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TomLikesPhysics
A rod with length l and mass m is standing at first perpendicular on a frictionless table, then he begins to fall. What is the velocity of the center of mass in terms of its position?
I thought I could work this out using the energies. mgH=mgh+0.5mv^2+0.5*I*w^2 v is the speed of the center of mass but I have no clue how I can express w (omega) in terms of v.
|dw:1354958088864:dw|
Yes, but what is the velocity of the upper end of the rod? I don´t think that omega=radius*velocity of the center of mass.
As the rod falls it begins to rotate as you know. The energy of rotation is calculated at the center of rotation i.e. center of mass. If you view this rotation from the center of mass and look at the end of the rod in contact with the table. you see it rising. How fast does it seem to be rising from this vantage point? You see the table coming toward you with velocity of the center of mass. Since the end of the rod is in contact with the table it too will seem to be rising as this same velocity straight up. When determining the angular velocity w=r*v , v is the velocity component perpendicular to the end of the rod not the actual velocity. To account for this you need in determine the angle between the velocity and the component perpendicular to the rod and establish the relation between them.
Ok, so if I use the notation from your (gleem) picture I got: sina=h/r with h = the height of the center of mass at some time cosa=v(perpendicular)/v(centerofmass) so v(perpendicular)=v(centerofmass)*cos(arcsin(h/r)) and therefore omega=v(perpendicular)/r Did I got that right?
Thank you, gleem. This "simple" problem really confused me, but you cleared the fog in my mind.