Angular speed of a falling rod

[dyn]

Homework Statement

A thin uniform rod of length 2a and mass m stands vertically on a frictionless horizontal table. The rod is slightly disturbed and falls down. What is its angular speed when it hits the table ?

Relevant Equations

Conservation of mechanical energy , PE = mgh , KE = (1/2)mv^2 + (1/2) Iw^2

I have seen the solution and understand it. The solution defines θ to be the angle between the falling rod and the table. It then equates initial PE to the final (PE+KE) where the final PE=0 and the final KE to be (1/2) Iω²+ (1/2) ma²ω² to finally obtain ω = √(3g/2a)

But i would like to know why method doesn't work. I took the Centre of Mass (CM) to only be acted upon by gravity so it descends in a straight line with constant velocity g , using "suvat" equations , first gives the time of descent as t = √(2a/g) which means the linear velocity of the CM when it hits is √(2ag). This gives a final linear KE which cancels with the initial PE and so gives ω=0 which is obviously wrong. What is wrong with my reasoning ?
Thanks

 

[kuruman]

The suvat equations are not applicable because the acceleration is not constant. The center of mass is not in free fall because it rotates about one end. It does not descend in a straight line. Just do the experiment with a pencil and see for yourself.

 

[Lnewqban]

The linear acceleration with which the Centre of Mass (CM) moves downwards must be less than g, because it is limited by the surface of the table and the angular inertia of the rod.
In other words, some of the potential energy that the rod had when in the vertical position had to be used in rotating the rod from vertical to horizontal before its side hit the table.

 

[dyn]

The suvat equations are not applicable because the acceleration is not constant. The center of mass is not in free fall because it rotates about one end. It does not descend in a straight line. Just do the experiment with a pencil and see for yourself.

The solution actually states that the CM descends in a straight line. The table is frictionless in this case so the rod just slides along it. I though motion could be thought of as the sum of the translation of the CM plus rotation about the CM. Is it because the normal reaction of the table acts against g thus reducing it ?

 

[kuruman]

The solution actually states that the CM descends in a straight line. The table is frictionless in this case so the rod just slides along it. I though motion could be thought of as the sum of the translation of the CM plus rotation about the CM. Is it because the normal reaction of the table acts against g thus reducing it ?

OK, I thought the rod was pivoted. Because all the forces are vertical, there is no horizontal acceleration of the center of mass. Nevertheless the acceleration of the CM is not $g$ because of the normal force. It would be $g$ if the table were not there.

 

[haruspex]

One thing that tends to get overlooked in such questions is the possibility that the rod loses contact with the table. In the specific case of this question, it does not, but I see no trivial proof.

 

[dyn]

Thanks for your replies. After thinking about this question the following questions have arisen in my mind

1 - As the table is frictionless , the rod just slides along it , so the normal force does no work. But this normal force is the force causing a torque and the angular momentum to increase. Is that right ?

2 - A different situation here ; a thin uniform rod has a frictionless pivot at its centre and rotates in a vertical plane. There are 2 forces acting ; gravity and the normal force but they both act at the CM and thus cause no torque resulting in a constant angular momentum about the pivot at all points in the vertical rotation. Is that right ?

3 - if the pivot is not frictionless there will still be no torque implying constant angular momentum. But friction should cause loss of energy resulting in angular momentum decreasing. How is this resolved ?

Thanks

 

[kuruman]

1. Angular momentum and torque about what point? For example, the angular momentum and torque about the point directly below the CM is zero. Yes, about the point directly below the CM the normal force results in a torque that increases the angular momentum about the CM but not the angular momentum of the CM.

2. It is right.

3. If the rod is just sitting there without spinning, no resolution is necessary. If you get it spinning with some initial angular speed, there will be a frictional torque at the pivot that will eventually reduce the angular momentum to zero. This is the rotational equivalent of a block on a rough floor. You leave it alone, it stays where it is. You give it a push, it slides across the floor until it stops.

 

[haruspex]

angular momentum and torque about the point directly below the CM is zero.

The normal force would have a torque about that point, surely?

 

[haruspex]

As the table is frictionless , the rod just slides along it , so the normal force does no work. But this normal force is the force causing a torque and the angular momentum to increase. Is that right ?

Yes, if you mean torque about the CM, or any point in a vertical line through it. Taking an instantaneous point of contact as the axis, the normal force has no torque about it, but mg does instead.
It is analogous to a wheel rolling down a ramp. The static friction does no work, but it does divert some of the lost GPE into rotational KE instead of linear KE.

 

[dyn]

The normal force would have a torque about that point, surely?

So the normal force produces a torque even though it does no work ?

 

[dyn]

3. If the rod is just sitting there without spinning, no resolution is necessary. If you get it spinning with some initial angular speed, there will be a frictional torque at the pivot that will eventually reduce the angular momentum to zero. This is the rotational equivalent of a block on a rough floor. You leave it alone, it stays where it is. You give it a push, it slides across the floor until it stops.

For the rod spinning about its CM the normal force and weight produce no torque. But the frictional force at the pivot also produces no torque. So zero torque equates to constant angular momentum about the CM. How is this compatible with energy dissipation leading to decreasing angular momentum ? One statement gives constant ang. mom. , the other one leads to decreasing ang. mom ?

 

[kuruman]

The normal force would have a torque about that point, surely?

Surely it would. I was thinking of something else at the time. I fixed the post and thanks for the heads up.

For the rod spinning about its CM the normal force and weight produce no torque.

Yes.

But the frictional force at the pivot also produces no torque.

Why not? The force of friction acts tangentially at some radius from the axis of rotation.

 

[dyn]

Why not? The force of friction acts tangentially at some radius from the axis of rotation.

Wouldn't the frictional force just act at the pivot , located at the CM ?

 

[haruspex]

Wouldn't the frictional force just act at the pivot , located at the CM ?

I would guess @kuruman's model is that the rod is mounted on an axle of some small radius r. Although we would commonly just think of the friction as applying an axial torque, in reality it consists of a tangential force at the surface of the axle.
An important consequence is that the magnitude of the frictional torque is proportional to the axle load. Another is that the greater the radius the greater the torque.

 

[kuruman]

I would guess @kuruman's model is that the rod is mounted on an axle of some small radius r. Although we would commonly just think of the friction as applying an axial torque, in reality it consists of a tangential force at the surface of the axle.
An important consequence is that the magnitude of the frictional torque is proportional to the axle load. Another is that the greater the radius the greater the torque.

That is my model. I cannot imagine a "point" pivot. The axle must have a non-zero radius and the force of friction acts tangentially at all points on the circumference. A mathematical point cannot rotate about itself because it has no extent.

 

[dyn]

So the normal force produces a torque even though it does no work ?

Thanks for all your replies. It's making more sense to me now. My last question is the one above ; in the case of the rod sliding on the frictionless table ; the normal force produces a torque but it does no work. Is that correct ?

 

[haruspex]

Thanks for all your replies. It's making more sense to me now. My last question is the one above ; in the case of the rod sliding on the frictionless table ; the normal force produces a torque but it does no work. Is that correct ?

Right. The point of contact of the rod does not move vertically, so the normal does no work.

 

[Lnewqban]

Thanks for your replies. After thinking about this question the following questions have arisen in my mind

1 - As the table is frictionless , the rod just slides along it , so the normal force does no work. But this normal force is the force causing a torque and the angular momentum to increase. Is that right ?
...

You are welcome.

Please, note that the normal force is just the reaction of the weight.
The lack of alignment of those parallel forces induces a moment which magnitude increases as the distance between those forces becomes greater.
Because of that, the rod is forced to rotate faster and faster (its angular acceleration can't be constant as moment increases with time) from vertical to horizontal positions.