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...
-
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...
-
As the title says. It is common sense that sharp things cut, but how do they work at the atomical level? Answer For organic matter, such a...
-
How can we know the order of a Feynman diagram just from the pictorial representation? Is it the number of vertices divided by 2? For exampl...
-
Literature states neutral pion decay by QED cannot occur directly because the pion is uncharged. However, I cannot see why Photons are not a...
-
Recently I was going through "Problems in General physics" by I E Irodov. In Electromagnetics chapter, there is a question how muc...
No comments:
Post a Comment