Atwood's Machine- Calc Based & Differentials

AI Thread Summary
The discussion focuses on using Atwood's machine to measure local acceleration due to gravity with a specified accuracy. The user successfully derived the expression for gravity in terms of the masses and distance, but struggles with the implications of measuring time and the relationship between uncertainties in measurements. They derived a differential equation but are uncertain about their approach to calculating the necessary mass m2 for accurate gravity measurement. The user questions the validity of their assumptions regarding the known value of gravity and the time measurement. Clarification on these points is sought to ensure accurate results.
bollocks748
Messages
10
Reaction score
0

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.)

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:
Physics news on Phys.org
Just didn't want to get bumped to page 2. :-)
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top