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 <firstname.lastname@example.org> on behalf of brian whatcott <email@example.com>
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
I considered two possibilities: solid sphere and hollow sphere.
I used two moments of inertia for these cases:2/5*m*r^2 & 3/5*m*r^2
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
What am I missing?