Pure rolling motion?

Question

A sphere of radius r is in pure rolling motion on a plane surface. If w is the angular velocity what is the angular momentum about the instantaneous point of contact of the sphere with the surface

Answer 1

mrjeffy, bad logic. The fixed point of contact is the very definition of a center of rotation.

The moment of inertia of a sphere of mass m about its own center is .4mr^2. About any other center one adds the mass*offset^2, the offset in this case being equal to the radius, so the total value of I about a point on the surface of the sphere is (1+.4)mr^2.

Since the angular velocity of an object is constant throughout it, the angular momentum is

L = Iw = 1.4m(r^2)w

Answer 2

I beg do differ. mrjeffy is making a mistake. Although the point of contact has zero angular velocity the rest of the sphere does have angular velocity about the point of contact. It looks like your gonna have to integrate over the volome of the sphere. Thinking...
Ok, if x is the distance from the point of contact, then at x = 0 the tangential velocity is zero and at x = 2r the tangential velocity is 2(rw) as stated by mrjeffy. The velocity as a function of x will increase linearly with x and is given by v(x) = xw. Now the angular velocity about the point of contact as a function of x will be v(x)/x = w, a constant for all points on the sphere. So the answer will be L = mr^2 w, the same as the angular momentum about the center of the sphere. I think that is right. Anyone else?

1 mistake. mr^2 is not the moment of inertia of a sphere about it's center but of a sphere about any point outside the sphere. The moment of inertia of a sphere about it's center is (2/5)mr^2.

2nd mistake. Steve got it right. I forgot the shift theorem. In many problems the moment of inertia of spheres on rods is given by MR^2 but by the shift theorem it is actually (2/5)Mr^2 + MR^2. This is approximately MR^2 if the radius is much less then the length of the rod, r << R. Thanks Steve.

Answer 3

As an object rolls without slipping, the point in contact with the surface is instantaneously at rest.
If the point of contact with the ground is at rest, then it has no angular velocity, therefore it has no angular momentum.

Angular Momentum (L) equals Angular velocity (ω) * Rotational Inertia (I)
L = ω * I
If ω = 0, then L = 0

Following this same logic, the point which is opposite the point of contact with the ground has an angular speed which is double the value corresponding to the overall sphere's angular velocity, giving it twice the angular momentum.
The two points which lie exactly in between the top and bottom of the sphere have an angular velocity equal to that over the overall sphere.


EDIT:
Okay, I concede I was incorrect in my original reasoning. I did not read the question as carefully as I needed to.
They word phrase I missed which threw me off was, "about [the point of contact]".
We would need to look at the sphere's angular momentum as if the sphere were to all of a sudden begin rotating around this contact point instead of its current center of rotation.