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Homework Help: Rotational Mechanics Problem (Rotating solid wooden disk)

  1. Nov 14, 2011 #1
    I have the working for part d) but I don't understand certain parts, so my problems lie with part d), f) and g)

    1. The problem statement, all variables and given/known data
    3. A solid wooden disk is suspended as shown: it has mass 7.70 kg and radius
    0.219 m . It is rotated from rest about its vertical axis, reaching an angular velocity of 3.69 rad/s in 15.1 s.


    (a) the average angular acceleration
    (b) the angular momentum.
    (c) the rotational kinetic energy

    A 0.17 kg metal block is gently placed on the disk 0.140 m from the rotational axis.
    (d) Show that the new angular velocity is 3.62 rad/s
    (e) Calculate the new rotational kinetic energy.

    The wooden cylinder is very slowly accelerated, and when the new angular velocity is 4.02 rad/s the block is just about to slide.

    (f) the frictional force on the block.
    (g) the coefficient of static friction μs between the block and the disk.

    2. Relevant equations
    a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t
    I(Disk) = MR^2/2
    Krot = 1/2I[itex]\omega[/itex]^2
    W2 = I/I2 x [itex]\omega[/itex]

    3. The attempt at a solution

    (a) a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t = 3.69/15.1 = 0.244 rad/s^2

    I(Disk) = MR^2/2 = (7.70x0.219^2)/2 = 0.185 kg m^2
    L=I[itex]\omega[/itex] = 0.185x3.69 = 0.681 kg m^2 s^-1

    (c) Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.185)(3.69)^2 = 1.257J

    I know L1 = L2, can we prove that this is the case if we didn't have given the value of angular velocity?
    [itex]\tau[/itex]NET = 0 (Not sure how I can prove this)

    I2 = I1 + MR^2 (Inertia of block)
    I2 = 0.185 + 0.17(0.14)^2 = 0.188 kg m^2

    [itex]\omega[/itex]2 = (I1/I2)x[itex]\omega[/itex] = (0.185/0.188)x3.69 = 3.62 rad/s

    Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.188)(3.62)^2 = 1.23J

    (f) & (g) no idea how to work out, can anyone please give me a hand?
  2. jcsd
  3. Nov 14, 2011 #2
    Although the wooden disk is rotating, the block's motion is circular, that means there must be a force (centripetal force) that keeps it going in circles. In this particular example, it's the friction.

    Fμ = v2 / R
    Last edited: Nov 14, 2011
  4. Nov 14, 2011 #3
    I tried using the formula given a few different ways, but can't seem to get the correct answer:

    so, Fμ = v^2 / R

    using v=ωr = 4.02x0.14 = 0.5628 ms^-1

    We have R = 0.14m from previous part

    Subbing that in we get

    Fμ = (0.5628)^2/0.14 = 2.26N

    The answer for this part is 0.385N

    Why am I not getting this answer? :frown:
  5. Nov 14, 2011 #4
    Sorry my fault..

    It's fμ = m v2/r

    I forgot to mention the mass. :tongue: Newton's law is F = ma after all, and the acceleration in this one is the centripetal acceleration that keeps the block moving in circles a = v^2/r

    Now that you know the friction, you should be able to calculate the coefficient of fiction
  6. Nov 14, 2011 #5
    Yep, I got both answers correct now. I'll post up my working tomorrow. Thanks alot for your help!
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