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Speed with which m must move for M to stay at rest

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    A 2.80 kg mass, m, on a frictionless table is moving in a circle with radius 0.440 m at a constant speed. m is attached to a 5.40 kg mass, M, by a cord through a hole in the table. Find the speed with which m must move for M to stay at rest.

    for this one i am completely lost.
  2. jcsd
  3. Oct 6, 2008 #2
    You can apply this for the force pulling down on the mass at the end of the cord under the table

    The other force is centripetal force caused by the rotational motion of the mass on the table.
    This is governed by [tex] F = \frac{mv^2}{r} [/tex]

    Let those two formulae equal each other and you have your answer.
  4. Oct 6, 2008 #3
    so it would be ma=mv^2/r? and then the m's would cancel out and you be be left with a=v^2/r. but then i only know r and not v or a so how do i find that?
  5. Oct 7, 2008 #4
    no... because the masses in the equations are not the same..
    Mass on the table = m
    Mass under table = M
    so you have Ma = mv^2/r
    where M does not equal m
  6. Oct 7, 2008 #5
    but what about having a and v since i dont have either of those, how do i find that
  7. Oct 7, 2008 #6
    Don't be so sure. The Ma side of the equation applies to the block hanging under the table.
    What force do you think is pulling it towards the ground? This gives you a.

    And I know you don't have v, this is what you're solving for, you want to re-arrange your equation such that it is in the form of v = [...]
  8. Oct 7, 2008 #7
    so would u use gravity for a? or like gravity times mass?
  9. Oct 7, 2008 #8
    gravity is a measure of acceleration (ms-2). If you multiply it by mass, you have ma = F.
  10. Oct 7, 2008 #9
    lol, sorry for all of the questions but i am still a little lost.
    so i could just do a/g=m2/(m1+m2) to get a right?
  11. Oct 7, 2008 #10
    Okay yes you are still a little lost.
    we have F = Ma and F = mv2/r
    We are given M, m and r and we are solving for v.
    we will let the two equations equal each other. The only thing that remains in a.

    Now notice that the left hand side of the equation represents the mass under the table. The acceleration affecting this mass IS gravity. Therefore a = g.
  12. Oct 7, 2008 #11
    thank you so much. hahaha i am def going to write this down so i will not have to ask about it again.
  13. Oct 7, 2008 #12
    Haha don't worry about it, it's what we're here for.
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