge force will develop if the rod net force is zero in horizontal direction already.I can't visualise. — HIMANSHU KUMAR, May 17 '21 at 10:17

Bob D · Accepted Answer · 2021-05-19T13:53:59.590

3

Why does rotation happens about hinge point in a bar when net force acting on it are zero?

The short answer is because there is a net torque about the hinge pin caused by two equal, opposite, and parallel forces which constitute a force-couple, or simply a "couple", resulting in no net force on the bar. From the free body diagram (FBD) of FIG 1 below, the sum of the horizontal forces is

$$\sum F_{H}=R_{H}+P-P=0$$ $$R_{H}=0$$

The sum of the moments (torques) about the hinge pin, taking counter-clockwise torque as positive is

$$\sum M=\frac{5}{6}PL-\frac{1}{4}PL=\frac{7}{12}PL$$

Notice that this same torque about the hinge could be produced by a single force $P$ applied a distance $\frac{7}{12}L$ from the pin. However, in order for there to be no net horizontal force on the bar, we would then require a horizontal reaction at the hinge of $R_{H}=-P$.

What makes a couple unique is that the moment about any point along the bar, including the COM and the hinge, is the same and equals $\frac{7}{12}PL$. For this reason a couple is called a "free vector", meaning it can be moved anywhere on the body, including the hinge, and have the same external effect on the body. See FIG 2. FIG 2 is equivalent to FIG 1 with the free moment vector moved to the hinge. To further understand why, see the accepted answer to this question: https://physics.stackexchange.com/questions/535963/force-couple-off-centre-of-mass-in-space#:~:text=2%20Answers&text=Net%20force%20is%20zero%20so,about%20the%20centre%20of%20mass.

CONCLUSION: The net torque at the hinge, assumed to be ideal (frictionless), produced by the couple (a free vector) causes the COM to rotate counter clockwise about the hinge. Because the net torque is produced by a couple, instead of a single force, there is no horizontal reaction at the hinge.

Hope this helps.

edited May 19 '21 at 13:53

answered May 18 '21 at 20:16

Bob D

71,527

So,final conclusion is that due to acceleration of center of mass in case of ideal hinge a horizontal net force will develop at the hinge and in case of real hinge a moment will develop and the door will not move. – HIMANSHU KUMAR May 19 '21 at 08:55
I have included the final conclusion in an update to my answer. Hope it helps. – Bob D May 19 '21 at 13:55
Please once again read what you are saying sir just shifting the couple forces does not change the fact that"when net force is zero the body can only rotate about its center of mass". – HIMANSHU KUMAR May 19 '21 at 15:20
@HIMANSHUKUMAR. What I am saying is the couple is a free vector than can moved anywhere along the body. But that doesn’t mean the bar can rotate anywhere along the body. That depends on whether or not the body is constrained so that rotation can ONLY occur at particular location. If the bar were not constrained by the hinge (and ignoring translational motion due to gravity) you would have pure rotation and that rotation can only occur about the COM, regardless of the fact that the mid-point between the P forces is not the COM. Instead you have pure rotation about the hinge. – Bob D May 19 '21 at 16:06
The link I gave you provides the proof. But this bar is not free to rotate about the COM because of the hinge restraint, and can only rotate about the hinge. I’m sorry, but I no more time to spend on this. Sorry it apparently hasn’t answered your question. Good luck. – Bob D May 19 '21 at 16:06
Sorry sir i think i have irritated you but the only thing i am not able to digest where is the net force which is causing the center of mass of this body to move – HIMANSHU KUMAR May 19 '21 at 18:28
@HIMANSHUKUMAR You did not irritate me. It just appears I am unable to help you. One last try: It is net torque that initiates rotation of the COM about the hinge, not a net force. Perhaps you are thinking about the centripetal force which is a net force responsible for maintaining the circular motion. The centripetal force is directed towards the center of rotation (hinge pin). For your example, ignoring gravity (as in outer space) the centripetal force on the COM is equal to: $$\frac{mv^2}{L/2}$$ Where $v$ is the speed of the COM. – Bob D May 19 '21 at 18:54

score 0 · Answer 2 · answered May 17 '21 at 07:53

0

Why won't rotation happen? That is dependent upon torque, not force. Translational and rotational equilibrium are independent of each other. Moreover, the object is not at translational equilibrium here because of because of a hinge force, the magnitude of which you can calculate this way:

Find out torque about hinge.
Using this information, find out angular acceleration of rod.
Hence, Find out acceleration of CM of rod.
Use Newton's Second law to find out horizontal component of hinge force.

answered May 17 '21 at 07:53

Ritam_Dasgupta

612
4
8

Please try to understand my question when the forces are balanced the center of mass is at rest as in this case then why would the hinge horizontal force come especially when the rod is already in equilibrium means can you tell the process how the horizontal force develops in hinge – HIMANSHU KUMAR May 17 '21 at 07:58
The forces are not balanced, you are ignoring hinge force. I already said why hinge force develops. There is torque, so com accelerates. That is impossible if forces balance. So hinge force acts. – Ritam_Dasgupta May 17 '21 at 08:00
I am ignoring because the rod is already balanced by forces so how will the force has hinge develop as this point cannot move if a force especially develop at that point a reaction force will also come by the wall – HIMANSHU KUMAR May 17 '21 at 08:03
If an unbalanced force is applied to a point it does not mean that that point has to move. It means the COM has to move. – Ritam_Dasgupta May 17 '21 at 08:15
@HIMANSHUKUMAR, equilibrium requires zero net translational force AND zero net torque. You have zero net translational force. However, your pivot point is at the hinge, not at the COM. This means that, given the definition of torque, you do NOT have zero net torque on the bar, and it will rotate counter-clockwise as a result. – David White May 17 '21 at 18:38

R.W. Bird · Answer 3 · 2021-05-21T13:35:07.730

0

Assuming the two P's represent equal horizontal forces acting at equal distances from the center of mass, then the associated torques would tend to rotate the rod about the CM without moving the CM. This rotation would require the ends of the rod to move relative to the CM. But, the upper end is constrained by the axle which would have to exert a horizontal force to prevent such motion. For rotation about the axle, the two torques are not equal.

edited May 21 '21 at 13:35

answered May 17 '21 at 14:50

R.W. Bird

12,139

Which two torques are you talking about in last line – HIMANSHU KUMAR May 17 '21 at 16:53
The torques from the two P's about the hinge. – R.W. Bird May 18 '21 at 12:57
Means you want to say that if the forces were as such that they create equal torque then the net torque would be zero about that hinge point and the rod will not rotate but here in this case the rod will rotate. – HIMANSHU KUMAR May 18 '21 at 13:51

Protein · Answer 4 · 2021-05-22T10:23:58.070

0

Imagine what would happen if the rod was not hinged, it would rotate about a the COM of rod (and not about mid point of forces (see Farcher's answer). The Hinge reaction develops to counter this rotation resulting in a net force on the COM of the rod.

The hinge reaction is the reason why the COM of the rod moves.

edited May 22 '21 at 10:23

answered May 17 '21 at 15:44

Protein

685

FYI Although intuitively it would seem that if the rod was not hinged it would rotate about the midpoint between the P forces (I used to think so too), it will, in fact, always rotate about the COM. See Farcher's answer in the link I provided. – Bob D May 19 '21 at 18:28
@Bob D Thank you very much – Protein May 22 '21 at 10:19

Rishab Navaneet · Answer 5 · 2021-05-19T19:24:10.673

the answer is pretty simple... :)

The thing is here the rod is not an isolated system. The rod is connected to a VERY MASSIVE wall through a hinge. So the net force due to $P$ equals zero means that the centre of mass of the whole system... including wall and rod remains stationary, which is true...

if the rod was isolated and free, then your argument would have been correct. It would rotate about rod's COM.

Note : By considering the reaction force from the hinge, we are essentially taking the rod out from the rod+wall system. And hence, all other answers here explain the same thing

score 0 · Answer 6 · answered May 22 '21 at 10:39

Here is a simple philosophy to understand the deal with forces and torques:

Action of forces of force vector at any point on a rigid body, given that it's in the same direction, causes the exact same acceleration but different torque. The idea is that the rotation experienced by the body due to the pushes and pulls you give it depends on the location of force.

For a more mathematical treatment, see Kleppner and Kolenkow.

Why does rotation happens about hinge point in a bar when net force acting on it are zero?

6 Answers6