# Falling stick(s) - initial force not equal to gravity?

1. Jun 21, 2013

### dreamLord

Suppose we have two sticks held in a vertical plane, pivoted at one point (check the diagram). If we let go, why is the initial acceleration of the center of the mass not equal to g? Why does the pivot exert some force at the beginning?

By contrast, if we keep the sticks in a horizontal plane, the initial acceleration is g.

This is a solved problem in Introduction to Classical Mechanics by Morin (8.11) where he has shown the proof quantitatively ; however I cannot understand the qualitative aspect of it.

The stick of length b is massless, and is glued perpendicular to the stick of length l.

PS : It is NOT a homework question! As I said, it is a solved question in Morin.

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2. Jun 21, 2013

### jbriggs444

As the system falls, the point where stick b touches the platform is fixed. That constraint means that the system must rotate as it falls.

If you consider an axis of rotation at the center of mass, that means that there is a change in angular momentum due to the rotation. That means that there must be an external torque. That torque cannot be from gravity since gravity aligns with the chosen axis. It must be from the point where b touches the platform.

Since stick b is horizontal, the torque must arise from a force with a vertical component. A vertical external force will affect the net vertical acceleration of the system.

If the moment of inertia of the system about the selected axis of rotation is zero, the required external torque is zero and the associated vertical external force on stick b is zero. So the key to the conundrum is that changing the orientation of the sticks changes the moment of inertia of the system about the chosen axis of rotation.

3. Jun 21, 2013

I'm a bit confused by this as well. Regardless of if we place stick L horizontally or vertically, won't there be rotation of the system as the stick falls down? Is the difference that the setup with stick L being horizontal has no initial rotation and simply free falls initially whereas the setup with stick L being vertical does have an initial rotation on top of being pulled down by gravity?

Also, am I right in assuming that since the stick b is massless, if there is some vertical pivot force at the endpoint where it is glued to the ground, this pivot force must be balanced out by other forces acting on stick b so as to not give it infinite acceleration?

4. Jun 21, 2013

### jbriggs444

Yes. The system must rotate in either case.

I do not follow this. In both cases the initial condition (starting from rest) is with no rotation and no vertical movement. In both cases there must be rotational acceleration about a horizontal axis because stick L is accelerating downward and the end of stick b is not free to pass through the platform.

If stick L is of negligible thickness and is horizontal, rotation about its long axis involves negligible angular momentum and can be attained with negligible torque.

Yes. Since the mass of stick b is zero, its linear momentum must be constant and zero. F = dp/dt = 0. Or, using your reasoning, F = ma. If m is zero and a is not infinite then F = 0.

Similarly, the net torque on stick b must also be zero.

Last edited: Jun 21, 2013
5. Jun 21, 2013

Thank you for the response. I just don't get why there wouldn't be a vertical pivot force on the glued end of stick b when the stick L is placed horizontally, one which could affect the initial acceleration of the center of mass, but when the stick L is placed vertically there is a vertical pivot force on the glued end of stick b which now does affect the initial acceleration of the center of mass. It seems like in both cases the fact that the center of mass is at the least being pulled down by gravity would cause an initial pivot force upwards on that end of stick b since it is constrained to stay glued.

6. Jun 21, 2013

### jbriggs444

"When stick L is placed horizontally", I take that to mean that it is horizontal so that one end sticks out of the drawing toward us and one sticks into the drawing away from us.

7. Jun 21, 2013

Yes indeed that is how it is depicted in the drawing, in the textbook mentioned by the OP; basically rotate the vertical stick by 90 degrees. The book says that this setup would have the CM free fall initially because the pivoted end of stick b would not yet "know" that the stick L has started to fall but I fail to see why this same argument couldn't be applied to the setup in which stick L is vertical. Thanks again for responding.

8. Jun 21, 2013

### jbriggs444

I notice that you have mentioned twice that stick b is glued to the ground. That does not match the givens of the problem. The OP wrote that:

The glue joint is where b meets the stick we have been calling L, not where b meets the platform. As I understand it, the tip of stick b is resting on the platform, but is not attached.

If stick L is horizontal and is of negligible thickness then the entire assembly has a very small moment of inertia. A significant force between b and the platform would amount to a net torque that would cause a high rotation rate. A high rotation rate would result in stick b rising free of the platform. This would quickly eliminate the force between b and the platform.

We can conclude that the force between b and the platform is negligible in the horizontal case.

9. Jun 21, 2013

Oops sorry yes I meant glued at the intersection point between stick b and stick L. I get your explanation but I don't get the books' explanation which is what I think is confusing the OP as well. The book's explanation is what I mentioned in post #7.

10. Jun 22, 2013

### dreamLord

Yes, James, that is my doubt. Why does the pivot 'know' that the intersection is moving in one case, but not in the other? I don't understand.

Also, jbriggs, if I consider my axis of rotation to be the pivot point, is the moment of Inertia not the same for both situations? The distance between all the particles is the same - just in different planes. No?

11. Jun 22, 2013

### jbriggs444

By "pivot point", I assume you mean the point at the tip of stick b where it rests on the platform. Yes, you can analyze the problem from that axis.

The moment of inertia of the assembly with respect to this axis of rotation can be broken down into two components. The first component is the mass of stick L multiplied by the distance of its center of gravity from the pivot point. The second component is the moment of inertia of stick L about its own center of gravity [in the same plane of rotation]. This second component is zero in the case of a horizontal stick and non-zero in the case of a vertical stick.

If this breakdown into components does not feel right to you, you can think about it in a different way...

If stick L is vertical then the midpoint on stick L is the closest point to the pivot axis. Every other point on stick L is farther from the pivot axis. So the moment of inertia of stick L about the axis must be larger than the mass of stick L multiplied by the square of the distance from the axis to its midpoint.

If stick L is horizontal then every point on stick L is just as close to the pivot axis as the midpoint. So the moment of inertia of stick L about the axis must be identical to the mass of stick L multiplied by the square of the distance from the axis to its midpoint.

The difference in planes does make a difference.

12. Jun 22, 2013

Thanks jbriggs but it is not the mathematical derivation of the result that is confusing, the issue is essentially what dreamLord mentioned in the first sentence of post #10. It is the book's physical explanation. What does it even mean for the endpoint of stick b that is initially on the cliff to not "know" that the center of mass of stick L has started to fall and why would it not "know" in the case where stick L is horizontal but somehow "know" when stick L is vertical?

13. Jun 23, 2013

### dreamLord

Ah, I understand now about the moment of inertias being different in the 2 cases ; thanks for the two lucid explanations jbriggs.

I am still, however, at a loss as to why the pivot exerts any sort of initial force at all. If, hypothetically, gravity was turned off, would the pivot still exert an initial force?

14. Jun 23, 2013

### springwave

If gravity was hypothetically turned off, the pivot would not exert any initial force.

Think of it this way:

Assume the pivot would not exert any force, then analyze what happens to the point of contact.

When gravity is present: if the pivot does not exert any force, then the only force acting on the body is the force of gravity. (the body is in free fall). All points on the body have the same downward acceleration of g. (since gravity acts via COM and causes no torque). So the point of contact has a downward acceleration of g, and would eventually gain a downward velocity.

But the pivot would not allow this to happen, so it would exert a force in order to keep this point of contact at rest (zero acceleration). You can figure out this force by using the equations of motion and by using the constraint that the point of contact should be at rest.

15. Jun 23, 2013

### dreamLord

^Fair enough, I get that.

But doesn't the exact same reasoning apply to the horizontal case?

16. Jun 27, 2013

### dreamLord

This question is still bothering me - anyone?