Subliminal Posted November 30, 2013 Share Posted November 30, 2013 L=vmr (angular momentum) If an object 'm' is moving and has a reference point 'p' that is a distance r from the object, is the angular momentum of the object its capacity to rotate a light plank fixed to point I know the torque is the change in angular momentum over time, but is L simply tangential velocity times mass times r or angular velocity times mass times r? I don't know what it is.... Angular velocity is theta/t = w, so is does L = wmr and not vmr? Link to comment Share on other sites More sharing options...
swansont Posted November 30, 2013 Share Posted November 30, 2013 L = mvr (as a vector it's r x p) or Iw where I is the moment of inertia for the rotation. wmr has the wrong units so we know it can't be right. 1 Link to comment Share on other sites More sharing options...
Endercreeper01 Posted November 30, 2013 Share Posted November 30, 2013 It would not be wmr and it would instead be wmr2 because w is the angular velocity, and the velocity is v=wr. Since L=mvr, then it is also L=wmr2 because v=wr, and if we plug that into the equation, we get L=(wr)mr, and then we get wmr2. Link to comment Share on other sites More sharing options...
Subliminal Posted November 30, 2013 Author Share Posted November 30, 2013 L = mvr (as a vector it's r x p) or Iw where I is the moment of inertia for the rotation. wmr has the wrong units so we know it can't be right. Link to comment Share on other sites More sharing options...
decraig Posted December 2, 2013 Share Posted December 2, 2013 (edited) L=vmr (angular momentum) If an object 'm' is moving and has a reference point 'p' that is a distance r from the object, is the angular momentum of the object its capacity to rotate a light plank fixed to point To be precise, r is the perpendicular distance to the ray traced by the velocity vector of the center of mass of the object (within Newtonian physics). Edited December 2, 2013 by decraig Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now