---
title: "Double pendulum angular momentum, m1 = 1, m2 = 2 kg"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:double-pendulum-angular-momentum-0-5-0-6"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/double-pendulum-angular-momentum-0-5-0-6
site: BCT Engineering
---

# Double pendulum angular momentum, m1 = 1, m2 = 2 kg

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

- Asked in: 2073 Shrawan, 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-0-5-0-6
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

A double pendulum as shown in figure below oscillates in the X-Y plane. As shown in figure below, $W_1 = 2$ rad/sec. CCW and $W_2 = 4$ rad/sec CCW. What is $\vec H_O$ at this instant if $m_1 = 1$ kg and $m_2 = 2$ kg. The lower pendulum is connected to mass $m_1$ by a pin joint and is free to rotate about this point. [Figure: upper arm of length 0.5 m from O at $30^\circ$ to the X axis (X down, Y to the right), carrying $m_1$; lower arm 0.6 m carrying $m_2$ at $60^\circ$ to the vertical.]

## Short answer

With a=0.5 m, b=0.6 m, ω1=2 rad/s, ω2=4 rad/s, calculate positions and velocities. Angular momentum sum gives H_O ≈ 7.50 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 ($b=0.6$ m at $60^\circ$ to the vertical):

$$
\begin{aligned}
\vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.433\,\hat i + 0.250\,\hat j)\ \text{m}\\
\vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.300\,\hat i + 0.520\,\hat j)\ \text{m}\\
\vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.733\,\hat i + 0.770\,\hat j)\ \text{m}
\end{aligned}
$$

#### Velocities
$$
\begin{aligned}
\vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.500\,\hat i + 0.866\,\hat j)\ \text{m/s}\\
\vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.578\,\hat i + 2.066\,\hat j)\ \text{m/s}
\end{aligned}
$$

#### Angular momentum
$$
\begin{aligned}
H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 1(0.433\times 0.866 - 0.250\times -0.500) = 0.50\ \text{kg m}^2/\text{s}\\
H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 2(0.733\times 2.066 - 0.770\times -2.578) = 7.00\ \text{kg m}^2/\text{s}
\end{aligned}
$$

$$
\vec H_O = H_1 + H_2 = 7.50\,\hat k\ \text{kg m}^2/\text{s}
$$

**Answer: $\vec H_O \approx 7.50\,\hat k$ 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. What is the basic expression used for H_O of the two masses?
   A. H_O = m1 v1 + m2 v2
   B. H_O = m1(r1 x v1) + m2(r2 x v2)
   C. H_O = (m1 + m2) r2 x v1
   D. H_O = m2(r2/1 x v2) only
   Answer: B. Each mass contributes m(r x v) about O and the contributions are added. (Key point: System angular momentum)

2. Which is the position of m2 from O (a = 0.5 m at 30 deg, b = 0.6 m at 60 deg to the x axis as set up)?
   A. (0.733 i + 0.770 j) m
   B. (0.433 i + 0.250 j) m
   C. (0.300 i + 0.520 j) m
   D. (0.433 i + 0.520 j) m
   Answer: A. r2 = r1 + r2/1 = (0.433 + 0.300) i + (0.250 + 0.520) j. (Key point: Position vectors)

3. Why is v2 = v1 + w2 k x r2/1 rather than just w2 k x r2?
   A. Because m2 rotates about O with w2
   B. Because the lower arm pivots on the moving mass m1, so m2 has the velocity of m1 plus rotation about m1
   C. Because w2 is relative to the vertical
   D. Because the lower arm does not rotate
   Answer: B. The lower pendulum rotates about the moving pin at m1, so v2 is v1 plus the relative velocity about m1. (Key point: Velocity of m2)

4. With H1 = 0.50 and H2 = 7.00 kg m^2/s, what is H_O?
   A. 6.50 k
   B. 7.50 k
   C. 3.50 k
   D. 7.00 k
   Answer: B. H_O = 0.50 + 7.00 = 7.50 k kg m^2/s counter-clockwise. (Key point: 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 = m2 = 1 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Adouble-pendulum-angular-momentum-0-5-0-7) — Applied Mechanics (Dynamics), chapter 3, asked 3×