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Angular Momentum | Study Mondo
Topics / Rotational Motion / Angular Momentum Angular Momentum Conservation of angular momentum, cross products, and applications
๐ฏ โญ INTERACTIVE LESSON
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Start Interactive Lesson โ Angular Momentum
Definition
For a particle:
L โ = r โ ร p โ = m ( r โ ร v โ ) \vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v}) L =
๐ Practice ProblemsNo example problems available yet.
Explain using: ๐ Simple words ๐ Analogy ๐จ Visual desc. ๐ Example ๐ก Explain
๐งช Practice Lab Interactive practice problems for Angular Momentum
โพ ๐ Related Topics in Rotational Motionโ Frequently Asked QuestionsWhat is Angular Momentum?โพ Conservation of angular momentum, cross products, and applications
How can I study Angular Momentum effectively?โพ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Regular review and active practice are key to retention.
Is this Angular Momentum study guide free?โพ Yes โ all study notes, flashcards, and practice problems for Angular Momentum on Study Mondo are 100% free. No account is needed to access the content.
What course covers Angular Momentum?โพ Angular Momentum is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Rotational Motion section. You can explore the full course for more related topics and practice resources.
๐ก Study Tipsโ Work through examples step-by-step โ Practice with flashcards daily โ Review common mistakes r
ร
Magnitude:
L = r m v sin โก ฮธ = r โฅ m v = m r v โฅ L = rmv\sin\theta = r_{\perp}mv = mrv_{\perp} L = r m v sin ฮธ = r โฅ โ m v = m r v โฅ โ
For rotation about fixed axis:
L = I ฯ L = I\omega L = I ฯ
Relationship to Torque ฯ โ = d L โ d t \vec{\tau} = \frac{d\vec{L}}{dt} ฯ = d t d L โ
(Rotational analog of F โ = d p โ d t \vec{F} = \frac{d\vec{p}}{dt} F = d t d p โ โ )
For fixed axis with constant I I I :
ฯ = I d ฯ d t = I ฮฑ \tau = I\frac{d\omega}{dt} = I\alpha ฯ = I d t d ฯ โ = I ฮฑ
Conservation of Angular Momentum When ฯ โ e x t = 0 \vec{\tau}_{ext} = 0 ฯ e x t โ = 0 :
L โ = constant \vec{L} = \text{constant} L = constant
I i ฯ i = I f ฯ f I_i\omega_i = I_f\omega_f I i โ ฯ i โ = I f โ ฯ f โ
Example: Figure Skater Skater pulls arms in, reducing moment of inertia:
I i ฯ i = I f ฯ f I_i\omega_i = I_f\omega_f I i โ ฯ i โ = I f โ ฯ f โ
If I f < I i I_f < I_i I f โ < I i โ , then ฯ f > ฯ i \omega_f > \omega_i ฯ f โ > ฯ i โ (spins faster)
Energy changes:
K E i = 1 2 I i ฯ i 2 , K E f = 1 2 I f ฯ f 2 KE_i = \frac{1}{2}I_i\omega_i^2, \quad KE_f = \frac{1}{2}I_f\omega_f^2 K E i โ = 2 1 โ I i โ ฯ i 2 โ , K E f โ = 2 1 โ I f โ ฯ f 2 โ
K E f = 1 2 I f ( I i I f ฯ i ) 2 = I i I f โ
1 2 I i ฯ i 2 = I i I f K E i KE_f = \frac{1}{2}I_f\left(\frac{I_i}{I_f}\omega_i\right)^2 = \frac{I_i}{I_f}\cdot\frac{1}{2}I_i\omega_i^2 = \frac{I_i}{I_f}KE_i K E f โ = 2 1 โ I f โ ( I f โ I i โ โ ฯ i โ = I f โ I i โ โ โ
2 1 โ I i โ ฯ i 2 โ = I f โ I i โ โ K E i โ
Energy increases! (Work done by internal forces)
Angular Momentum of System L โ t o t a l = โ i L โ i = โ i r โ i ร p โ i \vec{L}_{total} = \sum_i \vec{L}_i = \sum_i \vec{r}_i \times \vec{p}_i L t o t a l โ = โ i โ L i โ = โ i โ r i โ ร p โ i โ
About center of mass:
L โ = L โ c m + L โ o r b i t a l \vec{L} = \vec{L}_{cm} + \vec{L}_{orbital} L = L c m โ + L or bi t a l โ
L โ c m = I c m ฯ โ \vec{L}_{cm} = I_{cm}\vec{\omega} L c m โ = I c m โ ฯ (spin)
L โ o r b i t a l = r โ c m ร M v โ c m \vec{L}_{orbital} = \vec{r}_{cm} \times M\vec{v}_{cm} L or bi t a l โ (orbital)
Central Force Motion For central force (directed toward/away from fixed point):
ฯ โ = r โ ร F โ = 0 \vec{\tau} = \vec{r} \times \vec{F} = 0 ฯ = r ร F = 0
(because F โ \vec{F} F parallel to r โ \vec{r} r )
Therefore: L โ \vec{L} L = constant
Motion confined to a plane
Areal velocity constant (Kepler's second law)
r 2 ฮธ ห = L m r^2\dot{\theta} = \frac{L}{m} r 2 ฮธ ห = m L โ = constant
Areal Velocity Area swept out per unit time:
d A d t = 1 2 r 2 d ฮธ d t = 1 2 r 2 ฯ = L 2 m \frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} = \frac{1}{2}r^2\omega = \frac{L}{2m} d t d A โ = 2 1 โ r 2 d t d ฮธ โ = 2 1 โ r 2 ฯ = 2 m L โ
(Constant for central forces)
Collisions and Angular Momentum For collision, if ฯ โ e x t = 0 \vec{\tau}_{ext} = 0 ฯ e x t โ = 0 about some point, then L โ \vec{L} L conserved about that point.
Example: Putty Ball Hitting Rod Ball of mass m m m , speed v v v hits rod of length L L L , mass M M M at distance d d d from pivot.
Before: L i = m v d L_i = mvd L i โ = m v d (ball's angular momentum)
After: L f = I t o t a l ฯ L_f = I_{total}\omega L f โ = I t o t a l โ ฯ
where I t o t a l = 1 3 M L 2 + m d 2 I_{total} = \frac{1}{3}ML^2 + md^2 I t o t a l โ = 3 1 โ M L 2 + m d 2 (rod + stuck ball)
m v d = ( 1 3 M L 2 + m d 2 ) ฯ mvd = \left(\frac{1}{3}ML^2 + md^2\right)\omega m v d = ( 3 1 โ M L 2 + m d 2 ) ฯ
ฯ = m v d 1 3 M L 2 + m d 2 \omega = \frac{mvd}{\frac{1}{3}ML^2 + md^2} ฯ = 3 1 โ M L 2 + m d 2 m v d โ
Precession Spinning top with angular momentum L โ \vec{L} L tilted at angle ฮธ \theta ฮธ :
Gravitational torque:
ฯ = m g r sin โก ฮธ \tau = mgr\sin\theta ฯ = m g r sin ฮธ
This causes precession (axis rotates) with angular velocity:
ฮฉ = ฯ L sin โก ฮธ = m g r L = m g r I ฯ \Omega = \frac{\tau}{L\sin\theta} = \frac{mgr}{L} = \frac{mgr}{I\omega} ฮฉ = L s i n ฮธ ฯ โ = L m g r โ = I ฯ m g r โ
Gyroscopic Motion Gyroscope resists changes in orientation due to angular momentum conservation.
Applied torque ฯ โ \vec{\tau} ฯ causes change:
ฮ L โ = ฯ โ ฮ t \Delta\vec{L} = \vec{\tau}\Delta t ฮ L = ฯ ฮ t
Direction of ฮ L โ \Delta\vec{L} ฮ L perpendicular to both L โ \vec{L} L and ฯ โ \vec{\tau} ฯ , causing precession.
Angular Impulse โซ t 1 t 2 ฯ โ โ d t = ฮ L โ \int_{t_1}^{t_2} \vec{\tau} \, dt = \Delta\vec{L} โซ t 1 โ t 2 โ โ ฯ d t = ฮ L
Analog of linear impulse โซ F โ โ d t = ฮ p โ \int \vec{F} \, dt = \Delta\vec{p} โซ F d t = ฮ p โ
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