0
$\begingroup$

I am trying to understand conservation linear momentum with in angular momentum

Herewith i just explain what i am looking for. Imagine an object of mass(M), velocity(V), is undergoing circular motion
with radius(R) Angular momentum L = RMV (R-radius M-mass- V- tangential velocity)

Mass is constant here

Since L is conserved so when R is decreased tangential velocity(V) has to be increased to keep angular momentum constant.

Does this mean linear momentum not conserved when radius of rotation changes?

When linear momentum is not conserved then what is the force that increases the velocity of object when R decreases.

Could anyone clarify this. or could any one say is my basic understanding about momentum is wrong. for easy understanding i am only considered magnitude of momentum's not cross product of vectors

$\endgroup$
  • $\begingroup$ You cannot change the radius without applying an unbalanced force. The momentum is changing legitly without breaking any law. $\endgroup$ – Yashas Nov 11 '16 at 5:35
1
$\begingroup$

Two ways of answering your question.

Consider the mass as the system which is acted on by a gravitational force - a central force.
The central force does not apply a torque on the mass and this means that the angular momentum of the mass is constant.
However even if the radius stays the same the linear momentum of the mass changes (in direction) because there is a net force on the mass.
However the speed and the kinetic energy of the mass does not change because the displacement of the mass is at right angle to the line of action of the force and so the force does no work on the mass.

If the radius of the orbit decreases then to conserve angular momentum the mass starts to move faster and so there is an increase in the kinetic energy and the magnitude of the linear momentum of the mass.
This comes about because as the radius of the orbit is decreased work is done on the mass by the central force because there is a displacement of the mass along the line of action of that force.
That work done increases the kinetic energy of the mass and the magnitude of the linear momentum of the mass.

Now consider as the system your mass and another mass which is providing the central force on your mass.
N3L tells you that there must also be a force on the other mass due to your mass which is equal and opposite.

Again angular momentum is conserved because there are no external torques acting on the two masses but this time the angular momentum is carried by both masses because they rotate about their common centre of mass.
The linear momentum of the two masses also does not change because there are no external force acting on the two masses.
Any change in linear momentum of one mass is compensated for by an equal and opposite change in the linear momentum of the other mass.
The equal and opposite forces on the two masses act for exactly the same tine on the two masses.
If the separation of the two masses decreases then both masses have more kinetic energy that energy coming from the decrease in the gravitational potential energy of the two masses.

It is often the case that the mass of one object is much greater than the other and the assumption is made that the orbit of the less massive object is centre at the centre of mass of the more massive object. This can simplify calculations without much loss os accuracy.

$\endgroup$
2
$\begingroup$

The thing you are overlooking is that something else must be applying an equal and opposite force in order to pull something into the circle you mention.

Consider the case of two astronauts each holding an end of a rope in between them. If they are both moving in a circle around their common center of mass, then when they pull the rope in, their rotational motion increases, the momentum of each increases in magnitude, but the total momentum stays the same.

$\endgroup$
  • $\begingroup$ Not clear how the momentum of each astronaut increases but the total momentum stays the same. $\endgroup$ – sammy gerbil Jun 1 '18 at 4:38
  • $\begingroup$ The magnitude of the vector for the momentum increases, but each momentum vector has the opposite direction of the other momentum vector, giving us a vector sum that's zero. So, for instance, if one astronaut has a momentum of (300, -400) kg m/s and the other astronaut has a momentum of (-300, 400) kg m/s, their total momentum is (0, 0), but each astronaut's momentum has a magnitude of 500 kg m/s. $\endgroup$ – David Elm Jun 4 '18 at 14:27
  • 1
    $\begingroup$ Sorry, I see what you mean now. $\endgroup$ – sammy gerbil Jun 4 '18 at 14:34
2
$\begingroup$

It is simple to study conservation rules in the center of mass system. In the center of mass system of two rotating bodies if the attraction is increased or decreased the balance of angular momentum and momentum will correspondingly go to both bodies.

In a system of two planets orbiting each other in space and a third body impacts on one of them, the equations for gravitational attraction have to be solved including the extra momentum and energy entering the two body system.

It is not simple to change the radius of a rotating body on a surface, because energy has to be supplied and with it momentum. These have to be taken into account in the conservation equations, so your statement

Since L is conserved so when R is decreased tangential velocity(V) has to be increased to keep angular momentum constant.

is not accurate. It needs specific conditions.

$\endgroup$
  • $\begingroup$ Both links are broken $\endgroup$ – try-catch-finally Oct 3 at 15:44
  • $\begingroup$ @try-catch-finally thanks. I removed them $\endgroup$ – anna v Oct 3 at 18:05
0
$\begingroup$

I'm not quite sure what you mean by

Since L is conserved so when R is decreased tangential velocity(V) has to be increased...

In circular motion, the distance from the orbiter to the center is constant.

Let's say, it's an elliptical orbit, e.g. a planetary motion around a star. Then, it is true. The velocity of the planet is higher when it's closer to the star, and it's slower when it's farther away. However, in an elliptical orbit, the angular momentum is not conserved all the time, as the direction of the gravitational force is not always perpendicular to the direction of the tangential velocity, which would create torque. Similarly, the linear momentum is also not conserved, as force exists, which would change momentum.

$\endgroup$
  • $\begingroup$ L conserved means object Rotating at radius R1 will have one angular momentum. When the same object radius of rotation is decreased it will have new radius of rotation say R2. $\endgroup$ – Manikandan C Nov 11 '16 at 5:39
  • $\begingroup$ L conserved means: object Rotating at radius R1 will have one angular momentum. When the same object's radius of rotation is decreased it will have new radius of rotation say R2. but angular momentum will not change that is conserved. however the tangential velocity has to be increased (Ref :L=RMV) to keep L constant. Now how velocity will increase with out any force. or does the force applied perpendicular to object to move it inwards will change its linear momentum $\endgroup$ – Manikandan C Nov 11 '16 at 5:46
  • $\begingroup$ Angular momentum L is only conserved where there is no net torque on a body about a point. For circular motion, force is perpendicular to velocity, so there is no net torque about the center, so L is conserved. If you have two circular motions, one with R1, L1 and one with R2, L2, the L1 is conserved, and L2 is conserved, but L1 does not necessarily equal L2. $\endgroup$ – Vibius Vibidius Zosimus Nov 11 '16 at 6:13
  • $\begingroup$ I have watched online video about angular momentum, where the author of that video claims that L is constant(L1=L2) for R1 and R2. i have attached the link to the video here, at time t=7.40sec in the video author claims this statement. Could you please go through this video and say is his statement correct. it is a 11minute video link: khanacademy.org/science/physics/torque-angular-momentum/… $\endgroup$ – Manikandan C Nov 11 '16 at 8:18
-3
$\begingroup$

The statement in the Khan video is false; the tangential velocity (or linear Newtonian velocity) of a mass wrapping around a post will not increase. A pin can be set to interrupt the string of a rotating mass; When the pin forces the r to shorten the tangential velocity does not change. Angular momentum (in the lab) is not a conserved quantity.

$\endgroup$
  • 1
    $\begingroup$ "Angular momentum (in the lab) is not a conserved quantity." I've never heard that before. Can you back that statement up? $\endgroup$ – JMac Dec 27 '17 at 19:36
  • $\begingroup$ Yes: An object rotating on the end of a string; can and will wrap naturally around a post. It needs no outside force; only its own momentum. The string becomes shorter without the application of outside force. Some say the linear or tangent velocity increases; That would mean the energy increases. Your rules say this is impossible. I know velocity does not increase by doing experiments. $\endgroup$ – Delburt Phend Dec 27 '17 at 20:03

protected by Qmechanic Dec 27 '17 at 21:03

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.