# Atwood's Machine- Calc Based & Differentials

1. Oct 10, 2008

### bollocks748

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. Oct 10, 2008

### bollocks748

Just didn't want to get bumped to page 2. :-)