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Circular motion and energy

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=67417&stc=1&d=1394285257.jpg

    A massless spring of constant k is fixed on the left side of a level track. A block of mass m is pressed against the spring and compresses it a distance d, as shown in the figure. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. Given that the coefficient of kinetic friction between the block and the track along AB is k, and that the length of AB is L. The gravitational acceleration is g.

    a) Express the minimum speed of the mass at point C, vC, in terms of symbols given such that it can make through the loop-the-loop (i.e. remain in contact with the track).
    (b) Hence or otherwise, determine the minimum compression, d, of the spring for such situation.

    2. Relevant equations
    Fnet=ma,
    ac=v^2/R,
    Total energy = KE + PE + friction
    Spring energy = -kx

    3. The attempt at a solution

    a) Fnet=ma
    N+mg=m(ac)
    2mg=m(v^2/R)
    2g=(v^2/R)
    vc=sqt(2gR)

    b) -kd=1/2m(vc)^2 + mgR + k
    d= (1/2m(vc)^2 + mgR + k )/-k


    Are these the correct approaches, please?
     

    Attached Files:

  2. jcsd
  3. Mar 8, 2014 #2
    In (a), where does 2 in 2mg come from?

    In (b), what does the last term mean?
     
  4. Mar 8, 2014 #3
    In (a), i thought N + mg = m(ac) where N=mg
    therefore mg + mg = 2mg = m(ac)
    So... maybe I'm wrong?

    In (b), -kd=1/2m(vc)^2 + mgR + (μk)*mg*L
    d= (1/2m(vc)^2 + mgR + (μk)*mg*L )/-k
    Is this correct, please?
     
  5. Mar 8, 2014 #4
    In (a), why would N = mg?

    In (b), the right hand side has the dimension of energy. The left hand side, however, has the dimension of force. That cannot be correct.
     
  6. Mar 8, 2014 #5
    In (a), because I thought N is always equal to mg...
    But just now I've thought it twice.. it's not true, a simple example is the Lift Case..
    So N > 0 ; mg > m(ac) ; g > v^2/r ; v > sqrt(g*r) ?

    In (b), for both sides having the dimension of energy, I should use 1/2*k*x^2 instead of -kd then??

    thanks
     
  7. Mar 8, 2014 #6
    (b) is correct now.

    For (a), observe that you are asked about the minimum speed that would still keep the block in contact with track. Which means that if the speed is a just tiny bit less, it will just fall down. Would there be any reaction force from the track when the block is just short of losing contact with it?
     
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