# Experimental Design: Pulley and Mass Hangers

• uSee2
In summary, the conversation discusses an experiment involving a pulley, cord, and mass hangers to determine the value of gravity. The original approach was to calculate the rotational inertia of the pulley and use energy methods to determine acceleration due to gravity. However, the answer key provided a different method that involved creating an imbalance of masses and measuring the acceleration of each mass. This method is similar to using the equation F = ma, with a modified version to account for the inertia of the pulley. The conversation also addresses practical concerns such as accuracy and the historical context of the experiment.
uSee2
Homework Statement
Use the equipment shown in the diagram to design an experiment to determine the acceleration due to gravity. Use a total mass of 0.5 kg to 1 kg on the mass hangers. Standard lab equipment may also be used.

(Diagram Below)
Relevant Equations
##F = ma##

^ This is my personal drawing of the diagram, I couldn't take a picture of the actual one. The setup is a pulley wrapped with a cord and mass hangers attached to each end.

My first thought when approaching this problem was to first determine the rotational inertia of the pulley, then use some other method like energy to determine the acceleration due to gravity.

If only one mass was hung, then:
τ=Iα
τ=Frsin⁡(θ)
##F = F_{t1}## where ##F_{t1}## is the force of tension exerted from mass 1
r is just the radius of the pulley which can be measured experimentally.
And since ##\sin(\theta) = 1## in this scenario:
##\tau = F_{t1}r = I\alpha##

To determine ##F_{t1}##, the acceleration of mass 1 once dropped can be measured:
##F_{net} = F_g - F_{t1} = ma##
Acceleration can be measured using a motion sensor so:
##F_{t1} = m(g-a)##
Now that the torque done by ##F_{t1}## could be calculated, I could use rotational impulse to determine the rotational inertia.
##\Delta L = \tau_1r##
##\Delta L = I\Delta\omega##

Once rotational inertia is determined I then used energy:

##ME_i = ME_f## Where ME is mechanical energy
##ME_i = mgh##
##ME_f = KE_r = 0.5I\omega^2##
Since everything is known (after I measure them), you could then calculate g using the above equation. By making a graph of ##2m_1h## on the x axis, and ##I\omega^2## on the y axis, the slope is g.

However checking the answer key I found I was completely wrong. They first placed equal masses on each hanger, then removed some mass from one side to the other side. This created an inbalance in the net force, and since it was originally at rest with the balanced masses, transferring the mass essentially allowed them to know the net force acting upon the system. Which was the difference in weights of each hanging mass. From there they let the masses fall and recorded the acceleration of each mass. Each trial they increased the difference in masses. They then made a graph of ##\frac{m_1-m_2}{m_1+m_2}## on the x axis and on the y axis was the acceleration of the masses. The slope of this graph was the experimental value of gravity.

I'm still confused as to how they got their answer, but I did recognize some variables. If a graph of ##\frac{m_1-m_2}{m_1+m_2}## on x and ##a## on y gives a slope of g. Then ##g = \frac{a}{\frac{m_1-m_2}{m_1+m_2}}## which is ##g = \frac{a(m_1+m_2)}{m_1-m_2}## From this, I think that they got their equation from ##F = ma## since:

##g = \frac{a(m_1+m_2)}{m_1-m_2}##
Is the same as:
##(m_1-m_2)g = (m_1+m_2)a##
Which is pretty similar to F = ma, since the net force is indeed equal to ##(m_1-m_2)g## (I think)
However, couldn't you not use this since the pulley would need to be accounted for? I am assuming the pulley has non-negligible mass, since this is an experimental design.

My questions are:
Would my original strategy have worked?
How did they get their answer on the answer key?

You are right that the inertia of the pulley must be taken into account. Their equation should be ##g = \frac{a(m_1+m_2+I/r^2)}{m_1-m_2}##.
Your method is ok in principle but rather indirect. And this is not right:
##\Delta L = \tau_1r##
You mean ##\Delta L = \tau_1t##, which means you have to measure t as well as a.
If you are going to measure t, use the drop distance to calculate a and apply the corrected acceleration formula above..

uSee2
uSee2 said:
Would my original strategy have worked?
In theory: yes. But this is experimental physics, where considerable attention is paid to practical aspects, such as:
• How to make sure measurements can be done with some accuracy.
With a weight only on one side, the angular acceleration will be pretty high, so accurate measurements will be difficult, and the rope may well slip over the pulley.

uSee2 said:
How did they get their answer on the answer key?
The setup is that of an Atwood machine - you are re-creating a historical experiment!

uSee2 said:
I was completely wrong
No, you were not. Basically, you worked out ##m_2=0## and there is a possibility to improve on that.

##\ ##

uSee2 and Lnewqban
BvU said:
In theory: yes. But this is experimental physics, where considerable attention is paid to practical aspects, such as:
• How to make sure measurements can be done with some accuracy.
With a weight only on one side, the angular acceleration will be pretty high, so accurate measurements will be difficult, and the rope may well slip over the pulley.The setup is that of an Atwood machine - you are re-creating a historical experiment!No, you were not. Basically, you worked out ##m_2=0## and there is a possibility to improve on that.

##\ ##
So essentially they assumed that the mass of the pulley was negligible, just to get an answer with less degree of accuracy in exchange for an easier experiment? (Since they skipped getting the rotational inertia)

Apparently ! Haru gave the expression including moment of inertia, and it's absent from your book answer. But if I were grading this, there would be extra points for checking that pulley moment of inertia can be neglected .

##\ ##

uSee2
uSee2 said:
So essentially they assumed that the mass of the pulley was negligible, just to get an answer with less degree of accuracy in exchange for an easier experiment? (Since they skipped getting the rotational inertia)
If the combination of the hanging masses is right, the string is ligth and flexible, the diameter and mass of the pulley are small, and the axle is well lubricated, the angular acceleration of the pulley could be small enough as to induce a small error.

uSee2
uSee2 said:
So essentially they assumed that the mass of the pulley was negligible, just to get an answer with less degree of accuracy in exchange for an easier experiment? (Since they skipped getting the rotational inertia)
Or they simply overlooked it. Strikes me as an unwarranted assumption, likely to lead to a significant underestimate.
If, instead of keeping the total mass constant, you were to simply increase one mass incrementally then, in principle, you could take the pulley's inertia as being an extra unknown and deduce it from the set of measurements. But I don't see a way to do that by simple linear regression.

uSee2
Lnewqban said:
the diameter and mass of the pulley are small
Tricky to arrange in combination with large suspended masses.

uSee2
BvU said:
Apparently ! Haru gave the expression including moment of inertia, and it's absent from your book answer. But if I were grading this, there would be extra points for checking that pulley moment of inertia can be neglected .

##\ ##
haruspex said:
Tricky to arrange in combination with large suspended masses.
Do you think that this was an oversight on the maker's part? Or was there supposed to be a way for me to determine that the pulley had negligible rotational inertia?

If this was a test on a final and I got points deducted, do you think I would have a case for getting my points back?

uSee2 said:
Do you think that this was an oversight on the maker's part? Or was there supposed to be a way for me to determine that the pulley had negligible rotational inertia?

If this was a test on a final and I got points deducted, do you think I would have a case for getting my points back?
What did you get as an estimate for the moment of inertia ?

This is PF, not a court house ....

uSee2
BvU said:
What did you get as an estimate for the moment of inertia ?

This is PF, not a court house ....
Haha, youre right. This wasn't actually on a test, it was a practice test. I'm just curious as if there was a supposed way to determine that the pulley's rotational inertia was negligible, or if my solution is an alternative solution.

uSee2 said:
... I'm just curious as if there was a supposed way to determine that the pulley's rotational inertia was negligible, or if my solution is an alternative solution.
Could you perform and compare both tests, including and excluding the calculated rotational inertia of the pulley?

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html#dis

uSee2
Lnewqban
haruspex said:
Tricky to arrange in combination with large suspended masses.
To add detail to that, if we model the pulley as a uniform disc, mass M, and we want at most a 1% error then we need ##M<\frac{m_1+m_2}{50}##. But if the masses are vertical cylinders then to avoid collisions we need the pulley's diameter to be  more than the average of their diameters.

Last edited:
uSee2 and BvU
Of course you mean more, not less.

haruspex and uSee2
nasu said:
Of course you mean more, not less.
Why? If we are to ignore M but want to stay under a threshold for the resulting error then that puts an upper bound on M.

uSee2
Sorry, I mean the diameter of the pulley gas to be more, not less. Your last sentence in post 14.

uSee2
haruspex said:
To add detail to that, if we model the pulley as a uniform disc, mass M, and we want at most a 1% error then we need ##M<\frac{m_1+m_2}{50}##. But if the masses are vertical cylinders then to avoid collisions we need the pulley's diameter to be less than the average of their diameters.
If I may ask, how did you arrive at that equation?

nasu said:
Sorry, I mean the diameter of the pulley gas to be more, not less. Your last sentence in post 14.
Ah yes, thanks and corrected.

uSee2
uSee2 said:
If I may ask, how did you arrive at that equation?
The approximation substitutes a factor ##m1+m_2## for ##m1+m_2+I/r^2=m1+m_2+M/2##. For that to be less than 1% different, ##M<\frac{m_1+m_2}{50}##.

uSee2
haruspex said:
The approximation substitutes a factor ##m1+m_2## for ##m1+m_2+I/r^2=m1+m_2+M/2##. For that to be less than 1% different, ##M<\frac{m_1+m_2}{50}##.
haruspex said:
You are right that the inertia of the pulley must be taken into account. Their equation should be ##g = \frac{a(m_1+m_2+I/r^2)}{m_1-m_2}##.
Your method is ok in principle but rather indirect. And this is not right:
##\Delta L = \tau_1r##
You mean ##\Delta L = \tau_1t##, which means you have to measure t as well as a.
If you are going to measure t, use the drop distance to calculate a and apply the corrected acceleration formula above..
In your original equation, I don't understand why that it works. I do see that ##I/r^2## gives mass, since ##I = bmr^2## where b is some fraction, but shouldn't the mass of the pulley be included in some different way? Can you just include it in the mass of the system just like that?

uSee2 said:
Can you just include it in the mass of the system just like that?
If you work through the equations, that's how it turns out. Try it.
If you were to double the radius of the pulley (same mass) that doubles the torque the the forces exert on the pulley and halves the angular acceleration for the same linear acceleration. That all cancels out.

uSee2
haruspex said:
You are right that the inertia of the pulley must be taken into account. Their equation should be ##g = \frac{a(m_1+m_2+I/r^2)}{m_1-m_2}##.
Your method is ok in principle but rather indirect. And this is not right:
##\Delta L = \tau_1r##
You mean ##\Delta L = \tau_1t##, which means you have to measure t as well as a.
If you are going to measure t, use the drop distance to calculate a and apply the corrected acceleration formula above..
haruspex said:
If you work through the equations, that's how it turns out. Try it.
If you were to double the radius of the pulley (same mass) that doubles the torque the the forces exert on the pulley and halves the angular acceleration for the same linear acceleration. That all cancels out.

Very sorry for the 7 day late response, I was unable to respond temporarily. To prove your equation above, I tried to derive the tensions and then solve for g, but then I was unable get pass a road block. Here are my derivation:

My FBD's:

Assuming ##m_1## > ##m_2##
For the force of tension acting on Mass #1:
##F_{g1} - F_{t1} = m_1a##
##F_{t1} = m_1g - m_1a##
##F_{t1} = m_1(g-a)##

For the force of tension acting on Mass #2:
##F_{t2} - F_g = m_2a##
##F_{t2} = m_2(g+a)##

Then putting it into the torque of the pulley:
##\tau_{net} = I\alpha##
##\tau_{net} = F_{t1}r - F_{t2}r##
Substitution from above
##\tau_{net} = m_1r(g-a) - m_2r(g+a)##
##I\alpha= m_1r(g-a) - m_2r(g+a)##
Distribution
##I\alpha = m_1rg - m_1ra - m_2rg - m_2ra##
##I\alpha = g(m_1r - m_2r) - a(m_1r + m_2r)##
Adding and division to get:
##g = \frac{(I\alpha + a(m_1r + m_2r))}{(m_1r - m_2r)}##
And since ##\alpha = \frac{a}{r}##
##g = \frac{I\frac{a}{r} + a(m_1r + m_2r)}{m_1r - m_2r}##
Dividing r out
##g = \frac{I}{r^2}a + \frac{a(m_1 + m_2)}{m_1 - m_2}##
And then factoring out an ##a##
##a(\frac{I}{r^2} + \frac{(m_1 + m_2)}{m_1r - m_2}) = g##

I was able to get a really similar equations to yours in Post #2, but I'm unsure how to merge the fractions together. How did you get your equation from here?

uSee2 said:
##g = \frac{I\frac{a}{r} + a(m_1r + m_2r)}{m_1r - m_2r}##
Dividing r out
##g = \frac{I}{r^2}a + \frac{a(m_1 + m_2)}{m_1 - m_2}##
How did the I term escape from the numerator?
Use your ##I=bmr^2## formulation to see better how the masses combine.

haruspex said:
How did the I term escape from the numerator?
Use your ##I=bmr^2## formulation to see better how the masses combine.
Thank you so much!! I don't know how I made that mistake. So:
##g = \frac{I\frac{a}{r} + a(m_1r + m_2r)}{m_1r - m_2r}##
Divide out r (properly this time)
##g = \frac{a\frac{I}{r^2} + a(m_1+m_2)}{m_1-m_2}##
Then factor out an a to get your equation:
##g = \frac{a(\frac I r^2 + m_1 + m_2)}{m_1-m_2}##
Were my steps above correct?

And then to get your 1% approximation equation, you said that I/r^2 = 1/2 M where M is the mass of the pulley and 1/2 because that is the rotational inertia of a solid Cylinder object (assuming that the pulley can be modelled as such)

Then you solved for M in:

##m_1 + m_2 + \frac I {r^2} = m_1 + m_2 + \frac M 2##
Where the left hand side is the actual model, and the right hand side is what it would be if the pulley was modelled as a solid Cylinder.

However, I am still confused as to how you proceeded from there. If solved for M:
## M = \frac {2I} {r^2} ## And I'm not sure how to get 50 from this.

Last edited:
uSee2 said:
However, I am still confused as to how you proceeded from there.
You have two expressions for the acceleration, one considering I and one ignoring it. What is the ratio between the two expressions?

uSee2
haruspex said:
You have two expressions for the acceleration, one considering I and one ignoring it. What is the ratio between the two expressions?
##\frac {m_1 + m_2 + \frac {I} {r^2}} {m_1 + m_2 + \frac M 2} = 101/100## (The 101/100 since 1% error)
Cross multiplication
##101(m_1 + m_2 + \frac {M} {2}) = 100(m_1 + m_2 + \frac I {r^2})##
Distributing and Subtracting
##m_1 + m_2 + 101 \frac M 2 = 100 \frac I {r^2}##
From here, I'm still uncertain as to how to get rid of I within the equation. I know that ##\frac I {r^2}## is some fraction of mass ##M##, but since that fraction is not known we can't get rid of I?

uSee2 said:
##\frac {m_1 + m_2 + \frac {I} {r^2}} {m_1 + m_2 + \frac M 2} = 101/100## (The 101/100 since 1%)
Cross multiplication
##101(m_1 + m_2 + \frac {M} {2}) = 100(m_1 + m_2 + \frac I {r^2})##
Distributing and Subtracting
##m_1 + m_2 + 101 \frac M 2 = 100 \frac I {r^2}##
From here, I'm still uncertain as to how to get rid of I within the equation. I know that ##\frac I {r^2}## is some fraction of mass ##M##, but since that fraction is not known we can't get rid of I?
I did not mean those two expressions. I wrote, the one considering I (either ## {m_1 + m_2 + \frac {I} {r^2}}## or the uniform disc version ##{m_1 + m_2 + \frac M 2}##) and the one ignoring I, just ##m_1 + m_2##.

uSee2
haruspex said:
I did not mean those two expressions. I wrote, the one considering I (either ## {m_1 + m_2 + \frac {I} {r^2}}## or the uniform disc version ##{m_1 + m_2 + \frac M 2}##) and the one ignoring I, just ##m_1 + m_2##.
Oh I see now, thank you! So:
##\frac {m_1 + m_2 + \frac M 2} {m_1 + m_2} < \frac {101} {100}##
Cross multiplication
##100(m_1+m_2+\frac M 2) < 101(m_1 + m_2)##
##100m_1 + 100m_2 + 50M < 101m_1 + 101m_2##
##50M < m_1 + m_2##
##M < \frac {m_1 + m_2} {50}##

And then if ##\frac I {r^2}## was used instead
##100(m_1+m_2+\frac I {r^2}) < 101(m_1 + m_2)##
##100m_1 + 100m_2 + 100\frac I {r^2} < 101m_1 + 101m_2##
##100\frac I {r^2}< m_1 + m_2##
##\frac I {r^2} < \frac {m_1 + m_2} {100}##

All of this to get less than 1% of error right?

uSee2 said:
Oh I see now, thank you! So:
##\frac {m_1 + m_2 + \frac M 2} {m_1 + m_2} < \frac {101} {100}##
Cross multiplication
##100(m_1+m_2+\frac M 2) < 101(m_1 + m_2)##
##100m_1 + 100m_2 + 50M < 101m_1 + 101m_2##
##50M < m_1 + m_2##
##M < \frac {m_1 + m_2} {50}##

And then if ##\frac I {r^2}## was used instead
##100(m_1+m_2+\frac I {r^2}) < 101(m_1 + m_2)##
##100m_1 + 100m_2 + 100\frac I {r^2} < 101m_1 + 101m_2##
##100\frac I {r^2}< m_1 + m_2##
##\frac I {r^2} < \frac {m_1 + m_2} {100}##

All of this to get less than 1% of error right?
Yes.

uSee2
haruspex said:
Yes.
Thank you!

## 1. What is the purpose of using a pulley and mass hanger in an experiment?

The purpose of using a pulley and mass hanger in an experiment is to create a controlled and measurable system for studying the effects of different masses on a pulley. This allows for a more precise understanding of the relationship between force, mass, and acceleration.

## 2. How does a pulley and mass hanger affect the results of an experiment?

The pulley and mass hanger can affect the results of an experiment by providing a constant and consistent force to the system. This helps to reduce any external factors that may influence the results, making the data more reliable and accurate.

## 3. What is the difference between using a single pulley and a double pulley in an experiment?

A single pulley only changes the direction of the force, while a double pulley also reduces the amount of force needed to lift a mass. This means that a double pulley can be used to study the effects of smaller masses on a pulley system.

## 4. How do you set up a pulley and mass hanger for an experiment?

To set up a pulley and mass hanger, first attach the pulley to a fixed point, such as a stand or table. Then, add the desired mass to the hanger and thread the string through the pulley. Finally, attach the string to the hanger and adjust the height of the pulley as needed.

## 5. What are some potential sources of error when using a pulley and mass hanger in an experiment?

Potential sources of error when using a pulley and mass hanger in an experiment include friction in the pulley, variations in the string tension, and human error in recording data. It is important to minimize these sources of error through careful setup and repeated trials to ensure accurate results.

### Similar threads

• Introductory Physics Homework Help
Replies
15
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
928
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
68
Views
10K
• Introductory Physics Homework Help
Replies
26
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
8
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
5K
• Introductory Physics Homework Help
Replies
5
Views
1K