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Homework Help: Question regarding block and pulley system with inertia

  1. Nov 30, 2015 #1
    This problem is an example from Mosca and Tipler, 9-13, 6th edition. I believe the books equation for the acceleration of the system is incorrect according to my work... Anyway, here goes

    1. The problem statement, all variables and given/known data

    "Two blocks are connected by a string that passes over a disk pulley of radius R and moment of inertia I. The block of mass m1 slides on a frictionless, horizontal surface; the block of mass M2 is suspended from a string over the pulley. Find the acceleration of the blocks and the tensions. The string does not slip on the pulley."

    2. Relevant equations
    Non slip, so: a=r*α
    For pulley: r(T2-T2)=I*α
    For block 2: m2g-T2=m2*a m2=mass two
    For block 1: T1=m1*a m1=mass one

    3. The attempt at a solution

    Well, for the pulley: T2-T1=(I*α)r, and therefore: T2-T1=(I*a)/r^2 (substituting a/r=α).

    I added the above equation to the two others so the tensions would cancel:

    so: m2*g=(m2+m1+I/r^2)*a

    so: a=(m2/(m2+m1+I/r^2))*g
    The books says the the third term in the middle parentheses is (I/r)^2....how is this possible??? Shouldn't it be I/r^2??? Any input would be greatly appreciated! Thanks! :)
  2. jcsd
  3. Nov 30, 2015 #2


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    Check the dimensions - you can only add a mass to m1 and m2, so a term ##I / r^2## is acceptable, a term ##(I/r)^2## is surely not.
  4. Nov 30, 2015 #3
    Everything looks correct.

    You're right, from the torques acting on the pulley you get:

    T2-T1=[(Mass of the pulley)*R*α]/2 = (I*α)/R

    and from the no slip condition you know that α=a/R therefore

    T2-T1=[(Mass of the pulley)*a]/2 = (I*a) / R^2

    and solving for a, you get:
    a=(m2/(m2+m1+I/R^2))*g, which is what you did.
  5. Nov 30, 2015 #4
    BvU is right!
  6. Nov 30, 2015 #5
    Awesome, thanks guys! I will send Mosca and Tipler a "tip" on how they can rewrite their physics book haha. We use it for engineering here at UW Seattle :) Again, really appreciated :) :)
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