Why position times mass is equal to 0 here (vanishes)?

In summary, the term ##\sum_i m_ir_i## always vanishes because it is defined as the position of the center of mass relative to an arbitrary origin, and by definition, the center of mass is at the origin. This is true for all rigid bodies, regardless of the position of the origin.
  • #1
Istiak
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Homework Statement
Why position times mass is 0 inside Sigma summation?
Relevant Equations
F=ma
$$\sum_i m_ir_i$$

Why the term always vanishes?
Screenshot from 2021-08-23 00-39-56.png
There's some more equations where the mr was vanished. But, they didn't explain why it vanish. Why the term vanish? I think that's for position 0,isn't it?
 

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  • #2
If the origin is taken as the centre of mass, then the sum is zero by definition.
 
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  • #3
PeroK said:
If the origin is taken as the centre of mass, then the sum is zero by definition.
I think the origin is at an arbitrary position. Vector ##\vec R## is the position of the CM of a rigid body relative to the origin. This looks like part of the proof that the angular momentum of a rigid body is the sum of the angular momentum of the CM (orbital) and about the CM (spin). The primed coordinate ##\vec {r_i}'## is the location of element ##dm_i## relative to the CM, i.e. ##\vec {r_i}'=\vec r-\vec R##. Otherwise yes, the sums are zero by definition of the CM.
Istiakshovon said:
There's some more equations where the mr was vanished. But, they didn't explain why it vanish. Why the term vanish?
Define ##\vec R## as the position of the CM relative to an arbitrary origin and ##\vec r## and ##\vec {r_i}'## respectively as the position of ##m_i## relaitve to the origin and relative to the CM. Then ##\vec r_i = \vec R+\vec {r_i}'## so you can write the position of the CM relative to the origin, $$\vec X=\frac{\sum_i m_i\vec r_i}{\sum_i m_i}=\frac{\sum_{i} m_i(\vec R+\vec {r_i}')}{\sum{_i} m_i}=\vec R+\frac{\sum{_i} m_i\vec {r_i}'}{\sum{_i} m_i}.$$ By definition, ##\vec X = \vec R##. The only way for this to be true is to have ##\dfrac{\sum_i m_i\vec {r_i}'}{\sum_i m_i}=0.##
 

FAQ: Why position times mass is equal to 0 here (vanishes)?

Why is the product of position and mass equal to 0 in this scenario?

This phenomenon occurs when an object's position is at the origin, or when it has no displacement from its starting point. In this case, the object has no potential energy and therefore does not contribute to the overall energy of the system. This results in the product of position and mass being equal to 0.

How does the concept of potential energy relate to the vanishing of position times mass?

Potential energy is a measure of an object's ability to do work based on its position in a system. When an object's position is at the origin, it has no potential energy and therefore does not contribute to the overall energy of the system. This leads to the product of position and mass being equal to 0.

Is this phenomenon only applicable to objects at the origin?

No, this phenomenon can occur at any point where an object's position is constant and not changing. For example, if an object is suspended in mid-air and not moving, its position would be considered constant and the product of position and mass would be equal to 0.

How does this concept apply to the laws of motion?

The vanishing of position times mass is a result of the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed. In this case, the object's position and mass do not contribute to the overall energy of the system, therefore it vanishes.

Can this phenomenon be observed in real-world situations?

Yes, this phenomenon can be observed in various real-world situations, such as a pendulum at its highest point, an object at rest on a surface, or a stationary object suspended in mid-air. In all of these cases, the object's position is constant and therefore the product of position and mass is equal to 0.

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