1. Limited time only! Sign up for a free 30min personal 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!

Atwood's Machine- Calc Based & Differentials

  1. Oct 10, 2008 #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.)

    3. The attempt at a solution

    Okay, I got excellent help on one problem I struggled with, so hopefully I'll get some help on this one. Solving for the net force, I ended up with the equation that the acceleration downward is g * (m1-m2/m1+m2). Setting that equal to another expression for acceleration, L=1/2 a t^2, I end up with the function g= [2L(m1+m2)]/[t^2(m1-m2)]. Part 1 done.

    Then for part two, I derived dg/dt and formed the differential equation:

    dg= (-4L(m1+m2)/t^2(m1-m2))(dt)

    Since I need dg/g and dt/t, I divided both sides by g, and added a t to the right side:

    dg/g= [(-4*L*t*(m1+m2))/(g*t^3(m1-m2))](dt/t), which would finish part two.

    Part three is where I run into an issue. I used 9.8 m/s^2 to fill in for g in the equation, and .1s for t, in order to solve for m2. However, I don't think that's the right approach, as I'm supposed to be measuring the local gravity, which may or may not be exactly 9.8 m/s^2. As well as that, it doesn't make sense to me that I can set dt and t to .1s each. But, by doing that approach, I reached and answer of 1.000817kg for m2. That is so close to m1's value that it doesn't seem correct either. Am I approaching differentials incorrectly? Thanks in advance to anyone who can help me out with this.
    Last edited: Oct 10, 2008
  2. jcsd
  3. Oct 10, 2008 #2
    Just didn't want to get bumped to page 2. :-)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Atwood's Machine- Calc Based & Differentials
  1. Atwoods Machine (Replies: 4)

  2. Atwoods machine (Replies: 1)

  3. Atwood's machine (Replies: 12)

  4. The Atwood machine. (Replies: 1)