1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pulled box

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A wheel of radius 0.358 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.0421 kg·m2. A massless cord wrapped around the wheel is attached to a 3.21 kg block that slides on a horizontal frictionless surface. If a horizontal force of magnitude P = 4.31 N is applied to the block as shown in Fig. 10-54, what is the angular acceleration of the wheel? Take the clockwise direction to be the negative direction and assume the string does not slip on the wheel.

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c10/fig10_54.gif


    2. Relevant equations

    tnet = I(alpha)

    at [tangential acceleration] = (alpha)r

    3. The attempt at a solution

    I did the forces of the box and got that Ft = p [the horizontal force] - ma where m is the mass of the box.
    So, Ft = p - ma

    I found the net torque of the disk rotating at the center of its axis to be -Ftr and the rotational inertia of the disk to be I = 1/2Mr^2 where M is the mass of the disk.
    So, Tnet = -Ftr & I = 1/2Mr^2

    We know that Tnet = I[alpha] so therefore,
    -Ftr = 1/2Mr^2[alpha]

    We know that the string does not slip on the wheel, therefore the tangential acceleration is equal to the angular acceleration times it's radius.
    at = [alpha]r
    [alpha] = at/r

    So by substituting it to the previous equation, I get
    -Ftr = 1/2Mr^2(at/r)
    Cleaning up this equation I get
    Ft = -1/2Ma

    Before I go any further, I found the mass of the disk by using I = 1/2Mr^2. We know that I=0.0421 kg m^2.
    0.0421 = 1/2Mr^2 we know what r is so therefore,
    M = .657 kg

    So I then set the tension found using the disk and the free body diagram of the box equal to each other to get
    P - ma = -1/2Ma
    cleaning this up and having acceleration set on one side of the equation I get,
    a = 2p/[2m-M]
    a = 1.496 m/s^2

    And finally I substituted acceleration back into the tangential acceleration equation which was
    [alpha] = a/r
    [alpha] = 4.18 rad/s^2


    Yet, I got the answer wrong. Please help! I've been staring at this problem for too long.
     
  2. jcsd
  3. Nov 29, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Probably a mistake to assume I = .5mr².
    You can figure out the rotational part from
    torque = Iα
    Tr = I*a/r (where T = tension)
    without the assumption.
    If you sub this expression for T into the T = p - ma you can solve for acceleration and you'll get a smaller value.
     
  4. Nov 29, 2009 #3
    I substituted it for T = I[tex]\alpha[/tex]
    I used -Ftr = I [a/r]
    Them subbed in Ft = P - ma
    rma - rP = I a/r

    I cleaned it up and still got 4 m/s^2 which still gave me 20 rad/s^2
     
  5. Nov 29, 2009 #4

    Delphi51

    User Avatar
    Homework Helper

    Ah, I didn't have that minus sign. Must rethink the whole thing!
    The question specifies counterclockwise as positive, so the first equation should be Ft - p = ma
    And the circular one Ft*r = I*a/r
    So I*a/r² - p = ma
    (I/r² - m) a = p
    a = p/(I/r² - m)
    a = 4.31/(.0421/.358² - 3.21) = -1.496
    so we agree on the number except for the sign. Could the computer be upset over the missing minus sign? It clearly is correct - the p force makes it accelerate clockwise which is defined as negative.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pulled box
  1. Rope pulling a box (Replies: 2)

Loading...