High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize AP Physics C: Mechanics Textbook > Unit 5: Rotation (14-20%)

Angular Momentum

Rotational Work and TorquePrevious Section
Conservation of Angular Momentum Next Section
Popular Courses
Find My Course
0:00 / 0:00

Angular Momentum


Angular momentum is a vector that is defined as the cross product of the position vector (from the axis of rotation or a pivot point to the particle position) and the momentum vector.

L⃗=r⃗×P⃗\boxed{\vec{L}=\vec{r}\times\vec{P}}







It is also shown by J⃗\vec{J}.
  • Magnitude of this vector is equal to:

∣L⃗∣=∣r⃗∣∣p⃗∣sin⁡θ\boxed{|\vec L|=|\vec r||\vec p|\sin\theta}


where θ\thetais the angle between r⃗\vec{r}and p⃗\vec{p}vectors.
  • Its direction is obtained using the Right Hand Rule and it is perpendicular to the plane of position vector and momentum vector
  • For Rigid bodies rotating around an axis with the angular frequency of ω\omega, the magnitude of angular momentum in analogy to linear momentum can be represented by
L=Iω\boxed{ L=I \omega}
where I is the moment of inertia about the axis of rotation.

Wize Tip
  • The angular momentum is the maximum when θ=90∘\theta=90^{\circ}
∣L⃗∣=∣r⃗∣∣p⃗∣sin⁡θ=rmv|\vec{L}|=|\vec{r}||\vec{p}|\sin\theta=rmv

  • The angular momentum is zero when r⃗\vec{r}and p⃗\vec{p}vectors are parallel.

PAGE BREAK

Angular Momentum and Torque

The following relation can be obtained by looking at the torque definition:
τ⃗=r⃗×F⃗=r⃗×dP⟶dt=dL⟶dt\boxed{\vec\tau=\vec r\times \vec F=\vec r\times\frac{\stackrel{\longrightarrow}{dP}}{dt}=\frac{\stackrel{\longrightarrow}{dL}}{dt}}







This is similar to the statement of Newton's second law in terms of linear momentum.

Exam Tip
In a uniform circular motion L is constant. Because τ⃗=r⃗×F⃗=0\vec \tau=\vec r\times\vec F=0 .






0:00 / 0:00

Example: Angular Momentum of a Pendulum


An object of mass m and speed v0 strikes a bob pendulum with mass M and length of llwhich is hanging from the ceiling and sticks to it and they both move (oscillate) to the right. Find the size and direction of the angular momentum with respect to where the rope is attached to the ceiling .

Solution


First we need to find the speed of the combination of mass and pendulum after collision. Since they stick together after collision, we have a perfectly inelastic collision. So, we can use conservation of linear momentum:
Pbefore=PafterP_{before}=P_{after}

mv0=(m+M)vfvf=mm+Mv0\begin{aligned} mv_0&=(m+M)v_f\\ v_f&=\frac{m}{m+M}v_0 \end{aligned}

Now we can find magnitude of angular momentum using its definition:

L⃗=r⃗×(m+M)v⃗f∣L⃗∣=r(m+M)vfsin⁡90⏟1=l(M+m)vf=l(mv0)\begin{aligned} \vec L&=\vec r\times (m+M)\vec v_f\\ |\vec L|&=r(m+M)v_f\underbrace{\sin90}_{1}=l(M+m)v_f=l(mv_0)\\ \end{aligned}
Note that in the last line, I set the angle between position vector and linear momentum vector to be 90 degrees. The reason is that after collision, they start to oscillate about the pivot point in a circular path, so, their velocity vector is tangent to this circle and as a result perpendicular to the rope. Since linear momentum and velocity vectors are parallel, linear momentum vector is also perpendicular to the rope. Furthermore, position vector is defined from the pivot point to the rotating object, hence, it should be along the rope. Thus, position and linear momentum vectors are perpendicular.

In addition, the length of the position vector is the same as length of the rope and equal to ll.Finally, in the last step, I inserted the expression we found for vfv_f.

Now, let's think about direction of the angular momentum. We know that since it is defined as the cross product of position and linear momentum vectors, it is perpendicular to the plane containing these two vectors. So, it should be pointing either into the page or out of the page. We can use Right-Hand Rule for L⃗=r⃗×P⃗\vec{L}=\vec{r}\times \vec{P}to find direction of the angular momentum to be out of the page.

You're sitting on a rotating stool, initially rotating at 3.0 radians per second. You're holding a 5.0 kg dumbbell very close to you, 0.05 meters away from your center of rotation. You then extend your arms 1.0 meters away from you with the dumbbell. The moment of inertia of you and the stool is 4.0 kg m2. Neglect the mass of your arms and any rotational friction in the seat. What is your new angular velocity?