Mass whirling on table pulled by string through center

[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: $- r\dot{\theta}^2 = - v^2 / r$.

 

[voko]

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.

The article referenced, particularly the text under the "Co-rotating frame" heading, basically re-iterates what I have been saying all along. I understand that you can move one term in an equation from one side to another, flipping its sign as you go, and that you call that term different names depending on which side of the equation it is on, but I fail to see how that contradicts anything I have said.

Anyway, it seems the OP has long lost interest in this discussion, so I won't debate this any longer.