Pendulum: Kinetic vs Potential Energy

A swinging pendulum is the textbook demonstration of conservation of mechanical energy. At the top of each swing, all the energy is potential (gravitational, $mgh$). At the bottom, all of it is kinetic ($\tfrac{1}{2}mv^{2}$). In between, the two trade off — but their sum is always constant (in the absence of friction).

The Setup

Take the lowest point of the swing as the reference height ($h = 0$). For a pendulum of length $L$ released from rest at angle $\theta_0$, the height at any angle $\theta$ is:

$h(\theta) = L(1 - \cos\theta)$

At the release point, the energy is purely potential:

$E_{\text{total}} = mgL(1 - \cos\theta_0)$

At any other point in the swing, by conservation:

$\tfrac{1}{2}mv^{2} + mgh = E_{\text{total}}$

Solving for the speed:

$v(\theta) = \sqrt{2g\bigl[h(\theta_0) - h(\theta)\bigr]} = \sqrt{2gL(\cos\theta - \cos\theta_0)}$

Step-by-Step Solution

Given: Pendulum length $L = 2\;\text{m}$, mass $m = 1\;\text{kg}$, released from $\theta_0 = 30°$, $g = 9.81\;\text{m/s}^{2}$.

Find: The speed at the bottom of the swing, the maximum height, and the energy split at $\theta = 15°$.

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Step 1 — Compute the maximum height (release point).

$h_{\max} = L(1 - \cos 30°) = 2(1 - 0.8660) = 2(0.1340) \approx 0.2679\;\text{m}$

Step 2 — Compute the total mechanical energy.

At the release point $v = 0$, so all the energy is potential:

$E_{\text{total}} = mgh_{\max} = (1)(9.81)(0.2679) \approx 2.629\;\text{J}$

This number stays the same throughout the entire swing.

Step 3 — Find the speed at the bottom of the swing.

At the lowest point, $h = 0$, so all the energy is kinetic:

$\tfrac{1}{2}mv_{\max}^{2} = 2.629$

$v_{\max} = \sqrt{\dfrac{2(2.629)}{1}} = \sqrt{5.258} \approx 2.293\;\text{m/s}$

(Equivalent: $v_{\max} = \sqrt{2gh_{\max}} = \sqrt{2(9.81)(0.2679)} \approx 2.293\;\text{m/s}$.)

Step 4 — Find the energy split halfway up, at $\theta = 15°$.

$h(15°) = 2(1 - \cos 15°) = 2(1 - 0.9659) \approx 0.0682\;\text{m}$

$\text{PE}(15°) = mgh = (1)(9.81)(0.0682) \approx 0.669\;\text{J}$

$\text{KE}(15°) = E_{\text{total}} - \text{PE}(15°) = 2.629 - 0.669 \approx 1.960\;\text{J}$

The speed there:

$v(15°) = \sqrt{2(1.960)} \approx \sqrt{3.920} \approx 1.980\;\text{m/s}$

So at the halfway angle, the pendulum has about 75% of its maximum kinetic energy (since it's mostly accelerated already) and 25% of its peak potential energy.

Step 5 — Verify conservation throughout the swing.

At $\theta = 0°$ (bottom): KE = 2.629, PE = 0, total = 2.629 ✓ At $\theta = 15°$: KE ≈ 1.960, PE ≈ 0.669, total = 2.629 ✓ At $\theta = 30°$ (top): KE = 0, PE = 2.629, total = 2.629 ✓

The total is constant at every point — that's mechanical energy conservation in action.

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Answer: The pendulum has total mechanical energy $E \approx 2.629\;\text{J}$, reaches a maximum speed of $\boxed{v_{\max} \approx 2.293\;\text{m/s}}$ at the bottom of its swing, and at any angle $\theta < 30°$:

$\text{KE}(\theta) = mgL(\cos\theta - \cos\theta_0), \qquad \text{PE}(\theta) = mgL(1 - \cos\theta)$

The two add up to the same constant total at every point.

Try It

• Watch the bar graph on the right side — KE (cyan) and PE (pink) trade off as the pendulum swings.
• The green total bar stays the same height — that's energy conservation.
• Adjust the release angle — bigger angles give higher peak energy and faster bottom speeds.
• Adjust the length — longer pendulums swing more slowly but reach higher absolute heights.