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Homework Help: More Rotational Dynamics

  1. Jan 12, 2006 #1
    Ok I'm in desperate need of some direction! I have the following question in which I must answer twice once using rotational dynamics and then again using Conservation of Mechanical Energy.

    A massless string is wrapped around a solid cylinder as
    shown in the diagram at the right. A block of mass
    kg 2.0 = m hangs from the string. When released, the
    block falls a distance 82 cm in 2.0 s. Starting with a freebody
    diagram, calculate the mass of the cylinder.

    I have started with ma + 0.5m(cylinder)a= -mg
    That got me a mass of 43.2 kg for the cylinder

    determining a from the equation v^2 = v2^2 +2a (y-y2)
    a = 0.1025 m/s^2

    But I don't think that's right

    Now for Mechanical Energy I'm completely lost.
  2. jcsd
  3. Jan 12, 2006 #2
    Using S = ut + 1/2 a t^2, you can find acceleration a.

    V = u + at should give you the final velocity.

    For rotational part:

    Use, Torque = I Alpha.

    For mechanical energy:

    Use conservation of energy :

    Total energy of the system = potantial energy of the weight + Rotational KE of the cylinder + KE of the hanging mass.
  4. Jan 13, 2006 #3
    i don't understand what s and u represent?

    Also, for mechanical energy I tried:
    M=m1+m2 I = 0.5 m2 r^2 omega = v/r

    0.5Mv1^2 + 0.5Iomega1^2 + Mgh1 =
    0.5Mv2^2 + 0.5Iomega2^2 + Mgh2

    E initial = m2gh = 2.0kg x 9.80 x 0 = 0

    E final = m1gh + 0.5 m1v^2 + 0.5 Iomega^2 = m1gh + 0.5m1v^2 + 1/4 m2v^2

    m2 = .425 ????????

    What have I done wrong?
  5. Jan 13, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    In Gamma's equation, s stands for distance and u stands for initial speed.

    This is a bit confusing:
    The cylinder (mass = m2) only rotates (I presume) so the only relevant energy for the cylinder is rotational KE (its gravitational PE doesn't change). The block (mass = m1), on the other hand, falls, so it has both translational KE and gravitational PE. So rewrite your mechanical energy equation.
  6. Jan 13, 2006 #5
    KE = KE + PE

    0.5 m(cylinder) v^2 = 0.5 m(block) v^2 + m(block) gh

    v = .82m/s
    a = .41 m/s^2

    Is this correct?
  7. Jan 13, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    No. Think this way: Initially nothing is moving so the only energy is potential. As the block falls the potential energy decreases as the kinetic energy increases. The kinetic energy has two parts: The KE of the block ([tex]1/2 m_b v^2[/tex]) plus the rotational KE of the cylinder ([tex]1/2 I_c \omega^2[/tex]).
  8. Jan 14, 2006 #7
    Alright so let's see now, if there is only PE at first then the equation would be:

    PE = KE(block) + KE(Cylinder)
    mgy = [tex]1/2 m_b v^2[/tex] + [tex]1/2 I_c omega^2[/tex]

    Am I getting there?
  9. Jan 14, 2006 #8
    yes, You are getting there.
  10. Apr 27, 2008 #9
    I am confused about the KE(cylinder) in the above equation. Can someone please explain this? Thanks.
  11. Apr 30, 2008 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Can you be more specific? What exactly do you not understand about it?

    An object that rotates about its center of mass has a rotational KE equal to [itex]1/2 I \omega^2[/itex].
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