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Homework Help: Atwood Machine and gravity

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the Atwood's machine of Lecture 8. We wish to use this machine to measure our local acceleration of gravity with an accuracy of 5% [i.e. (Delta g)/g = 0.05]. To begin, suppose we let the mass m_1 fall through a distance L.

    3.1 Find an expression for the acceleration of gravity, g, in terms of m_1, m_2, L and t.

    3.2 Now suppose we are able to measure time with an accuracy of (Delta t) = 0.1 s. Assuming that, for example, (Delta t)/t can be approximated by the differential dt/t, write the relationship between (Delta g)/g and (Delta t)/t. You can do this by starting with the derivative dg/dt determined from the equation in the previous part.

    3.3 If L = 3 m and m_1 = 1 kg, determine the value of m_2 required to determine g to 5%. If we want to measure g to 1% would the mass m_2 increase or decrease - why? (On your own, you might want to consider the effect of the uncertainty in the masses of m_1 and m_2 on the determination of g.)

    2. Relevant equations

    F=ma
    x-x_0=v_o*t+.5*a*t^2

    3. The attempt at a solution

    3.1
    First I summed up the forces on the two masses and solved for the acceleration of the blocks.

    I ended up with...
    a=g(m1-m2)/(m1+m2)

    Then I used x-x_0=v_o*t+.5*a*t^2 and solved for the acceleration.
    a=2L/t^2

    Combing the two equations and solving for gravity I got...
    g=(2L(m1+m2))/((m1-m2)t^2)

    3.2

    I took the derivative of both sides of the equation with respect to t and got...
    dg/dt=(-4L(m1+m2))/((m1-m2)t^3)

    My attempt to relate dg/g to dt/t

    dg/g=(-4L(m1+m2))dt / ((m1-m2)gt^3)

    And that's where I'm stuck. I'm not sure if I did everything right and I can't figure out what I have to plug in to do part 3.3

    Any help is appreciated. Thank you!!!
     
  2. jcsd
  3. Mar 30, 2010 #2

    kuruman

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    Science Advisor
    Homework Helper
    Gold Member

    It seems that you got tangled up in all those symbols. You can make your life simpler if you look at it this way. The expression for g is of the form

    g = C t2, where C is a constant.

    Now take the derivative and see what you get. BTW, the derivative of t2 does not go as t3.
     
  4. Mar 30, 2010 #3

    Doc Al

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    Staff: Mentor

    Good. Now simplify by plugging your earlier expression for g into the right hand side.
     
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