- #1

uSee2

- 57

- 7

- 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?