EQUILIBRIUM STATICS: Stationary and Moving Center of Mass Derivation

In physics, concepts like force, energy, and motion go hand-in-hand with one another. An unbalanced force will cause an object to accelerate. Equal and opposite force pairs will cause an object to remain at rest or maintain a constant velocity if it is already in motion. When an applied force causes an object or system to twist, the effort that transpires is call torque:

𝝉 = Fr sin θ

Mathematically speaking, torque ( 𝝉 ) is a product of a force ( F ) acting upon a lever arm ( r ). When a system is motionless despite forces being applied to it, it is in static equilibrium; thus, the sum of all torques acting upon a static structure must be zero. When interactions between a force and a lever arm occur at a 90^o angle, the torque that is exerted will be at a maximum. Consider the torques that are exerted on the balance beam below:

For the sake of simplicity, let’s assume that the masses are the same while keeping in mind that this need not be the case. Additionally, the lever arm will be represented henceforth with an ( x ) variable:

m₁ = m₂

Let’s also assume that x₂ is twice the length of x₁:

x₂ = 2x₁

We are now ready to derive a value for the center of mass ( x_cm ) of the system. A center of mass is a hypothetical location that exists between actual masses; it is a place where the mass of a system can be assumed to exist when convenient. When dealing with torques, it is customary to assign a negative ( – ) value to the clockwise direction and positive ( + ) value to the counterclockwise direction. As we will see, the center of mass in the diagram above is a distance in the -x-direction that would fall between the ( m₁ ) and ( m₂ ) masses of the balance beam if it was superimposed on the +x-axis. Notwithanding, let’s assign a positive value to the clockwise direction and designate the counterclockwise direction as being negative and see whether we can still obtain the desired derivation:

F₁x₁ + F₂x₂ – F_cmx_cm = 0

Let’s look at the diagram once again before moving forward. The masses ( m₁ ) and ( m₂ ) would cause the system to move in the clockwise ( + ) direction if the ( M_cm ) mass on the left-hand side of the balance beam is removed; however, if either ( m₁ ) or ( m₂ ) are removed from the right-hand side of the system, it would rotate in the counterclockwise ( – ) direction. Let’s now progress forward with our derivation:

F₁x₁ + F₂x₂ = F_cmx_cm

Please recall that force is a product of mass and acceleration. Within this system, the acceleration constant is the gravitational constant of acceleration ( g ) for objects that are close to the earth’s surface:

F = ma

a = g

F = mg

We are now ready to make appropriate force substitutions for our previously derived expression:

F₁ = m₁g

F₂ = m₂g

F_cm = M_cmg

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m₁gx₁ + m₂gx₂ = M_cmgx_cm

Note: The ( M_cm ) mass is equal to the sum of masses on the left side of the equation. The center of mass ( x_cm ) distance has a value somewhere between the values of the ( x₁ ) and ( x₂ ) distances in the diagram. Fortunately, the gravitational constant of acceleration can be factored out of the expression with ease:

M_cmx_cm = m₁x₁ + m₂x₂ 

x_cm = ( m₁x₁ + m₂x₂ / M_cm )

M_cm = m₁ + m₂

x_cm = ( m₁x₁ + m₂x₂ / m₁ + m₂ )

Further substitution enables us to solve for ( x_cm ) for the diagram:

m₁ = m₂

x₂ = 2x₁

x_cm = ( m₁x₁ + m₁2x₁ / 2m₁ )

x_cm = ( m₁ )( x₁ + 2x₁ / 2m₁ )

x_cm = ( m₁ )( 3x₁ / 2m₁ )

x_cm = ( 3 / 2 )( x₁ )

This answer is indeed reasonable. If ( x₁ ) is 1m and ( x₂ ) is 2m in length, the center of mass will be located at the 1.5m location on the x-axis just to the right of mass ( m₁ ). Although a balance beam was used in the derivation, any system of masses will suffice. For example, a system that consists of two objects speeding towards one another at a constant velocity ( v ) will have a center of mass between them, and it moves with a constant velocity as well. The direction in which the center of mass travels will depend on the net momentum of the system:

velocity ( v ) = meters / second

To convert that static circumstance above to one in which the masses at hand are moving, we simply divide each x-term in the expression by seconds ( s ):

v = x / s

x_cm = ( m₁x₁ + m₂x₂ / m₁ + m₂ )

v_cm = ( m₁v₁ + m₂v₂ / m₁ + m₂ )

Alas!!!

We now have a center of mass equation that can be applied to momentum problems that involve collisions.