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*From*: brian whatcott <betwys1@sbcglobal.net>*Date*: Mon, 15 Jun 2020 13:53:48 -0500

Dear Jeffrey & David,

Please forgive my tardy thanks for explaining this dilemma.

My mail handler is now updated to move some replies into a local save folder - so they escape my attention.

In particular, thanks for spotting my error for Ishell.

Sorry

Brian W

On 5/20/2020 4:37 PM, David Bowman wrote:

When treating the rigid motion of a solid body involving both rotation and translation the analysis is often simplest and cleanest when one makes use of orthogonal degrees of freedom which have cancelling interfering cross terms. This means it is preferred to analyze the composite motion by simply breaking it down as a translation of the body's center of mass and a rotation about an axis through the body's center of mass because such a breakdown has the infinitesimal translations orthogonal to infinitesimal rotations in their corresponding subspaces of the body's overall configuration space. (This is especially so when treating total kinetic energy of motion and not analyzing local torques about eccentric pivot points.) In such a situation the relevant moment of inertia is the one for an axis *through* the center of mass. So for the problem BW considers things work out simplest (for the energy conservation analysis) if one uses the moment of inertia about an axis through the sph

ere's center, and *not* the moment of inertia about an axis through the sphere's contact point with the incline.

BTW, the moment of inertia for an axis through the center of a uniform infinitely thin spherical shell of radius R is (2/3)*M*R^2. If the moment of inertia of a hollow sphere of outer radius R is (3/5)*M*R^2 then the sphere has a shell thickness of 0.10773447...*R, assuming the shell has a constant mass density.

David Bowman

On 5/20/2020 3:29 PM, Jeffrey Schnick via Phys-l wrote:

In calculating the kinetic energy of an object that is rolling without slipping, you can treat the object as rotating about an axis through its center of mass while at the same time translating through space. If you do it that way, you use the moment of inertia with respect to the center of mass and the kinetic energy is .5*I_cm*omega^2 + .5*m*v^2 where v is the speed of the center of mass. Another option is to treat the object as being in pure rotation about its instantaneous center of zero velocity, which, for the case of an object that is rolling without slipping is the point of contact of the object with the surface upon which it is rolling. In that case, the moment of inertia is to be taken with respect to the axis through the instantaneous center of zero velocity and the kinetic energy is just .5*I_iczv*omega^2. Both methods have to give you the same answer.

________________________________________

From: Phys-l <phys-l-bounces@mail.phys-l.org> on behalf of brian whatcott <betwys1@sbcglobal.net>

Sent: Wednesday, May 20, 2020 3:18 PM

To: prefered phys-l address

Subject: [Phys-L] Rotation of a Rolling Ball.

I was considering a Galilean problem on Quora of a ball moving at a

given velocity climbing a ramp of given height without slipping.

The question asked about its final velocity after the ascent.

This depends upon its kinetic energy, and how much is converted to

potential energy.

I considered two possibilities: solid sphere and hollow sphere.

I used two moments of inertia for these cases:2/5*m*r^2 & 2/3*m*r^2 XX/NOT/ *3/5*m*r^2* XX

These happen to be the values used by Dan MacIsaac in his 1996 video

tutorial at Buffalo, and by John Yelton at Oxford U., when upvoting a

similar calculation recently.

Then I happened on a list of Moments:

1) a sphere spinning on a central axis 2/5*m*r^2

2) a sphere rolling on a surface 7/5*m*r*2

And I was taken aback.

The academic sources seem to be juggling the measure for rotation to

make the lower value of Moment work, ignoring the axis of rotation

offset correction.

What am I missing?

Brian W

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