Experimental Design: Pulley and Mass Hangers

[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?

 

[haruspex]

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..

 

[BvU]

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.

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!

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]

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)

 

[BvU]

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 .

$\$

 

[Lnewqban]

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.

 

[haruspex]

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.

 

[haruspex]

the diameter and mass of the pulley are small

Tricky to arrange in combination with large suspended masses.

 

[uSee2]

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 .

$\$

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?

 

[BvU]

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]

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.

 

[Lnewqban]

... 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]

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

I don't think I am able to do this, as this was a test done on paper.

 

[haruspex]

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 [edit] more than the average of their diameters.

 

[nasu]

Of course you mean more, not less.

 

[haruspex]

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.

 

[nasu]

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

 

[uSee2]

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?

 

[haruspex]

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.

 

[haruspex]

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]

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, $M<\frac{m_1+m_2}{50}$.

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?

 

[haruspex]

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]

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..

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?

 

[haruspex]

$g = \frac{I\frac{a}{r} + a(m_1r + m_2r)}{m_1r - m_2r}$
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.

 

[uSee2]

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.

 

[haruspex]

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]

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?

 

[haruspex]

$\frac {m_1 + m_2 + \frac {I} {r^2}} {m_1 + m_2 + \frac M 2} = 101/100$ (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]

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?

 

[haruspex]

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]

Yes.

Thank you!