Moment of inertia changes during rotation—Calculate work?

[Gian_ni]

Hi everyone, i have a question

Moment of inertia changes during rotation. Calculate the work done that changes kinetic energy?
Angular moment (along the axis of rotation) L = I * w
A point mass M rotates along an axis attached to a mass-negligible rod, of length r.
If someone moves the mass M at distance r / 2, the angular moment must conserve ( so
L1 = I2 w2 -> w2 = 4w1) , but kinetic energy is changed: ΔK = 0.5M (w2 ^ 2 * (r / 2) - w1 ^ 2 * r) = 0.5M * w1 ^ 2 * 7r
Since the work performed by the internal force (?) has increased, ΔK = W is positive.

- But what force in this case did the work and during which displacement?
- Is there a way to calculate the Work W without the work-energy theorem? Calculations?

Thank you

The direction of the force as it spirals inward when changing r may be centipetal but i have doubts.. also in the final position the point is in a fixed distance ( r/2 ) so there must be another force that stops it at that fixed distance from the axis...

 

[jbriggs444]

But what force in this case did the work and during which displacement?

As I visualize the problem, we have a rod. One end of the rod is anchored perpendicular to an axle about which it is free to rotate. There is a mass that is otherwise free to slide up and down the rod. But there is an internal mechanism that can "crank the mass in" or "crank the mass out".

 

[Gian_ni]

there is an internal mechanism

There is an internal mechanism or a person move the mass to the r/2 distance, the problem i do not understand is how to calculate that Work without the work-energy theorem but with the definition of Work=Force* displacement or better
W= ∫F*ds ?

 

[jbriggs444]

There is an internal mechanism or a person move the mass to the r/2 distance, the problem i do not understand is how to calculate that Work without the work-energy theorem but with the definition of Work=Force* displacement or better
W= ∫F*ds ?

Sure, taking the integral works.

Assume that the mass is cranked in slowly. It all works out the same if you crank it in rapidly, but then you have to worry about the initial inward acceleration and the final outward acceleration. It's easier to pick a simple path to integrate over with negligible [additional] radial acceleration throughout.

Now notice that all the tangential movement during the cranking in process is irrelevant. You are integrating the dot product of radial force times the incremental path element traversed along a segment of the spiral path.

Can you compute the required radial force to crank the mass in slowly when it is currently at radius x?

 

[haruspex]

internal force

"Someone" moves it inwards. It is an external force. An internal force would do no work on the system.

The direction of the force as it spirals inward when changing r may be centipetal but i have doubts..

What doubts?

there must be another force that stops it at that fixed distance from the axis...

No, you only have to reduce the force applied below the centripetal force needed to maintain it at the radius it has reached. That will be enough to slow its inward migration.

negligible radial acceleration throughout.

Negligible additional radial acceleration.

 

[jbriggs444]

It is an external force. An internal force would do no work on the system.

That is not correct. An internal force can most certainly do work on a system. It cannot do "center-of-mass" work, but it can do work.

Edit: Note that the distinction between an "internal" force where one point of action is on a portion of the structure that is already anchored in place by an external attachment and an "external" force where the point of action is simply external is subtle to the point where it is a distinction without a difference. One can readily adopt the viewpoint of @haruspex and treat the force as external.

Negligible additional radial acceleration.

Yes, thank you for that correction.

 

[haruspex]

That is not correct. An internal force can most certainly do work on a system. It cannot do "center-of-mass" work, but it can do work.

Yes, thank you for that correction.

And thankyou for yours.

 

[Dr.D]

Wouldn't a picture help here?

 

[Gian_ni]

Thank you for your replies

If i calculate the work of the centripetal force F necessary to set the point mass at r/2 distance it gives W = Fr/2 , but in this case W =/= ΔK = 0.5M * w1 ^ 2 * 7r
:/

 

[jbriggs444]

Thank you for your replies
If i calculate the work of the centripetal force F necessary to set the point mass at r/2 distance it gives W = Fr/2 , but in this case W =/= ΔK = 0.5M * w1 ^ 2 * 7r
:/

Your plan in post #3 was to integrate. What happened to that plan?

 

[Gian_ni]

W= ∫F*ds
I think F is centripetal so F ds sin(π/2) = F ds, W= ∫F*ds=W= ∫Fds . I don't know the intensity of that force though, but if F is constant W= F∫ds= F r/2

 

[jbriggs444]

W= ∫F*ds
I think F is centripetal so F ds sin(π/2) = F ds, W= ∫F*ds=W= ∫Fds.

The path you are integrating over... You are considering that to be a purely radial path? So you are adopting the rotating frame of reference in which the rod is stationary? That's a fine choice -- I just want to be sure that we are on the same page.

Now draw a free body diagram. In this frame, the mass has zero acceleration. It is subject to an unknown real centripetal force F which is being provided by the mechanism. This force F is what we wish to determine.

Since the mass is subject to one non-zero force and is not accelerating, there must be another force acting on it.

What other force acts on the mass?

 

[Gian_ni]

The path you are integrating over... You are considering that to be a purely radial path? So you are adopting the rotating frame of reference in which the rod is stationary? That's a fine choice -- I just want to be sure that we are on the same page.

Now draw a free body diagram. In this frame, the mass has zero acceleration. It is subject to an unknown real centripetal force F which is being provided by the mechanism. This force F is what we wish to determine.

Since the mass is subject to one non-zero force and is not accelerating, there must be another force acting on it.

What other force acts on the mass?

In the rotating frame of reference on the mass acts that force F provided by the mechanism and perhaps friction force but that would be an external force... :-/

 

[jbriggs444]

In the rotating frame of reference on the mass acts that force F provided by the mechanism and perhaps friction force but that would be an external force... :-/

No friction. Note the key word "rotating frame". What kind of forces can you think of that are associated with rotating frames?

If you prefer we can switch back to the inertial frame and discuss a spiral trajectory instead.

 

[Gian_ni]

No friction. Note the key word "rotating frame". What kind of forces can you think of that are associated with rotating frames?

If you prefer we can switch back to the inertial frame and discuss a spiral trajectory instead.

It acts the centripetal force of intensity mv ^ 2 / r towards the center and the force provided by the mechanism
Sorry I don't have any further idea...

 

[jbriggs444]

It acts the centripetal force of intensity mv ^ 2 / r towards the center and the force provided by the mechanism

I may be misunderstanding. You said that you did not know how to calculate F. But here you correctly calculate it as $\frac{mv^2}{r}$.

That's perfect. Now all we need to do is to express that r and that v in terms of "s". Where "s" is your current displacement along the path.

The use of "v" instead of "$\omega$" is a bit troublesome if we've adopted the rotating frame -- where velocity is purely radial and is slow. But let's shift back to the inertial frame for a bit and see what we can do.

Let R be the full length of the rod and V be the velocity of the mass when it is at radius R.
In order to conserve angular momentum, what must the velocity, v, of the mass be when it is at radius r instead?

$v(r) =\ ?$

 

[Gian_ni]

the velocity, v, of the mass be when it is at radius r

The velocity, v, of the mass when it is at radius r is v(r)=wr ,
w = constant angular velocity because it's a uniform circular motion

 

[jbriggs444]

The velocity, v, of the mass when it is at radius r is v(r)=wr ,
w = constant angular velocity because it's a uniform circular motion

No, it's not.

Let me rephrase... I interpret the problem to be speaking of a rod on a frictionless axle so that there can be no external torque. Conservation of angular momentum demands that $\omega$ not be constant.

 

[Gian_ni]

No, it's not.

Let me rephrase... I interpret the problem to be speaking of a rod on a frictionless axle so that there can be no external torque. Conservation of angular momentum demands that $\omega$ not be constant.

Thank you
So when I changes for the conservation of angular momentum:
m * r^2 * w1 = m * x^2 * wf where x = r final, wf = w final --> v( rf=x )=w*r= w1* r^2 / x ?

 

[jbriggs444]

Thank you
So when I changes for the conservation of angular momentum:
m * r^2 * w1 = m * x^2 * wf where x = r final, wf = w final --> v( rf=x )=w*r= w1* r^2 / x ?

Sounds right. There are various ways in which that could also be expressed. I'll adopt your notation.

Since $v(r) = \omega_1 r$, it follows that $\omega_1 = \frac{v(r)}{r}$. We can substitute that into
$$v(x) = \omega_1\frac{r^2}{x}$$ to get $$v(x) = v(r)\frac{r}{x}$$

One could obtain the same result from Kepler's "equal areas in equal times" premise or from the definition of angular momentum as $\vec{r} \times m\vec{v}$.

With that in mind, can you now compute the centripetal force F at a distance x from the axis?

 

[Gian_ni]

Sounds right. There are various ways in which that could also be expressed. I'll adopt your notation.

Since $v(r) = \omega_1 r$, it follows that $\omega_1 = \frac{v(r)}{r}$. We can substitute that into
$$v(x) = \omega_1\frac{r^2}{x}$$ to get $$v(x) = v(r)\frac{r}{x}$$

One could obtain the same result from Kepler's "equal areas in equal times" premise or from the definition of angular momentum as $\vec{r} \times m\vec{v}$.

With that in mind, can you now compute the centripetal force F at a distance x from the axis?

In my question there was an error: ΔK = 0.5M (w2 ^ 2 * (r / 2)^2 - w1 ^ 2 * r^2) = 0.5M * 3(w1*r) ^ 2

Yes from v(x) = v(r) * r / x, F = M * v(x)^2 / x , ∫ from r to r/2 W = ∫Fdx = ∫ - M * v(r)^2 * r^2 / x^3 dx = - M * v(r)^2 * r^2 ∫ x^-3 dx = - M * v(r)^2 * r^2 * - 0.5 (4/r^2-1/r^2) = 0.5M * 3(w1*r) ^ 2
This works! In this case this Work comes from the Internal energy of the mechanism or from what ''source''?

 

[jbriggs444]

This works!

Good.

In this case this Work comes from the Internal energy of the mechanism or from what ''source''?

It could be almost anything. Perhaps one has a motor turning a pinion on a rack-and-pinion arrangement powered by a battery. Perhaps one is riding the contraption and cranks in on a fishing line. Perhaps one runs a hydraulic pump attached to a double-acting hydraulic cylinder using energy harvested from a solar cell. One common arrangement for presenting a similar problem is to put the mass on top of a frictionless table and attach it to a string that runs down through a hole in the table top. Some one or some thing pulls on the string and you attempt to reason about the resulting motion of the mass.

If the force is internal... then the key thing that must hold is that the two points of action for that force (for instance, the place where you sit on the contraption and the place where the other end of your fishing line hooks on) must be in relative motion. In order for energy to be supplied, the force on each end must act in the direction of that end's motion relative to the other end. So this could be tension if the points are moving together. Or an expansion force if they are moving apart.

Note that a common way of harvesting work energy from an internal force is to put a mixture of gasoline and air into a cylinder and ignite the mixture with a spark plug. That's an example of an expansion force with the points of action moving apart.​

If the force is external... then one should first be sure that it applies no net torque. Then to supply energy, the single point of action of that force on the system must move in the direction of the applied force (for instance, the tension in the string must draw the mass in toward the hole in the center of the table top).

I hope that answers the question you were trying to ask.

 

[Gian_ni]

I hope that answers the question you were trying to ask.

Yes, very good answer. Thank you again for the help and clarity of explanation