I have seen an expression for the angular momentum of a rotating sphere calculated from outside the sphere as $$L = I\omega + mvr,$$ where $v$ is the velocity of the center of mass, $m$ is the mass of the sphere, and $r$ is the distance of center of mass of the sphere from the point of calculation. My concern is if $v=0$, then $L = I \omega$, which is the angular momentum of the sphere when calculate through the center of mass. How can the angular momentum be the same when the point of calculation is changing?
Subscribe to:
Post Comments (Atom)
Understanding Stagnation point in pitot fluid
What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...
-
Why can't we use fissions products for electricity production ? As far has I know fissions products from current nuclear power plants cr...
-
A rook stands in the lower left corner of an $m\times n$ chessboard. Alice and Bob alternately move the rook (horizontally or vertically, th...
-
Recently I was going through "Problems in General physics" by I E Irodov. In Electromagnetics chapter, there is a question how muc...
-
Yesterday, I understood what it means to say that the moon is constantly falling (from a lecture by Richard Feynman ). In the picture below ...
-
I am having trouble understanding how centripetal force works intuitively. This is my claim. When I have a mass strapped on a string and spi...
-
Work done is defined as the dot product of force and displacement. However, intuitively, should it not be the product of force and the time...
-
What shape does the viewer in a reference frame with $v=0$ perceive? I suppose that since the sphere moves in one direction only (oX only, n...
No comments:
Post a Comment