anonymous
  • anonymous
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?
Physics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
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.
1 Attachment
anonymous
  • anonymous
|dw:1354958088864:dw|
anonymous
  • anonymous
Yes, but what is the velocity of the upper end of the rod? I donĀ“t think that omega=radius*velocity of the center of mass.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

cj7529
  • cj7529
|dw:1354974582693:dw|
anonymous
  • anonymous
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.
1 Attachment
anonymous
  • anonymous
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?
anonymous
  • anonymous
You got it.
anonymous
  • anonymous
Thank you, gleem. This "simple" problem really confused me, but you cleared the fog in my mind.

Looking for something else?

Not the answer you are looking for? Search for more explanations.