A question regarding momentum

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In summary, the conversation discusses a system with the sum of momentum at equilibrium. The first drawing shows this equilibrium, while the second drawing shows a temporary increase in force on point A, causing an angle of 20°. When the force is stopped, the system returns to its previous position after some oscillation. This is due to the weight of the bar and its lack of perfect rigidity. The distortion of the bar can upset the balance of the system.
  • #1
Pnume
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There's something I struggle to understand.

In the picture I attached to this post , the first drawing is a system with the sum of momentum = 0 (equilibrium).

In the second drawing I increase temporarily the force on point A in order to get and angle of 20° compare to the last position and maintain the angle stable for a moment.

Why when I stop to force the angle of 20° does the system come back to the previous position (after some oscillation) since from my understading the sum of momentum is also at equilibrium?http://img199.imageshack.us/img199/7277/momentumg.th.jpg [Broken]

For information I'm not a student

Thanks for your help
 
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  • #2
A trivial point perhaps but I think you meant moments rather than momentum.Are you ignoring the weight of the bar itself or is it a non uniform bar where one of the marked forces is the weight?.Anyway, if you ignore the weight of the bar and assume it is perfectly rigid then when you push it down then the sum of the moments is still zero and it should, in theory, balance.However with a real, not perfectly rigid bar the distortion of both sides of the bar will be different to what it was previously, for example the force on the right side has a greater compression(squashing) component and the force on the left side left has a greater tension(stretching) component so these changes can upset the balance.
 
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  • #3
Dadface said:
A trivial point perhaps but I think you meant moments rather than momentum.Are you ignoring the weight of the bar itself or is it a non uniform bar where one of the marked forces is the weight?.Anyway, if you ignore the weight of the bar and assume it is perfectly rigid then when you push it down then the sum of the moments is still zero and it should, in theory, balance.However with a real, not perfectly rigid bar the distortion of both sides of the bar will be different to what it was previously, for example the force on the right side has a greater compression(squashing) component and the force on the left side left has a greater tension(stretching) component so these changes can upset the balance.

It is the moment and I was indeed ignoring the weight of the bar itself since it shouldn't have influence on the system equilibrium.

Thanks a lot for that explanation. it's very helpful.
 

1. What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is calculated as the product of an object's mass and velocity.

2. How is momentum measured?

Momentum is typically measured in units of kilogram-meters per second (kg*m/s).

3. What is the difference between momentum and velocity?

Momentum is a vector quantity that takes into account both an object's mass and its velocity, while velocity is only a measure of an object's speed and direction.

4. How is momentum conserved?

Momentum is conserved in a closed system, meaning that the total momentum before and after a collision or interaction remains constant. This is known as the law of conservation of momentum.

5. What are some real-world applications of momentum?

Momentum is an important concept in many fields, including physics, engineering, and sports. It is used to describe the motion of objects, such as cars, airplanes, and projectiles, and is also important in understanding collisions and explosions. In sports, momentum is often used to analyze the performance of athletes and their movements.

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