Go Back   Physics Help Forum > College/University Physics Help > Advanced Mechanics

Advanced Mechanics Advanced Mechanics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Aug 30th 2012, 10:03 PM   #1
Member
 
Join Date: Jul 2012
Posts: 67
angular momentum prob

Two point masses, with masses of m and 2m, are connected through a string that has a length of L. The two bodies are put on an horizontal frictionless table as in the figure, so that m is at the origin. The body 2m is above it (on the y axis) at a distance of L/2 m. At a certain moment they give 2m a velocity of v0 in the positive x axis direction.

What's the kinetic energy of the system after the string is streched and m starts to move?

I realize that conservation of both linear and angular momentum applies here - but I don't understand something: Am I suppose to understand that the system starts to rotate about an axis which passes through the center of mass just as the string is streched? And if so - was there any "mathmatical way" for me to know the system starts rotating? Because it seems from the solution that when they calculated the angular momentum of the system as the string was streched - they used Icm*w to describe its angular momentum, while when they calculated the angular momentum of the system at the beginning (The picture phase) they used L=mvrsin(q) to describe the angular momentum as the sum of the angular momentum of 2m and m with respect to the center of mass...
Attached Thumbnails
angular momentum prob-clipboard23.jpg  
assaftolko is offline   Reply With Quote
Old Aug 31st 2012, 08:43 AM   #2
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,322
You asked about the kinetic energy of the system, so I don't understand why you're trying to find angular momentum. Conservation of energy says that the initial KE is 1/2 mv_0^2, and so that should be the system's KE just as m starts to move (at that point the string hasn't stretched, so there's no PE in it yet).

Last edited by ChipB; Aug 31st 2012 at 09:17 AM.
ChipB is online now   Reply With Quote
Old Aug 31st 2012, 09:01 AM   #3
Member
 
Join Date: Jul 2012
Posts: 67
Originally Posted by ChipB View Post
You asked about the kinetic energy of the system, so I don't understand why you're trying to find angular momentum. Conservation of energy says that the initial KE is 1/2 mv_0^2, and so that should be the system's KE just as m starts to move (at that point the strung hasn't stretched, so there's no PE in it yet).
According to the solution there's no conservation of mechanical energy. I don't fully understand the reason because as you said the string isn't streched yet but according to their solution, that follows KE=0.5MVcm^2+0.5Icm*w^2, you get in the final stage that KE=0.75mv0^2, while the inital is mvo^2 (notice that 2m has the speed of vo and not m like you calculated)

edit: the question asks us to give the KE after the string is streched and m starts to move from the origin. What happens is that just when the string becomes streched and is not loose anymore - a kind of plastic collision occures between the two bodies that now move "together" and thats why theres no conservation of energy

Last edited by assaftolko; Sep 1st 2012 at 12:27 AM.
assaftolko is offline   Reply With Quote
Old Sep 1st 2012, 09:05 AM   #4
Member
 
Join Date: Jul 2012
Posts: 67
Chip don't give up on me
assaftolko is offline   Reply With Quote
Old Sep 2nd 2012, 06:35 AM   #5
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,322
It seems as if the solution you posted assumes that the string somehow becomes rigid once it reaches full extension, and hence the system starts to rotate about its center of mass as it moves. This would indeed be a form of inelastic collison, and hence KE is not constant. But as the problem is written, I don't see how this could happen. A string is assumed to always be flexible. In this problem the impulse received by the mass m is equal and opposite to the impulse provided by the 2m mass, and consequently the m mass starts to move faster than the 2m mass. The result is the string goes slack as the smaller mass starts to catch up to the larger. During this phase both objects are moving along straight paths and there is no angular momentum. So again, I don't get it. Sorry.

Last edited by ChipB; Sep 2nd 2012 at 08:50 AM.
ChipB is online now   Reply With Quote
Old Sep 2nd 2012, 06:50 AM   #6
Member
 
Join Date: Jul 2012
Posts: 67
Smile

Originally Posted by ChipB View Post
It seems as if the solution you posted assumes that the string somehow becomes rigid once it eaches full extension, and hence the system starts to rotate about its center of mass as it moves. This would indeed be a form of inelastic collison, and hence KE is not constant. But as the problem is written, I don't see how this could happen. A string is assumed to always be flexible. In this problem the impulse received by the mass m is equal and opposite to the impulse provided by the 2m mass, and consequently the m mass starts to move faster than the 2m mass. The result is the string goes slack as the smaller mass starts to catch up to the larger. During his phase both objects are moving along straight paths and there is no angular momentum. So again, I don't get it. Sorry.
Angular momentum is a quantity that exist even where there's no rotational movement - like in the start phase: 2m has the velocity of v0 and so it has angular momentum, so I don't see why you are saying there's no angular momentum...

You are correct that as the string streches the body that we get from 2m, m and the string is now considered as rigid - I think since this question appears in the dynamic of rigid body chapter, they assumed I'd understand it by myself. About what you said about the string - I always thought we assumed the string is not flexible so it doesn't get shorter or longer if you apply force on it...

By the way how do you know the system starts to rotate about it's center of mass? isn't it possible that the rotation axis will pass through another point than the cm?

Also chip I've posted another "logic" question that deals with conservation of angular momentum with respect to the cm while the cm position changes in space - I'd be very grateful if you could find the time to look at it :-)
assaftolko is offline   Reply With Quote
Old Sep 2nd 2012, 09:15 AM   #7
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,322
Originally Posted by assaftolko View Post
Angular momentum is a quantity that exist even where there's no rotational movement - like in the start phase: 2m has the velocity of v0 and so it has angular momentum, so I don't see why you are saying there's no angular momentum...
Angular momentum exists only if the object is moving in a curved path. If it is moving in a straight line it has linear momentum, but not angular momentum. An essential difference is that momentum is a vector, which for linear momentum is in the direction of motion (p= mv) and for angular momentun is in the direction w, which is perpendicular to the arc of the object (p=Iw).

Originally Posted by assaftolko
About what you said about the string - I always thought we assumed the string is not flexible so it doesn't get shorter or longer if you apply force on it...
I meant that the string can bend, and hence cannot carry any torque or sideways force to either object. I did not mean that the string can stretch.

Originally Posted by assaftolko
By the way how do you know the system starts to rotate about it's center of mass? isn't it possible that the rotation axis will pass through another point than the cm?
An unconstrained object has motion components consisting of translation of its cm plus rotation about the cm. If there are constraints involved (such as rotation about a pivot point) then the center of rotation is controlled by the constraint and would not necessarily be the center of mass.

Last edited by ChipB; Sep 4th 2012 at 12:36 PM.
ChipB is online now   Reply With Quote
Old Sep 2nd 2012, 10:47 PM   #8
Member
 
Join Date: Jul 2012
Posts: 67
Originally Posted by ChipB View Post
Angular momentum exists only if the object is moving in a curved path. If it is moving in a straight line it has linear momentum, but not angular momentum. An essential difference is that momentumis a vector, which for linear momentum is in the direction of motion (p= mv) and for angular momentun is in the direction w, which is perpendicular to the arc of the object (p=Iw).



I meant that the string can bend, and hence cannot carry any torque or siways force to either object. I did not mean that the string can stretch.



An unconstrained object has motion components consisting of translation of its cm plus rotation about the cm. If there a constraints involved (such as rotation about a pivot point) then the center of rotation is controlled by the constraint and would not necessarily be the center of mass.
Chip thanks a lot! I did read that angular momentum has to do with rotational movement - that's an understatement, but so why does the system have angular momentum in the start phase where 2m moves in a straight line along the x direction?
assaftolko is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Advanced Mechanics

Tags
angular, momentum, prob



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Another Angular Momentum Prob assaftolko Advanced Mechanics 12 Sep 18th 2012 11:46 AM
angular momentum Marley1503 Kinematics and Dynamics 2 Dec 5th 2009 12:41 PM
angular momentum amiv4 Advanced Mechanics 1 Mar 25th 2009 11:20 PM
angular momentum khumri64 Advanced Mechanics 0 Mar 5th 2009 10:39 PM
help with angular acceleration prob alexito01 Advanced Mechanics 1 Oct 15th 2008 11:30 AM


Facebook Twitter Google+ RSS Feed