Register to reply

Mass whirling on table pulled by string through center

by PhizKid
Tags: mass, pulled, string, table, whirling
Share this thread:
voko
#19
Feb4-13, 12:28 AM
Thanks
P: 5,687
Quote Quote by ehild View Post
It would be useful to read http://www.ecourses.ou.edu/cgi-bin/e....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
#20
Feb4-13, 12:59 AM
HW Helper
Thanks
P: 10,383
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
#21
Feb4-13, 01:13 AM
Thanks
P: 5,687
Quote Quote by ehild View Post
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
#22
Feb4-13, 03:08 AM
HW Helper
Thanks
P: 10,383
PhizKid said that the azimuthal component of acceleration was zero, that was, why he did not include that term in the integral.


ehild
ehild
#23
Feb4-13, 03:45 AM
HW Helper
Thanks
P: 10,383
Quote Quote by PhizKid View Post
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 [itex]a_{\theta}=2\dot {r} \dot {\theta}+r \ddot {\theta}=\frac{1}{r} \frac{d}{dt}\left(r^2\dot {\theta}\right)=0[/itex],so [itex]\frac{d}{dt} \left(r^2\dot {\theta}\right)=0 [/itex]→ [itex] r^2\dot {\theta}=const[/itex] which you can know without learning about angular momentum. (You only happen to derive its conservation in central force field :) )

ehild
voko
#24
Feb4-13, 05:28 AM
Thanks
P: 5,687
Quote Quote by ehild View Post
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
#25
Feb4-13, 08:00 AM
HW Helper
Thanks
P: 10,383
Quote Quote by voko View Post
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
#26
Feb4-13, 09:06 AM
Thanks
P: 5,687
Quote Quote by ehild View Post
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
#27
Feb4-13, 09:36 AM
HW Helper
P: 7,049
Quote Quote by PhizKid View Post
The change in kinetic energy is given by [itex]\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}[/itex].
Quote Quote by voko View Post
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.

Quote Quote by rcgldr View Post
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:

http://www.physicsforums.com/showthread.php?t=328121
voko
#28
Feb4-13, 11:28 AM
Thanks
P: 5,687
Quote Quote by rcgldr View Post
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
#29
Feb4-13, 11:43 AM
HW Helper
Thanks
P: 10,383
Quote Quote by voko View Post
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
#30
Feb4-13, 11:56 AM
PhizKid's Avatar
P: 471
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
#31
Feb4-13, 02:19 PM
HW Helper
P: 7,049
Quote Quote by rcgldr View Post
I don't see any vector notation in PhizKid posts.
Quote Quote by voko View Post
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
#32
Feb4-13, 03:31 PM
HW Helper
Thanks
P: 10,383
Quote Quote by PhizKid View Post
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
#33
Feb5-13, 03:45 AM
Thanks
P: 5,687
Quote Quote by rcgldr View Post
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
#34
Feb5-13, 05:14 AM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,690
Quote Quote by voko View Post
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_c...Coriolis_terms.
rcgldr
#35
Feb5-13, 08:11 AM
HW Helper
P: 7,049
Quote Quote by voko View Post
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
#36
Feb6-13, 02:20 AM
Thanks
P: 5,687
Quote Quote by vela View Post
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_c...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.


Register to reply

Related Discussions
Whirling a mass on a string General Physics 2
Puzzled about circular motion (whirling a bung on a string) General Physics 21
Related: Center of Mass, Torque, and Tension on a String. Find θ, m1, and m2. Introductory Physics Homework 3
String tension with an attached whirling mass Introductory Physics Homework 1
Object on whirling string Introductory Physics Homework 1