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Homework Help: Inertia block on inclined surface problem? help with solution?

  1. Nov 12, 2012 #1
    A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. A massless cord wrapped around the wheel is attaches to a 2.0 kg block that slides on a horizontal frictionless surface inclined at angle θ = 20 degrees with the horizontal. The box accelerates down the surface at 2.0 m/s^(2). What is the rotational inertia of the wheel about the axle?

    Solution and picture is on page 32-33

    I understand all the steps except the last one where

    m1(gsinθ - a1)r2 / a1 / r2 = m1(gsinθ - a1)r2^(2) / a1

    why did they divide the denominator a1 by r2? and how did the r2 in the numerator become r2^(2)??

  2. jcsd
  3. Nov 12, 2012 #2
    Note that linear acceleration of the string wound round the pulley is the same as a[itex]_{1}[/itex] since the string is assumed that does not extend.

    Now linear velocity = radius x angular velocity.

    Similarly linear acceleration = radius x angular acceleration.

    i.e. a[itex]_{2}[/itex] = r[itex]_{2}[/itex]α[itex]_{2}[/itex]

    but a[itex]_{2}[/itex] = a[itex]_{1}[/itex]

    therefore a[itex]_{1}[/itex] = r[itex]_{2}[/itex]α[itex]_{2}[/itex]

    i.e. α[itex]_{2}[/itex] = [itex]\frac{a_{1}}{r_{2}}[/itex]
  4. Nov 12, 2012 #3

    Doc Al

    User Avatar

    Staff: Mentor

    1/(1/x) = x

    x/(1/x) = x^2
  5. Nov 12, 2012 #4
    Then how did r2 become r2 ^ (2)?
  6. Nov 12, 2012 #5
    As Doc Al told you.

    The denomenator of a denomenator is a numerator.
  7. Nov 12, 2012 #6
    Whoops I thought the alpha was a regular a.

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