Mass whirling on table pulled by string through center

[PhizKid]

Homework Statement

Homework Equations

Energy

The Attempt at a Solution

So I'm not exactly sure what this question even wants considering the statement is true by the work - energy theorem. Does it want me to show it explicitly for this case? If so, then, since there are no forces present along $\hat{\theta }$, $W = m\int_{a}^{b}<\ddot{r} - r\dot{\theta }^{2}, 0>\cdot <dr,d\theta > = m\int_{l_{1}}^{l_{2}}\ddot{r} - r\dot{\theta }^{2}dr$ so $W = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} - \int_{l_{1}}^{l_{2}}mr\dot{\theta }^{2}dr$. The change in kinetic energy is given by $\Delta T = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} + \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}$. If we choose a coordinate system centered around the hole, then the position vector to the mass is parallel to the net force so the torque is identically zero thus angular momentum is conserved so $r^{2}\dot{\theta }$ is a constant of motion which implies $r^{4}\dot{\theta }^{2}$ is also a constant so $-\int_{l_1}^{l_2}mr\dot{\theta }^{2}dr = -\int_{l_1}^{l_2}m\frac{r^{4}\dot{\theta }^{2}}{r^{3}}dr =-mr^{4}\dot{\theta }^{2}\int_{l_1}^{l_2}\frac{dr}{r^{3}} = \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}$ so $W = \Delta T$ but I'm not sure if my solution is valid and I'm also not satisfied with this long winded solution, if it is correct, because the chapter this problem is from was before angular momentum was introduced so there must be a solution that doesn't use the concept but I can't think of it so if anyone could guide me towards that it would be great. Thanks!

 

[voko]

"Slowly" means that you can ignore the radial velocity and the acceleration related to the change in the radial velocity. Under this assumption, what is the acceleration of the mass?

 

[PhizKid]

Why can we assume all those things? Can you give a mathematically precise reason? Seems like we are assuming a lot just from the word slowly. Anyways if the radial acceleration is indeed small then $\ddot{r} << r\dot{\theta }^{2}$ so that $a_{r} = -r\dot{\theta }^{2} + O(\frac{\ddot{r}}{r\dot{\theta }^{2}})$. Also, could you please tell me if my solution above was correct or not? Thanks

 

[voko]

Why can we assume all those things? Can you give a mathematically precise reason?

There is no "mathematically precise" reason, it is a physical simplification.

Also, could you please tell me if my solution above was correct or not?

Your derivation does not take the Coriolis effect into account, which is precisely what the "slowly" simplification eliminates.

 

[PhizKid]

I don't see why we care about the Coriolis forces. They show up in the angular component of the force which is zero here so it doesn't even show up in the line integral for work due to the dot product. Also I am really not convinced about the whole "slowly" approximation without being able to see how to express that mathematically. Otherwise it seems like hand waving to me.

 

[rcgldr]

If you know that angular momentum is conserved, then you'd also know that the final energy state of the whirling mass is independent of the path. The net work related to radial acceleration and deceleration would be zero, since the whirling mass begins and ends with zero radial speed, and its angular momentum is conserved.

So it seems that for this problem, the "slowly" pulled in means you ignore the work related to radial acceleration and deceleration, rather than relying on knowledge of conservation of angular momentum, or in effect, deriving the equivalent.

 

[PhizKid]

But I don't get where my solution went wrong. I don't see how Coriolis forces would factor into the line integral the way I did it. The dot product takes care of that. Thanks for the response rcgldr!

 

[voko]

In your derivation of work, the acceleration vector is constant in direction. This is only possible when the vector is taken in a co-rotating frame. In a co-rotating frame, you have to take the Coriolis effect into account - unless you can assume that the velocity is "slow".

From a purely mathematical point of view, a lot of physics may seem like hand waving. Which is OK, because physics is not mathematics. Learn the difference.

 

[PhizKid]

If you can't quantify a result mathematically, approximation or not, then why should I accept it as fact? You're literally translating a simple phrase in English to a powerful physical assumption. There's a difference between being given an explicit approximation and taking something out of that and loosely interpreting an english word to mathematical results without justifying it mathematically. Anyways, can you point out where in the work integral it was assumed the direction of the radial acceleration was constant? Do you mean to say the direction of the radial acceleration is not the same as the direction of the radial displacement? Thanks.

 

[ehild]

The problem asks the work of the person who pulls the string W=∫Fdl. You calculated the work of the string on the mass, which is the same now, I did not find any error in your derivation. You only did not take the condition "pulling slowly" into account.

"Pulling slowly" means that you pull with a force which is almost equal to the centripetal force, it can be taken equal to it. So F=mrω². That means a constant radial speed, so the radial speed does not contribute to the change of KE.
Conservation of angular momentum is necessary to take into account.

ehild

 

[voko]

If you can't quantify a result mathematically, approximation or not, then why should I accept it as fact? You're literally translating a simple phrase in English to a powerful physical assumption. There's a difference between being given an explicit approximation and taking something out of that and loosely interpreting an english word to mathematical results without justifying it mathematically. Anyways, can you point out where in the work integral it was assumed the direction of the radial acceleration was constant? Do you mean to say the direction of the radial acceleration is not the same as the direction of the radial displacement? Thanks.

The acceleration vector you had was $(\ddot{r} - r\theta^2, 0)$. Its direction is constant. Which is only possible in a co-rotating frame. In which case there must be a Coriolis term.

You are free to debate whether "slow" means "absence of radial velocity", but then you must take it into account.

 

[PhizKid]

Hi ehild. Thanks for the response. So are you saying if I take into account the radial acceleration not contributing any work then I will get the same answer I have now except the change in .5m(dr/dt)^2 from l1 to l2 terms will go away in the work expression? Why will the term go away in the kinetic energy expression where radial velocity shows up without reference to the radial acceleration? Also, if I don't take the approximation into account, is my final expression/ solution for work and for kinetic energy still correct? I just don't get where any reference to Coriolis forces come up. Thanks!

 

[ehild]

Your derivation is correct proving that the work done on the mass is equal to the change of kinetic energy in general, but the problem asks to derive it when the radial velocity is "slow" ; that excludes the term connected to the change of the radial velocity from the KE. And I think the problem wants you to present the result in the usual form of the KE: 1/2 mv²,so expand the last expression and give the work in terms of the initial and final speeds.

The Coriolis force is irrelevant here.
Centrifugal force and Coriolis force are inertial forces, fictitious forces, experienced in a rotating frame of reference. You observe the motion from a rest frame.

ehild

 

[voko]

You observe the motion from a rest frame.

He may be observing it from a rest frame, but his derivation is an co-rotating frame.

 

[ehild]

He may be observing it from a rest frame, but his derivation is an co-rotating frame.

No. Why do you think so?

ehild

 

[voko]

No. Why do you think so?

See #11.

 

[ehild]

It is the radial acceleration written in terms of polar coordinates in a rest frame of reference. It would be useful to read http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=dy&chap_sec=01.6&page=theory, foe example.

ehild

 

[voko]

$(\ddot{r} - r\theta^2, 0)$ is a vector. Vectors are invariant. You can't have one and the same vector have constant and non-constant direction just because you change coordinates.

 

[voko]

It would be useful to read http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=dy&chap_sec=01.6&page=theory, foe example.

Yes, it would be useful. The article stresses multiple times that "the coordinate system is moving", and the final formula for acceleration includes, among other things, the Coriolis acceleration.

 

[ehild]

A coordinate system is not the same as the frame of reference.

Describing a motion with a polar coordinate system, there are both radial and azimuthal components of acceleration. The azimuthal component includes both the "Coriolis term" and the angular acceleration. As the string can exert only radial force, the azimuthal component of acceleration is zero. That does not mean that the acceleration or the force has constant direction as you said in #11. It means that the force is "central", acts along the radius, and the acceleration is radial.

ehild

 

[voko]

A coordinate system is not the same as the frame of reference.

Which is way a vector is either rotating or not, independently of the coordinate system.

The expression given for the acceleration vector implies its direction is constant. Which should not be the case for a rotating mass in an inertial frame.

Consequently, the expression is an non-inertial frame.

That does not mean that the acceleration or the force has constant direction as you said in #11.

I did not say that. PhysKid's expression for the acceleration vector implies that. Which is so obvious that I do not even understand why we are having this discussion.

The article that you linked has the correct expression for acceleration at the end. It should just be taken from there.

 

[ehild]

PhizKid said that the azimuthal component of acceleration was zero, that was, why he did not include that term in the integral. ehild

 

[ehild]

but I'm not sure if my solution is valid and I'm also not satisfied with this long winded solution, if it is correct, because the chapter this problem is from was before angular momentum was introduced so there must be a solution that doesn't use the concept but I can't think of it so if anyone could guide me towards that it would be great. Thanks!

You said that the azimuthal component of acceleration must be zero as the force is central. The azimuthal component is $a_{\theta}=2\dot {r} \dot {\theta}+r \ddot {\theta}=\frac{1}{r} \frac{d}{dt}\left(r^2\dot {\theta}\right)=0$,so $\frac{d}{dt} \left(r^2\dot {\theta}\right)=0$→ $r^2\dot {\theta}=const$ which you can know without learning about angular momentum. (You only happen to derive its conservation in central force field :) )

ehild

 

[voko]

PhizKid said that the azimuthal component of acceleration was zero, that was, why he did not include that term in the integral.

Which has no physical basis, unless one assumes that the motion is planar and that the rope is a straight line. And why should that be assumed? The whole point of this discussion is that PhizKid believes that "slowly" has no physical significance, while in fact that's why those assumptions could be made.

 

[ehild]

Which has no physical basis, unless one assumes that the motion is planar and that the rope is a straight line. And why should that be assumed?

The mass moves on a plane, pulled by a string. A table is usually plane and a stressed massless string is straight.

ehild

 

[voko]

The mass moves on a plane, pulled by a string. A table is usually plane and a stressed massless string is straight.

You are missing my point entirely. PhizKid's derivation has flaws. One of those is using the rotating frame of reference and assuming that tangential acceleration is absent there. This assumption must either be proved or reduced to another assumption, such as "pulling slowly".

"Pulling slowly" means that the motion is circular and uniform at any instant, which automatically proves the assumption. Whether this is the case with any other kind of pulling must be proved.

 

[rcgldr]

The change in kinetic energy is given by $\Delta T = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} + \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}$.

In your derivation of work, the acceleration vector is constant in direction.

I don't see any vector notation in PhizKid posts. ΔT represents the integral for tension in the string from radius L1 to L2 (force x distance). In his equations, the absolute direction of the tension doesn't matter, only that it's radial.

If you know that angular momentum is conserved, then you'd also know that the final energy state of the whirling mass is independent of the path. The net work related to radial acceleration and deceleration would be zero, since the whirling mass begins and ends with zero radial speed, and its angular momentum is conserved.

Since angular momentum is conserved, than at any time when the radial speed is zero (such as the initial and final states), the tension is proportional to (1/r)^3. This is shown in posts #3 and #7 of the previous thread linked to below. The comment about no net work related to radial acceleration (path doesn't matter) wasn't mentioned in that previous thread, so I've quoted my previous post from this thread that explains that aspect:

https://www.physicsforums.com/showthread.php?t=328121

 

[voko]

I don't see any vector notation in PhizKid posts.

The very first equation, the work integral, is a scalar product of two vectors.

 

[ehild]

You are missing my point entirely. PhizKid's derivation has flaws. One of those is using the rotating frame of reference and assuming that tangential acceleration is absent there. This assumption must either be proved or reduced to another assumption, such as "pulling slowly".

It is not assumed that the tangential acceleration is absent. You mix it with the azimuthal acceleration in the polar coordinate system, which is zero, as the force is radial.

The tangential and the centripetal acceleration are defined in the instantaneous system of coordinates, moving together with the particle along its trajectory. The velocity points in the direction tangent to the trajectory, and the acceleration can have both tangential and normal components. The tangential acceleration is equal to the time derivative of the speed. The normal (centripetal) component of acceleration is v² /R where R is the curvature of the trajectory.

In this problem, the trajectory is a spiral. The acceleration has components both tangential and normal to the path .
In polar coordinates, the velocity has both radial and azimuthal components, but the acceleration has only radial component, as the force is radial. ehild

 

[PhizKid]

Oh gosh I'm so confused. I don't know if my solution is right or not. I never took into account the slow approximation so does that void it's correctness? Or does applying the slow approximation simply get rid of the radial velocity terms in my final solution? I'm not seeing any reason to introduce Coriolis forces into this and I have no idea where voko got the idea that I was working in a frame co - rotating with the mass so that its trajectory in such a frame would be purely radial (this is what would happen in such a frame correct? If someone could answer that even though its unrelated =D). Thanks a ton!

 

[rcgldr]

I don't see any vector notation in PhizKid posts.

The very first equation, the work integral, is a scalar product of two vectors.

OK, but that dot product <... , 0> . <dr, dθ>, eliminates the vector and the direction (θ) component (it's multiplied by zero) and the result is a scalar function based on the radius and speed of the whirling mass, and the radial acceleration, none of which are dependent on direction (θ). For this problem, there's no need to consider a rotating frame or the associated fictitious centrifugal and coriolis effects.

Getting back to the issue of the work related to radial acceleration, there should be a way to show that since the initial and final state of the whirling mass has no radial speed, the net work related to radial acceleration is zero. This could be done by showing that angular momentum is conserved, since the time derivative of angular momentum is torque, and in this case, the torque is zero. Once it's shown that angular momentum is conserved (and that the initial and final state of the whirling mass has no radial speed), then it doesn't matter if the string is pulled slowly or quickly, the net work done by radial acceleration and deceleration is zero.

wiki article on angular momentum (this may help):

http://en.wikipedia.org/wiki/Angular_momentum

 

[ehild]

Oh gosh I'm so confused. I don't know if my solution is right or not. I never took into account the slow approximation so does that void it's correctness? Or does applying the slow approximation simply get rid of the radial velocity terms in my final solution? I'm not seeing any reason to introduce Coriolis forces into this and I have no idea where voko got the idea that I was working in a frame co - rotating with the mass so that its trajectory in such a frame would be purely radial (this is what would happen in such a frame correct? If someone could answer that even though its unrelated =D). Thanks a ton!

Have you read my posts #10 and #23?

Your derivation is correct, the work of the net force is equal to the change of KE, and you just arrived to that result. But I think you should expand the expression for KE and give it in terms of the initial and final speeds. And ignore radial speeds as that was said in the problem.
You are right, you work in inertial frame of reference, using polar coordinates. No centrifugal force, no Coriolis force, no other pseudo-forces. In a frame of reference, rotating together with the mass it would move in radial direction.

ehild

 

[voko]

OK, but that dot product <... , 0> . <dr, dθ>, eliminates the vector and the direction (θ) component (it's multiplied by zero) and the result is a scalar function based on the radius and speed of the whirling mass, and the radial acceleration, none of which are dependent on direction (θ). For this problem, there's no need to consider a rotating frame or the associated fictitious centrifugal and coriolis effects.

I am not sure why my point is not understood. I said: (A) the acceleration vector is expanded in a rotating frame, which is manifest in the equation; (B) there is no analysis in the solution why that expansion is valid and why, for example, the Coriolis effect may be ignored.

The statement, by you and ehild, that the associated fictitious forces can be ignored entirely, is manifestly wrong, because the $r\dot{\theta}^2$ component, which is retained in the analysis, is the centrifugal term.

 

[vela]

I am not sure why my point is not understood. I said: (A) the acceleration vector is expanded in a rotating frame, which is manifest in the equation; (B) there is no analysis in the solution why that expansion is valid and why, for example, the Coriolis effect may be ignored.

Your point is not understood because, to put it bluntly, you're wrong, which is what the others have been trying to convince you of. In the original post, the notation simply means $\vec{a} = a_r \hat{r} + a_\theta \hat{\theta}$ where the unit vectors are functions of r and θ. Since r and θ vary with time, the direction of the unit vectors do as well, so your assertion that PhizKid is saying the acceleration's direction does not change isn't correct.

Also, note that in his expression for the radial component of the acceleration, the $r\dot{\theta}^2$ term appears with a minus sign. It corresponds to the centripetal acceleration, not the centrifugal acceleration. In a rotating frame, that term moves to the other side of F=ma and corresponds to the centrifugal force. See, for example, http://en.wikipedia.org/wiki/Polar_coordinate_system#Centrifugal_and_Coriolis_terms.

 

[rcgldr]

The statement, by you and ehild, that the associated fictitious forces can be ignored entirely, is manifestly wrong, because the $r\dot{\theta}^2$ component, which is retained in the analysis, is the centrifugal term.

As mentioned in the above post, it's the centripetal force term (minus sign). For anyone reading this thread that may not have caught this, converting from polar to cartesian coordinates: $- r\dot{\theta}^2 = - v^2 / r$.