---
title: "Double pendulum angular momentum, m1 = m2 = 1 kg"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:double-pendulum-angular-momentum-0-5-0-7"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/double-pendulum-angular-momentum-0-5-0-7
site: BCT Engineering
---

# Double pendulum angular momentum, m1 = m2 = 1 kg

Applied Mechanics (Dynamics) · Chapter 3: System of particles

- Asked in: 2079 Baishakh, 8 marks
- Video: [Dynamics - Finding angular momentum | Applied Mechanics 2 | double pendulum | Numerical problem](https://www.youtube.com/watch?v=qrDdf9OMakY) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/double-pendulum-angular-momentum-0-5-0-7
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

A double pendulum shown in figure oscillates in the xy plane. At the instant shown $\omega_1 = 2$ rad/s ccw and $\omega_2 = 3$ rad/s ccw. Take $a = 0.5$ m and $b = 0.7$ m. What is the angular momentum ($H_O$) at this instant of $m_1 = m_2 = 1$ kg? It is given that the lower pendulum is connected to mass $m$ by pin joint and is free to rotate about this point. [Figure: pendulum arm $a$ from O at $30^\circ$ to the x-axis (vertical, downward) carrying $m_1$; lower arm $b$ carrying $m_2$ at $60^\circ$ to the vertical; x down, y to the right.]

## Short answer

Compute position vectors from given lengths and angles, velocities via cross products ω×r, then sum m(r×v) for both masses. Angular momentum about O ≈ 3.99 k̂ kg·m²/s counter-clockwise.

## Answer

Take the origin at the fixed pivot $O$, $x$ vertically downward, $y$ to the right and $\hat k$ out of the plane ($+\hat k$ = counter-clockwise). Assume both arms lie on the same (right-hand) side of the vertical, and that $\omega_2$ is the absolute angular velocity of the lower arm (as is usual in this problem). Angular momentum of the system about $O$:

$$
\vec H_O = m_1(\vec r_1 \times \vec v_1) + m_2(\vec r_2 \times \vec v_2)
$$

#### Position vectors
Upper arm ($a=0.5$ m at $30^\circ$ to the $x$-axis), lower arm ($b=0.7$ m at $60^\circ$ to the vertical):

$$
\begin{aligned}
\vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.433\,\hat i + 0.250\,\hat j)\ \text{m}\\
\vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.350\,\hat i + 0.606\,\hat j)\ \text{m}\\
\vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.783\,\hat i + 0.856\,\hat j)\ \text{m}
\end{aligned}
$$

#### Velocities
$$
\begin{aligned}
\vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.500\,\hat i + 0.866\,\hat j)\ \text{m/s}\\
\vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.319\,\hat i + 1.916\,\hat j)\ \text{m/s}
\end{aligned}
$$

#### Angular momentum
$$
\begin{aligned}
H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 1(0.433\times 0.866 - 0.250\times -0.500) = 0.50\ \text{kg m}^2/\text{s}\\
H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 1(0.783\times 1.916 - 0.856\times -2.319) = 3.49\ \text{kg m}^2/\text{s}
\end{aligned}
$$

$$
\vec H_O = H_1 + H_2 = 3.99\,\hat k\ \text{kg m}^2/\text{s}
$$

**Answer: $\vec H_O \approx 3.99\,\hat k$ kg m²/s (counter-clockwise, about the axis perpendicular to the plane).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is H_O of the two masses obtained?
   A. m1 v1 + m2 v2
   B. H_1 only, since m2 hangs from m1
   C. Sum of m_i (r_i x v_i)
   D. (m1 + m2) r v
   Answer: C. Angular momentum about O is the sum of the moments of each mass's momentum about the fixed pivot. (Key point: Definition of H_O)

2. How is r2 found?
   A. r2 = r_2/1 only
   B. r2 = r1 + r_2/1
   C. r2 = r1 - r_2/1
   D. r2 = b only
   Answer: B. m2 is located relative to O by the upper arm vector plus the lower arm vector: (0.783, 0.856) m. (Key point: Position of m2)

3. How is v2 found?
   A. v2 = omega_2 k x r2
   B. v2 = v1
   C. v2 = omega_1 k x r_2/1
   D. v2 = v1 + omega_2 k x r_2/1
   Answer: D. v2 = v1 + omega_2 x r_2/1, taking omega_2 as absolute angular velocity of the lower arm, giving (-2.319, 1.916) m/s. (Key point: Velocity of m2)

4. What is H_O?
   A. 3.99 k kg m^2/s counter-clockwise
   B. 0.50 k kg m^2/s
   C. 3.49 k kg m^2/s
   D. -3.99 k kg m^2/s
   Answer: A. H_1 = 0.50 and H_2 = 3.49, so H_O = 3.99 k kg m^2/s. (Key point: Final result)

## Close variants set in other exams

- [Double pendulum angular momentum, m1 = 3, m2 = 4 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Adouble-pendulum-angular-momentum-0-4-0-5) — Applied Mechanics (Dynamics), chapter 3, asked 3×
- [Double pendulum angular momentum, m1 = 1, m2 = 2 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Adouble-pendulum-angular-momentum-0-5-0-6) — Applied Mechanics (Dynamics), chapter 3, asked 3×