ENERGY AND MOMENTUM: The Potential Energy of a Pendulum

Q: A pendulum of unknown mass ( m ) is rotated through an angle ( θ ) until it is vertically displaced by a distance ( Δh ). As a consequence, it has a gain in potential energy ( PE ) within the gravitational field that is directly proportional to its vertical displacement. If the pendulum is eventually released, what general expression can be derived and then used to determine its velocity ( v ) as it swings past its point of origin?

A: Not much information has been provided, but we have all that is needed to derive an expression that appropriately addresses the circumstance at hand. 

As the pendulum is rotated off its axis, its arm becomes the hypotenuse of a right triangle:

This is fortunate, because we now have an angle ( θ ) between the pendulum’s arm and the original y-axis along which it was suspended. This allows us to use trigonometry to determine the adjacent side of the triangle that has been formed:

cos θ = ( adj / hyp )

( hyp )( cos θ ) = adj

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hyp = Total Length ( l )

adj = Length 2

Length 2 = l cos θ

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We may now use that Principle of Conservation of Energy to determine what the pendulum’s velocity will be as it swings past its origin. The gain in potential energy within the gravitational field is a consequence of a force ( F ) working against the gravitational force of attraction over some distance ( d ):

Energy ( E ) = Work ( W ) = Fd

F = mg

d = height ( Δh )

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The value of vertical displacement moved is obtained in the following manner:

Δh = Total Length – Length 2

Δh = l – l cos θ

Δh = ( l )( 1 – cos θ )

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Thus, the pendulum’s potential energy can be determined:

ΔPE = mgΔh

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We must now relate the total energy of the system prior to releasing the pendulum with the system as the pendulum swings past the origin. The total energy is the sum of the potential energy ( PE ) and kinetic energy ( KE ) of the system. Potential energy, as discussed, is energy related to position, and kinetic energy is the energy of motion:

PE_i + KE_i = PE_f + KE_f

Since the pendulum is motionless prior to being released, its kinetic energy is originally zero. To the contrary, all of this potential energy is converted to kinetic energy in the lattermost circumstance, thus making the final potential energy value zero. This allows us to simplify the expression at hand with convenience:

PE → KE

PE_i = KE_f

All that is missing is an expression for kinetic energy:

KE = ½ mv²

Now that we have an expression for ( KE ), we set the initial ( PE ) of the system equal to the final ( KE ) therein:

mgΔh = ½ mv²

gΔh = ½ v²

2gΔh = v² 

v = √ 2gΔh

v = √ 2gl( 1 – cos θ )

Notice that the mass canceled out of the equation. This is due to the fact that mass is a measure of inertia. If the pendulum had been more ( or less ) massive, it would have fallen under the influence of a greater ( or lesser ) force; however, any amount of mass will accelerate at the same rate ( neglecting wind resistance ) as it travels from a higher to lower elevation towards the earth’s surface. If we glue two bricks together, we’ll have an object with twice the inertia that will fall right alongside a single brick, no more, and no less.

Finally, what if an appropriate kinematic equation had been used to address the question? Would the outcome be the same? Well, let’s see. Let’s use a kinematic equation that relates final velocity to the height that an object descends as it moves towards the earth’s surface:

v_yf² = v_yo² + 2ay

a = – g

Note: If the upward direction is designated as being positive, the acceleration due to gravity has a vector pointing in the opposite direction; therefore, it must be given a negative sign designation. Additionally, the y-direction of travel is treated as if it exists independent of the horizontal component of motion that the pendulum has, and y-subscripts have been assigned appropriately:

v_yf² = v_yo² – 2gy

The initial velocity of the pendulum is zero, and our equation simplifies accordingly:

v_yf² = – 2gy

It may seem as if the negative sign will give us mathematical trouble, but it doesn’t. The direction of travel is now in the ( – y ) direction:

v_yf² = ( – 2g )( – y )

v_yf² = 2gy

The conclusion we reach is satisfying in that it is consistent with the derivations already seen:

v_yf = √ 2gy

y = Δh

v_yf = √ 2gΔh