Angular momentum of rolling sphere A sphere of uniform density $\rho$ and radius $r$ is rolling without slipping on a perfectly flat surface. It is moving in a perfectly straight line and its axis of rotation is parallel to the plane of the surface. Its centre of mass is moving at a constant velocity $v$ and the distance between its line of motion and a reference point $A$ is $r$. Find the angular momentum of the sphere about point $A$, taking into account the rolling. 
My AP Mechanics teacher told me to treat the sphere as a point particle, but I don't want to do that. Besides, any approximation will most certainly be off if the sphere gets too large. Is there an exact solution to this problem? If so, provide it. 
What about a cylinder doing the same? 
 A: I think that what your teacher has told you is that the angular momentum of a body can be split into two components:  


*

*The spin angular momentum which is an intrinsic property of the body
and is independent of the point about which you wish to find the
angular momentum.   $L_{\text{spin}} = I_{\text{cm}} \omega = \frac v
   r $ where $I_{\text{cm}}$ is the moment of inertia of the sphere about its centre of mass and $\omega$ is the angular speed of the sphere.  

*The orbital angular momentum about a point which is the angular
momentum of the centre of mass of the body about that point.  In your
example $L_{\text{orbital}} = Mvr$ where $M$ is the mass of the sphere.  


Without using vector notation and integration my answer is as follows.
Consider two point masses $A$ and $B$ of mass $m$ which are at a distance $a$ from the centre of mass of the sphere as shown on the diagram below.  

The velocity of each mass is the vector sum of a translational velocity of magnitude $v$ which is the same as the velocity of the centre of mass and a velocity due to rotation of the sphere about its centre of mass and is of magnitude $a\omega$.  The directions of these velocities are shown in the diagram.
Taking the rotational component first.
The clockwise angular momentum about point $X$ is $ma\omega(2a+x) – ma\omega x = 2 m a^2 \omega$.
Note that this angular momentum does not depend on the position of point $X$.     
Since this angular momentum only depends of the distance the masses are from the centre of mass of the sphere this is also the angular momentum of a ring of particles of mass $m_{\text{ring}}$ all at a distance $a$ from the centre of mass.
So the angular momentum of the ring is $ m_{\text{ring}} a^2 \omega = I_{\text{ring}} \omega$ where $I_{\text{ring}}$ is the moment inertia of the ring.   
Summing over all the rings which make up the sphere gives $I_{\text{cm}} \omega = I_{\text{cm}} \frac v
   r $ and this is the spin angular momentum of the sphere which is independent of the position of $X$.
Now looking at the translational velocity of the two masses.
Their angular momentum is $mv(r+a \sin \theta) + mv(r-a \sin \theta) = 2mvr$ and is independent of the positions of the masses within the sphere.   
So the total angular momentum for the whole sphere is $Mvr$ and this is the orbital angular momentum that your teacher was referring to.
You take the momentum of the centre of mass $Mv$ and multiply it by the distance between the line of motion and the point.
Note that this can be extended to any point $Y$ which is at distance $y$ from the line of motion of the sphere and then the orbital angular momentum is $Mvy$.
And yes, it is also true for a cylinder.
A: There is an exact solution to the problem. You can consider each point on the sphere by specifying two parameters ($r$ and $\theta$). Now since you know each points velocity vector you can calculate its contribution to the angular momentum by using $dL$ = $dm$($r$ x $v$). Now integrate over $\theta$(0 to $2\pi$) and $r$(0 to R). After doing all this, you will see that treating the sphere as a point particle gives the same result. You can treat the cylinder as a point particle too.
