# Atwood Machine and gravity

• Johnny0290
In summary: You should be able to get a simple expression for dg/g in terms of dm/m, which is what you want.In summary, we discussed using Atwood's machine to measure local acceleration of gravity with 5% accuracy. By summing forces and using equations for acceleration and position, we derived an expression for g in terms of the masses, distance, and time. Then, by taking the derivative and simplifying, we related dg/g to dt/t. Finally, we determined that to measure g with 5% accuracy, we would need to adjust the mass m_2, and that increasing accuracy to 1% would require a larger mass.

## 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!

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.

Johnny0290 said:
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.