---
title: "Double pendulum angular momentum, m1 = 3, m2 = 4 kg"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:double-pendulum-angular-momentum-0-4-0-5"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/double-pendulum-angular-momentum-0-4-0-5
site: BCT Engineering
---

# Double pendulum angular momentum, m1 = 3, m2 = 4 kg

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

- Asked in: 2074 Asoj, 4 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-0-4-0-5
- 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 X-Y plane. At the instant shown, $w_1 = 4$ rad/sec CCW and $w_2 = 5$ rad/sec CCW. What will be the angular momentum about 'O' at this instant, if $m_1 = 3$ kg and $m_2 = 4$ kg? Note that 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 0.4 m from O at $30^\circ$ to the vertical (Y down, X to the right), carrying $m_1$; lower arm 0.5 m carrying $m_2$ at $60^\circ$ to the vertical.]

## Short answer

Using r1, r2, v1, v2 derived from lengths (0.4 m, 0.5 m), angles (30°, 60°) and angular speeds 4 and 5 rad/s, angular momentum is H_O = Σ m(r×v). Result: H_O ≈ 15.72 k̂ kg·m²/s CCW.

## 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 ($a=0.4$ m at $30^\circ$ to the $x$-axis), lower arm ($b=0.5$ m at $60^\circ$ to the vertical):

$$
\begin{aligned}
\vec r_1 &= a(\cos 30^\circ\,\hat i + \sin 30^\circ\,\hat j) = (0.346\,\hat i + 0.200\,\hat j)\ \text{m}\\
\vec r_{2/1} &= b(\cos 60^\circ\,\hat i + \sin 60^\circ\,\hat j) = (0.250\,\hat i + 0.433\,\hat j)\ \text{m}\\
\vec r_2 &= \vec r_1 + \vec r_{2/1} = (0.596\,\hat i + 0.633\,\hat j)\ \text{m}
\end{aligned}
$$

#### Velocities
$$
\begin{aligned}
\vec v_1 &= \omega_1\hat k \times \vec r_1 = (-0.800\,\hat i + 1.386\,\hat j)\ \text{m/s}\\
\vec v_2 &= \vec v_1 + \omega_2\hat k \times \vec r_{2/1} = (-2.965\,\hat i + 2.636\,\hat j)\ \text{m/s}
\end{aligned}
$$

#### Angular momentum
$$
\begin{aligned}
H_1 &= m_1(x_1 v_{1y} - y_1 v_{1x}) = 3(0.346\times 1.386 - 0.200\times -0.800) = 1.92\ \text{kg m}^2/\text{s}\\
H_2 &= m_2(x_2 v_{2y} - y_2 v_{2x}) = 4(0.596\times 2.636 - 0.633\times -2.965) = 13.80\ \text{kg m}^2/\text{s}
\end{aligned}
$$

$$
\vec H_O = H_1 + H_2 = 15.72\,\hat k\ \text{kg m}^2/\text{s}
$$

**Answer: $\vec H_O \approx 15.72\,\hat k$ 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 the angular momentum of the double pendulum about the fixed pivot O found?
   A. H_O = m1(r1 x v1) + m2(r2 x v2), with r measured from O
   B. H_O = (m1 + m2)(r2 x v2)
   C. H_O = m1 v1 + m2 v2
   D. H_O = m1(r1 x v1) + m2(r2/1 x v2) using the lower arm vector only
   Answer: A. Angular momentum about O is the sum over particles of m(r x v), with each position vector measured from O. (Key point: Angular momentum of a system)

2. How is the position vector r2 of mass m2 from O obtained?
   A. Use only the lower arm vector r2/1
   B. r2 = r1 + r2/1, the sum of the upper and lower arm vectors
   C. r2 = r1 - r2/1
   D. Use the upper arm length 0.4 m only
   Answer: B. m2 sits at the end of the lower arm which starts at m1, so r2 = r1 + r2/1 = (0.596 i + 0.633 j) m. (Key point: Position of m2)

3. Which expression gives the velocity of m2 when w2 is the absolute angular velocity of the lower arm?
   A. v2 = w2 k x r2
   B. v2 = v1 + w2 k x r2/1
   C. v2 = w2 k x r2/1 only
   D. v2 = v1 + w1 k x r2/1
   Answer: B. m2 moves with m1 and also rotates about m1, so v2 = v1 + w2 k x r2/1. (Key point: Velocity of m2)

4. With H1 = 1.92 and H2 = 13.80 kg m^2/s, what is H_O and its sense?
   A. 11.88 kg m^2/s clockwise
   B. 15.72 kg m^2/s counter-clockwise
   C. 13.80 kg m^2/s counter-clockwise
   D. 15.72 kg m^2/s clockwise
   Answer: B. H_O = 1.92 + 13.80 = 15.72 k kg m^2/s, positive k meaning counter-clockwise. (Key point: Final result)

## Close variants set in other exams

- [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×
- [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×