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Angular acceleration and newton's 2nd law

  1. Jun 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A solid cylindrical disk has a radius of 0.16 m. It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a 45 N force is applied tangentially to the disk, perpendicular to the radius, the disk acquires an angular acceleration of 110 rad/s2. What is the mass of the disk?

    F=45 N
    r=0.16 meters

    2. Relevant equations

    torque = force * lever arm

    net external torque=mr^2*a
    3. The attempt at a solution
    So far, I've gotten torque = 45*.16 = 7.2
    7.2=m(0.16)^2 * 110
    and solving for m gives me 2.556
    but that is not correct. I am absolutely stumped.

    Thanks for any help.
  2. jcsd
  3. Jun 1, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's not true in general, and certainly not true for a cylinder. Look up the rotational inertia of standard shapes.
  4. Jun 1, 2007 #3

    D H

    Staff: Mentor

    What makes you think this? What is the moment of inertia of a solid cylinder about its axis of rotation?

    Edited to add:

    Dang Doc, you type too fast.
  5. Jun 1, 2007 #4
    My book doesn't have anything about different moments of inertia for different shapes, so I figured that was right.

    I looked up the rotational inertias :http://en.wikipedia.org/wiki/List_of_moments_of_inertia

    and it says for a thin solid disk, it is I=(mr^2)/4.

    I tried this:
    net torque=I*a, so
    7.2=((mr^2)/4)*110, and solving for m did not give me the right answer.

    oh no.
  6. Jun 1, 2007 #5
    nevermind, I=mr^2/2.
    thanks for the help, everybody.
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