How Accurate Can an Atwood Machine Measure Gravity?

[Johnny0290]

Homework Statement

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.)

Homework Equations

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

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 into do part 3.3

Any help is appreciated. Thank you!

 

[kuruman]

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.

 

[Doc Al]

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)

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