How Do You Calculate the Angular Momentum of a Rolling Sphere?

[Reshma]

A solid sphere of mass m and radius 'a' is rolling with a linear speed v on a flat surface without slipping. The magnitude of the angular momentum of the sphere with respect to a point along the path of the sphere on the surface is?

Moment of inertia of the sphere along any of its diameters is:{2\over 5}ma^2

By parallel axis theorem:
Moment of inertia of the sphere rolling with respect to a point on the surface is:{2\over 5}ma^2 + ma^2 = {7\over 5}ma^2

So the angular momentum will be:L = {7\over 5}ma^2\omega ={7\over 5}mav

The options given to me are:
{2\over 5}mav
{7\over 5}mav[/tex]<br /> mav<br /> {3\over 2}mav<br /> <br /> Is my answer correct?

 

[dextercioby]

What's the center of rotation...?

Daniel.

 

[Reshma]

What's the center of rotation...?

Daniel.

The diameter of the sphere.

 

[dextercioby]

The diameter is a line. I asked about the center, which is a point...

Daniel.

 

[Reshma]

The ball is rolling, so the motion is linear as well as rotational. Centre...you mean circumferential motion?

 

[dextercioby]

Yes, it has to be center if the ball, right...? So why would the angular momentum be calculated wrt to an axis around which the ball doesn't rotate...?

Daniel.

 

[Reshma]

ok...so the ball has only circumferential momentum: So L = mav.
Thank you very much for your help .

 

[arunbg]

Yes, it has to be center if the ball, right...? So why would the angular momentum be calculated wrt to an axis around which the ball doesn't rotate...?

Daniel.

We are talking of rotation and not circular motion.The locus of the points of contact of the sphere and the ground is a straight line and there is definitely an angular momentum associated with an axis through this line.
For eg, consider any point on the sphere.Even from this axis, u can still see that the point is executing pure rotation.
This shift in axis is actually useful in analysing rolling without slipping as it doesn't involve the translation of axis which is the case with an axis at the centre.
So the first answer is probably right.

 

[Reshma]

You guys are confusing me...what is the actual answer here?

 

[topsquark]

So the angular momentum will be:L = {7\over 5}ma^2\omega ={7\over 5}mav

I'm still trying to figure this line out. You seem to be saying a\omega=v. Among other problems, the units are wrong.

Come to think of it:

ok...so the ball has only circumferential momentum: So L = mav.

mav doesn't have the right units either.

-Dan

 

[arunbg]

You guys are confusing me...what is the actual answer here?

I am pretty sure that the answer is
\frac{7}{5}mav.

Ps:Even if you were considering the "circumferential"(don't think the term exists !) angular momentum about an axis through the centre of the circular cross section of the sphere you wouldn't end up with mav.

In the eqn, 'a' stands for radius of the sphere and not acceleration.
So the dimensions do indeed match

 

[Doc Al]

So the angular momentum will be:L = {7\over 5}ma^2\omega ={7\over 5}mav
...
Is my answer correct?

Yes, your answer is correct. At any instant, the sphere can be considered to be in pure rotation about the point of contact.

You can also consider the sphere's motion to be a combination of (a) rotation about its center, plus (b) translation of its center. You'll get the same answer, of course.

 

[topsquark]

I am pretty sure that the answer is
\frac{7}{5}mav.

Ps:Even if you were considering the "circumferential"(don't think the term exists !) angular momentum about an axis through the centre of the circular cross section of the sphere you wouldn't end up with mav.

In the eqn, 'a' stands for radius of the sphere and not acceleration.
So the dimensions do indeed match

Oops! My apologies!

-Dan

 

[Reshma]

Thanks for the confirmation, Doc Al and Arunbg for the clarification. So my answer is indeed correct!