The Indeterminacy of Problems with Multiple Friction Interfaces

[brotherbobby]

Homework Statement

The problem consists of a bar of weight $W$ to which a sphere of also of weight $W$ is fitted at one end. The conbination is released from rest, with the bar horizontal. How will the combination fall if air resistance can be neglected?

Relevant Equations

$c.g. = c.m.$ (in constant gravity)

For convenience, I have marked the C.G. of the sphere as $G_S$ and that of the bar only as $G_B$. The C.G. of the combination is halfway between them, shown by the green $G_C$. My answer is that the combination will rotate about $G_C$ (centre of the bar) as it falls.

I wonder if I am right.

Many thanks.

 

[PeroK]

Problem Statement: The problem consists of a bar of weight $W$ to which a sphere of also of weight $W$ is fitted at one end. The conbination is released from rest, with the bar horizontal. How will the combination fall if air resistance can be neglected?
Relevant Equations: $c.g. = c.m.$ (in constant gravity)

View attachment 244214
For convenience, I have marked the C.G. of the sphere as $G_S$ and that of the bar only as $G_B$. The C.G. of the combination is halfway between them, shown by the green $G_C$. My answer is that the combination will rotate about $G_C$ (centre of the bar) as it falls.

I wonder if I am right.

Many thanks.

What would happen if the bar and the ball were separated and dropped independently?

 

[brotherbobby]

Both would fall freely without rotating about their respective centers of gravity.

 

[PeroK]

Both would fall freely without rotating about their respective centers of gravity.

But, the ball faster than the bar?

 

[brotherbobby]

No, both would fall identically, independently of their masses (i.e. even if the ball was heavier/lighter than the bar)

 

[PeroK]

No, both would fall identically, independently of their masses (i.e. even if the ball was heavier/lighter than the bar)

But, if you stick them together, then the ball falls faster?

 

[brotherbobby]

No. If the ball fell faster, they would not remain stuck.

 

[PeroK]

No. If the ball fell faster, they would not remain stuck.

Or, the combination would rotate.

 

[brotherbobby]

Or, the combination would rotate.

If the ball and the bar fell identically (even if their masses were unequal - Galileo's principle), they should fall the same way if they were glued. The tendency to rotate would exist if there was a torque on the system due to the combined weight. I don't understand how can that be.

 

[PeroK]

If the ball and the bar fell identically (even if their masses were unequal - Galileo's principle), they should fall the same way if they were glued. The tendency to rotate would exist if there was a torque on the system due to the combined weight. I don't understand how can that be.

Perhaps I misunderstood your OP. I thought you said that the system would rotate. (Presumable anti-clockwise in your diagram). In which case, the ball would hit the ground first.

I don't understand how the system can rotate unless one component is falling faster than another.

 

[PeroK]

My answer is that the combination will rotate about $G_C$ (centre of the bar) as it falls.

To give you a straight answer: the system will not rotate. There is no external torque on any free-falling object (symmetric or asymmetric) due to the force of gravity.

 

[brotherbobby]

Perhaps I misunderstood your OP. I thought you said that the system would rotate. (Presumable anti-clockwise in your diagram). In which case, the ball would hit the ground first.

I don't understand how the system can rotate unless one component is falling faster than another.

You were right - indeed I thought it wuld rotate (counterclockwise). Your careful questioning made me reconsider that.

 

[PeroK]

You were right - indeed I thought it wuld rotate (counterclockwise). Your careful questioning made me reconsider that.

It's easy to go wrong with torque. If you are interested I can post something on this?

 

[brotherbobby]

It's easy to go wrong with torque. If you are interested I can post something on this?

Yes please do so and many thanks. I suppose we haven't had the last word on this. Rotational motion is far from straightforward.

 

[brotherbobby]

To give you a straight answer: the system will not rotate. There is no external torque on any free-falling object (symmetric or asymmetric) due to the force of gravity.

Yes I suppose that's true. But if it is, imagine a rod inclined at $60 ^{\circ}$ to the vertical and then dropped. Would it fall as it is without rotating?

 

[PeroK]

Yes please do so and many thanks. I suppose we haven't had the last word on this. Rotational motion is far from straightforward.

Imagine any object in free fall. A sphere will do. All the force can be considered through the centre.

Look at the left edge of the sphere. There is a torque about this point of $mgr$, where $r$ is the radius of the sphere. So, the sphere should rotate about the left edge?

But, look at the right edge. There is also a torque about that point, so the sphere should rotate about the right edge as well?

Of course, the sphere does not rotate as it falls, so how do you explain the torque?

Hint: when are Newton's laws not directly valid?

 

[PeroK]

Yes I suppose that's true. But if it is, imagine a rod inclined at $60 ^{\circ}$ to the vertical and then dropped. Would it fall as it is without rotating?

Yes. The acceleration due to gravity is equal for all points on a body.

 

[brotherbobby]

Imagine any object in free fall. A sphere will do. All the force can be considered through the centre.

Look at the left edge of the sphere. There is a torque about this point of $mgr$, where $r$ is the radius of the sphere. So, the sphere should rotate about the left edge?

But, look at the right edge. There is also a torque about that point, so the sphere should rotate about the right edge as well?

Of course, the sphere does not rotate as it falls, so how do you explain the torque?

Hint: when are Newton's laws not directly valid?

$m$ being the mass of the left edge?

 

[PeroK]

$m$ being the mass of the left edge?

No. The whole sphere. All the mass of the sphere is to the right of the left edge. So there should be a torque about the left edge.

 

[brotherbobby]

No. The whole sphere. All the mass of the sphere is to the right of the left edge. So there should be a torque about the left edge.

True. One way to answer that is to see both the left and the right edges. Surely the sphere can't rotate both ways, which would explain that the sphere has to fall down straight without rotating.

But the other way you suggested is more interesting. Newton's (second) law is valid for an extended body only when we consider the net external force ($\Sigma \vec{F}$) and the resulting acceleration of the center of mass ($\vec{a}_{cm}$). As for its rotational analogue : $\vec{\tau} = I_{cm} \vec\alpha$. There would be an angular acceleration only about the center of mass provided there was a torque acting on it. But gravity acts at the cm, and therefore fails to provide any torque about it.

Is this an instance where Newton's law is not generally valid?

 

[PeroK]

True. One way to answer that is to see both the left and the right edges. Surely the sphere can't rotate both ways, which would explain that the sphere has to fall down straight without rotating.

But the other way you suggested is more interesting. Newton's (second) law is valid for an extended body only when we consider the net external force ($\Sigma \vec{F}$) and the resulting acceleration of the center of mass ($\vec{a}_{cm}$). As for its rotational analogue : $\vec{\tau} = I_{cm} \vec\alpha$. There would be an angular acceleration only about the center of mass provided there was a torque acting on it. But gravity acts at the cm, and therefore fails to provide any torque about it.

Is this an instance where Newton's law is not generally valid?

The key point is that the ball is accelerating. So, the physical left edge provides an accelerating reference frame.

To use Newton's laws in an accelerating reference frame you need to add an additional "fictitious" or "inertial" force. It's this force that cancels out the apparent torque.

The moral is to be very careful applying torque about points on an accelerating object: e.g. an object in free fall or rolling down a slope. It's safer to look at torque about a fixed point in space.

 

[brotherbobby]

The key point is that the ball is accelerating. So, the physical left edge provides an accelerating reference frame.

To use Newton's laws in an accelerating reference frame you need to add an additional "fictitious" or "inertial" force. It's this force that cancels out the apparent torque.

The moral is to be very careful applying torque about points on an accelerating object: e.g. an object in free fall or rolling down a slope. It's safer to look at torque about a fixed point in space.

"It's this force that cancels out the apparent torque. ". You mean the torque due to the fictitious force cancels out the torque due to gravity taking the left edge as a the point of rotation?

 

[PeroK]

"It's this force that cancels out the apparent torque. ". You mean the torque due to the fictitious force cancels out the torque due to gravity taking the left edge as a the point of rotation?

Yes, exactly.

 

[brotherbobby]

Yes, exactly.

Yes got you. I am currently solving bodies in equlibrium (statics) and they are far from easy. I suppose, for a thorough approach, I'd have to go over to see how mechanical engineers handle ideas of torques and forces that collectively keep a body at rest.

Which reminds me : am sure you must have heard that the ordinary ladder problem cannot solved by rigid body considerations if we imagine the "wall to be rough" along with the floor. I have verified that indeed it cannot be solved. What is the key behind the problem them? I am sure, since both walls and floors have friction, that a ladder remains at rest under the combined action of the friction at both surfaces (along with the normal reactions at them).

 

[PeroK]

Yes got you. I am currently solving bodies in equlibrium (statics) and they are far from easy. I suppose, for a thorough approach, I'd have to go over to see how mechanical engineers handle ideas of torques and forces that collectively keep a body at rest.

Which reminds me : am sure you must have heard that the ordinary ladder problem cannot solved by rigid body considerations if we imagine the "wall to be rough" along with the floor. I have verified that indeed it cannot be solved. What is the key behind the problem them? I am sure, since both walls and floors have friction, that a ladder remains at rest under the combined action of the friction at both surfaces (along with the normal reactions at them).

You've got me there. All I know is that there are some statics problems where these forces are indeterminate.

 

[brotherbobby]

You've got me there. All I know is that there are some statics problems where these forces are indeterminate.

So it is here. There is no way to find the four forces (two frictions, two normal reactions), given the weight and length of the resting ladder along with the angle at which it does.

It's a bit baffling : why not? But I suppose you should work out the problem for convince yourself that it cannot be done to begin with. Try changing the point of rotation (from bottom to top rest edge of the ladder) and you'd find you're no wiser.

 

[haruspex]

Which reminds me : am sure you must have heard that the ordinary ladder problem cannot solved by rigid body considerations if we imagine the "wall to be rough" along with the floor. I have verified that indeed it cannot be solved. What is the key behind the problem them? I am sure, since both walls and floors have friction, that a ladder remains at rest under the combined action of the friction at both surfaces (along with the normal reactions at them).

Problems where more than one interface provides friction can be indeterminate because they depend on history, i.e. how set-up was created.

Consider a right angled apex in a ceiling, so each side slopes at 45 degrees. I place a ball weight W in the apex from below and let go - it falls of course. But what if I jam it into the apex to create a pair of normal forces, N, and the coefficient of static friction is $\mu$?
It will stay in place if $W<2N\mu/\sqrt 2-2N/\sqrt 2$. This is possible if $\mu>1$ and I jam it hard enough.

Likewise with the ladder, how the forces achieve balance when I walk up it might depend on whether my heavier friend walked up and down it first.