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Suppose we have a disk of radius $r$ and mass $m$ travelling at velocity $v$. I want to calculate the instantaneous angular momentum with axis through the edge of the disc (on the circumference).

Angular momentum $= I \omega$. $I = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$ by the parallel axis theorem. $\omega = \frac{v}{r}$. Therefore, angular momentum $= \frac{3mrv}{2}$.

Alternatively, angular momentum $=p\times r= m r \times v = mrv$.

Why do these two methods differ? Which, if any, are correct?

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    $\begingroup$ Is the disk rolling? If not then there is no angular velocity specified and angular momentum has only the $r \times m v$ term. $\endgroup$ – ja72 Apr 1 '14 at 20:34
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From your description, I assume the disk only only translating, not rotating. Is this correct? If so, read on. If not, I'll delete.

I'm uncomfortable with the first method that uses $L=I\omega$. In this equation, it is assumed that every point on the rigid body can be characterized by the same angular velocity $\omega$. From your description of the motion of the disk, it seems like this doesn't apply here. The disk is only translating, not rotating about a point; thus, each point will have a different angular velocity. I don't have a problem with your expression for the moment of inertia $I$, but that would only be applicable if the object were rotating about a point on its edge.

I believe your second method $\vec{L}=m\vec{r}\times\vec{v}$ makes an assumption that the object is a point particle. You can see this because you treat each point in the body as being characterized by the same position vector $\vec{r}$. This may or may not lead to the correct answer. As another poster stated, the form you want to use is $L=\int \vec{r}\times d\vec{p} = \int_A \vec{r} \times \sigma \vec{v}\ dA$, where I've used a two-dimensional integral since you are treating the disk as two-dimensional. The term $\sigma$ is the two dimensional areal mass density $m/(\pi R^2)$. Let's continue with this integral to see where it leads. $$L=\sigma\int_A \vec{r} \times \vec{v} \ dA = \sigma\int_A (\vec{r}_{CM} + \vec{r}_{body}) \times \vec{v} \ dA = \sigma\int_A \vec{r}_{CM} \times \vec{v} \ dA + \sigma\int_A \vec{r}_{body} \times \vec{v} \ dA $$ I've separated the position vector $\vec{r}$ into the sum of the vector to the center of mass and the vector from the center of mass to a general point on the body. $$ L = \sigma\int_A \vec{r}_{CM} \times \vec{v} \ dA + \sigma \underbrace{\int_A \vec{r}_{body} \times \vec{v} \ dA}_{=0?} \stackrel{?}{=} \sigma A \vec{r}_{CM}\times \vec{v}$$ I believe the integral with the underbrace is zero by symmetry. (Someone want to chime in?) If so, this method produces your second result.

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  • $\begingroup$ By definition for a rigid body, all points share the same angular velocity vector. Only the linear velocity vector changes from location to location. $\endgroup$ – ja72 Apr 1 '14 at 20:35
  • $\begingroup$ If the rigid body were rotating, I would agree with that. But compare $d\theta/dt$ for an object close to the reference point and far from the reference point. In some time $dt$, both points travel the same distance $dl=v\,dt$, but the closer point's path would subtend a larger $d\theta = dl/r=v\,dt/r$, where I've assumed the motion is perpendicular to the reference point. $\endgroup$ – BMS Apr 1 '14 at 20:39
  • $\begingroup$ I think we're imagining two different scenarios. I assumed from the description that the disk was only translating, not rotating. $\endgroup$ – BMS Apr 1 '14 at 20:42
  • $\begingroup$ If the body is not rotating, all points share the same rotational velocity of zero. You make it sound like some points are rotating while others do not, which is incorrect. $\endgroup$ – ja72 Apr 1 '14 at 20:42
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The first method is correct.

The second method is wrong because the equation you use only applies to point particles, not continuous masses with volume (such as a disk). You're incorrectly treating the disk as a point particle located at the disk center. If you want to use the second method, you'll need to use this equation for angular momentum of continuous masses:

$\vec{L} = \int_V dV \, \vec{r} \times \rho(\vec{r})\vec{v}$

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    $\begingroup$ Is the first method actually correct? The equation $L=I\omega$ assumes every point on the rigid body can be described by the same angular velocity $\omega,$ which is true for an object rotating about a point. Not sure about this case with the object moving with uniform motion. $\endgroup$ – BMS Apr 1 '14 at 18:47
  • $\begingroup$ If you assume that v is in a direction perpendicular to the vector r from the center to the edge of the disk, then the first method is correct because the disk is moving, at a given instant in time, in a rotational direction with respect to the axis through the edge of the disk. $\endgroup$ – Izzhov Apr 1 '14 at 18:53
  • $\begingroup$ Actually $\vec{H} = \vec{r} \times m \vec{v} + I \vec{\omega}$ is the full expression for angular momentum not about the c.m. so the second method is not wrong. They just describe different problems. $\endgroup$ – ja72 Apr 1 '14 at 20:38
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Case a)

Body with uniform motion (no rotation), with C the center of the disk and A a point on the edge (for example below, at a distance $R$).

$$ \begin{aligned} \vec{v}_A & = (v,0,0) \\ \vec{\omega} & = (0,0,0) \\ \vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (v,0,0) \\ \vec{L} & = m \vec{v}_C = (m v,0,0) \\ \vec{H}_A & = I \vec{\omega} + (0,-R,0)\times \vec{L} = (0,0,R m v) \\ \end{aligned} $$

Where $\vec{L}$ is linear and $\vec{H}_A$ is angular momentum about point A.

Case b)

Body rolling with edge point A motionless, but with rotational speed $\Omega$

$$ \begin{aligned} \vec{v}_A & = (0,0,0) \\ \vec{\omega} & = (0,0,\Omega) \\ \vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (\Omega R,0,0) \\ \vec{L} & = m \vec{v}_C = (m \Omega R,0,0) \\ \vec{H}_A & = I \vec{\omega} + (0,-R,0)\times \vec{L} = (0,0,\Omega (I_C+m R^2)) \\ \end{aligned} $$

The angular momentum here uses the parallel axis theorem and it expands to..

$$ v_C = \frac{\Omega}{R} \\ I_C = \frac{m}{2} R^2 $$

$$ \begin{aligned} L & = (m v_C,0,0) \\ H_A & = \frac{3}{2} m v_C R \\ \end{aligned} $$

So you see your two solutions correspond to two different problems.

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