# When to apply conservation of angular momentum?

• gracy
In summary: You can now conserve angular momentum.In summary, the conversation discusses the concept of applying conservation of angular momentum, specifically in the case of tension in a string. The key points are that in order for a torque to be produced, there must be a force with a component perpendicular to the axis of rotation, and that tension, being a force that is always radial, does not produce torque. However, if tension is increased gradually, the object will move in a spiral path and tension will have both radial and tangential components, making conservation of angular momentum invalid. Therefore, in this case, conservation of angular momentum can only be applied when tension is gradually increased and the object moves in an approximately circular path.
gracy
So my question is when to apply conservation of angular momentum?When there is no external force ,right?But in the case below
A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown.
The tension in the string is increased gradually and finally m moves in a circle of radius R0/2 . The final value of the kinetic energy is

A) mv20
B) 1/4 mv20
C) 2 mv20
D) 1/2 mv20
My teacher applied conservation of angular momentum.But I don't know why?As tension force is external force the angular momentum should not be conserved.
Note:I am not posting this question in homework sectio because I know it's solution .I don't want method to solve rather I am having a conceptual doubt.

gracy said:
So my question is when to apply conservation of angular momentum?When there is no external force ,right?
You can apply conservation of angular momentum whenever there is no external torque. The tension produces no torque about the axis.

DaleSpam said:
The tension produces no torque about the axis.
Because tension force is the only force,and for torque we require a pair of force separated by perpendicular distance,right?

Torque requires you to choose an origin
gracy said:
Because tension force is the only force,and for torque we require a pair of force separated by perpendicular distance,right?
Nop.

In magnitude the torque of a given force about certain point is given by
##
\mathbf{\tau}=rFsin\theta##

where r measured to point in question. In this case choose to measure r about the center of the circle, I mean choose the center of the circle as the origin of the coordinates.

In this case r is the radius of the circle, note that the tension and r are parallel, so the angle is 0 so the sin(theta)=0. Then, the torque associated to the tension is null, therefore the angular momentum is conserved.

gracy
gracy said:
Because tension force is the only force,and for torque we require a pair of force separated by perpendicular distance,right?

The force of tension in the string always points perpendicular to the direction the object is moving, so its component in the tangential direction is zero, and thus it does no torque. In order for a force to exert torque on an object, the component of the force in the tangential direction has to be non-zero.

Drakkith said:
The force of tension in the string always points perpendicular to the direction the object is moving,

This is not true, in this example the motion have a tangential and a radial component.

Drakkith said:
... so its component in the tangential direction is zero.

This is true and is the reason there is no torque.

andresB said:
This is not true, in this example the motion have a tangential and a radial component.

Yes, but the force from tension only has a radial component, right?

Drakkith said:
Yes, but the force from tension only has a radial component, right?
Yes this is correct. But the object also has a radial component of speed (it is not moving in a circle, it is spiraling inward) and so the tension is not perpendicular to the motion.
This is why the energy is larger at the end of the problem: the tension did some work.

Roger.

gracy said:
we require a pair of force separated by perpendicular distance
This pair of force is called "couple",I think it is required in calculation of moment.
andresB said:
In magnitude the torque of a given force about certain point is given by
τ=rFsinθ \mathbf{\tau}=rFsin\theta
So,torque doesn't need two forces.It just needs a force and some angle theta with r which is distance between point of application of force and axis of rotation.

gracy said:
Because tension force is the only force,and for torque we require a pair of force separated by perpendicular distance,right?
A torque requires a force that has a component perpendicular to the axis of rotation some distance from the axis of rotation. In this case, the tension force is through the axis of rotation and therefore it's torque is 0.

The important information given in the question is that tension is increased gradually . The increasing tension causes the block to move closer to axis of rotation.

The block then moves towards the center in a spiral path , such that tension has both radial and tangential components , of which the radial component can produce a torque making conservation of angular momentum invalid .

But because tension increases gradually the block moves towards the center in an approximately circular path and tension will have only a tangential component . Tension doesn't produce torque .

You can now conserve angular momentum .

Qwertywerty said:
The block then moves towards the center in a spiral path , such that tension has both radial and tangential components ,
The tension in the string is always purely radial, towards the center. You might mean normal and tangential components, perpendicular and parallel to the blocks velocity.

Qwertywerty said:
of which the radial component can produce a torque
The string tension cannot produce any torque around the center, no matter how you decompose it into components. The torques from the two components will simply cancel.

Qwertywerty said:
But because tension increases gradually the block moves towards the center in an approximately circular path and tension will have only a tangential component .
Now you seem to confuse tangential with radial.

Qwertywerty said:
Tension doesn't produce torque .
Yes, because it is only radial. No matter how quickly you increase it.

gracy

## 1. What is conservation of angular momentum?

Conservation of angular momentum is a fundamental principle in physics that states that the total angular momentum of a system remains constant unless an external torque (a force that causes rotation) is applied.

## 2. When should conservation of angular momentum be applied?

Conservation of angular momentum should be applied whenever there is rotational motion involved in a system, such as the spinning of a top or the orbiting of planets around a star.

## 3. How is conservation of angular momentum calculated?

Conservation of angular momentum is calculated using the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

## 4. What are some real-life applications of conservation of angular momentum?

Conservation of angular momentum has many real-life applications, including the stability of satellites in orbit, the motion of spinning gyroscopes, and the behavior of figure skaters during spins.

## 5. Are there any exceptions to the conservation of angular momentum?

In most cases, conservation of angular momentum holds true. However, there are some exceptions, such as when external torques are applied to a system or when there is friction present, which can cause a decrease in angular momentum.

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