Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Translational and Rotaional motion

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. h' is how far up the smooth side the marble will go, measured vertically from the bottom. Find h', express h' in terms of h.

    2. Relevant equations



    3. The attempt at a solution

    mgh = mgh' + 1/2 Iw^2 (w = angular velocity)

    mgh = mgh' + 1/2 *(2/5)(mR^2)(w^2)

    gh = gh' + v^2/5 -- v is the rotational speed of the marble

    h' = h - v^2/g -- I can't seem express h' just in terms of h, what am i doing wrong?
     
  2. jcsd
  3. Apr 5, 2010 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The marble will retain the angular speed ω that it has at the bottom of the bowl all the way up to maximum height. In other words, the marble will retain all the rotational kinetic energy that is has at the bottom of the bowl and convert only its translational kinetic energy to gravitational potential energy.

    So how will you find the translational kinetic energy at the bottom of the bowl?
     
  4. Apr 5, 2010 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Figure out w by analyzing the first half of the motion separately.
     
  5. Apr 5, 2010 #4
    Thanks, got it now.
     
  6. Apr 5, 2010 #5
    Another problem: A small 15.0 g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 55.0 g and is 90 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 cm/s relative to the table. What is the angular speed (w) of the bar just after the frisky insect leaps?

    I thought I should try to figure out the force that the bug applied to the bar, so then i could figure out the torque, but the problem only gives mass and velocity of the bird, which is not enough information to find the force. So I thought I would figure out its kinetic energy.

    KE of the bug = 1/2 (0.015 kg)(.2 m/s)^2 = 3 x 10^-4 J

    The bar's kinetic energy would equal the kinetic energy of the bird, right?

    3 x 10^-4 J = 1/2 (.055 kg)(v)^2

    v = .1044 m/s = r*w = (.9)(w) ==> w = .116 rad/s, which is not the right answer... what am I doing wrong?
     
  7. Apr 5, 2010 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Nope. Why should it? (You mean bug, not bird, I presume.)

    Hint: What's conserved? Hint2: The fact that the bar can freely rotate about the nail is important.
     
  8. Apr 6, 2010 #7
    Momentum? But when i use conservation of momentum, i still don't get the right answer.
     
  9. Apr 6, 2010 #8

    Doc Al

    User Avatar

    Staff: Mentor

    What kind of momentum? Linear momentum is not conserved.
     
  10. Apr 6, 2010 #9
    Oh, Angular momentum. But still not getting the right answer:

    I1*w1 = I2*w2

    (.015)(.9^2)(0.2/0.9) = (0.055)(.9^2)(w2)

    w2 = 0.061 rad/s....what am i doing wrong now?
     
  11. Apr 6, 2010 #10

    Doc Al

    User Avatar

    Staff: Mentor

    You have the wrong rotational inertia for the rod.
     
  12. Apr 6, 2010 #11
    Thanks
     
  13. Nov 7, 2010 #12
    What is the correct rotational inertia for the bar? Would it be the mass multiplied by half of the bar length squared?
     
  14. Nov 7, 2010 #13
    1/3 m L^2
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook