Use work/energy throem to derive an expression for moment of inertia?

[mattpd1]

Homework Statement

Today I had a physics lab. It involved a hanging mass on a string that hung off the table. It went through a pulley and the string was then spooled around a rotating spool on the table. We first measure the radius of the spool. Then we figured out the mass it took to counter act the friction of the spool. Call this Mf. I then calculated the work done by friction to be Mf*g*r (is this correct?) We then added a set mass to the hanger, and timed how long it took to hit the ground. From this, we calculated the final velocity of the falling mass Vf. The ultimate goal is to calculate the experimental value for the I, the moment of inertia of the spool, and compare this to the theoretical values (for a disk). I don't need any help with that part, but I do need help with some questions at the end!

The question verbatim is "Use Work and Mechanical Energy to derive the expression for the experimentally determined moment of inertia."

Homework Equations

mf=mass it took to overcome friction(this mass is always present)
m=extra mass added to hanging weight
(m+mf)=the entire falling apparatus
h=height the mass fell
r=radius of the spool
t=time it took to fall
vf=velocity of falling mass the instant it hit the ground
wf=work done by friction

I think this is all that was necessary...

The Attempt at a Solution

The formula for I was given to us as:

$$I = r^2[m(\frac{gt^2}{2h}-1)-m_{f}]$$

So I basically need to derive that formula from the work energy:

$$W_{f}=\Delta E=E_{f}-E_{i}$$

I know that:

$$E_{i}=mgh$$
$$E_{f}=\frac{1}{2}(m+m_{f})v_{f}^{2}+\frac{1}{2}I\omega _{f}^{2}$$

So how in the world do I go from work/energy -> moment of inertia?

 

[collinsmark]

Then we figured out the mass it took to counter act the friction of the spool. Call this Mf. I then calculated the work done by friction to be Mf*g*r (is this correct?)

Not r, but rather h. The work done by a force is W = Fd, where d is whatever distance that the force is applied. So if the block m_f falls a distance h at a constant velocity (constant velocity because it is counteracted by friction -- i.e. the acceleration is zero), then work done is W = Fd = m_f gh

The formula for I was given to us as:

$$I = r^2[m(\frac{gt^2}{2h}-1)-m_{f}]$$

So I basically need to derive that formula from the work energy:

$$W_{f}=\Delta E=E_{f}-E_{i}$$

I know that:

$$E_{i}=mgh$$
$$E_{f}=\frac{1}{2}(m+m_{f})v_{f}^{2}+\frac{1}{2}I\omega _{f}^{2}$$

So how in the world do I go from work/energy -> moment of inertia?

Set your initial and final energies equal to each other (conservation of energy*) and solve for I. 'Pretty simple really.

*Note that you have already taken frictional losses into account in your E_i part, so you don't need to worry about frictional losses any more. Technically, the initial energy is (m + m_f)gh, but if we subtract off the frictional losses from the beginning this becomes (m + m_f)gh - m_f gh = mgh.

There is still one little complication however. Your work energy equations have the variable v_f in them, but the given formula does not and has the variable t in it instead. But don't worry. You can convert it. You need to express v_f in terms of t and h. You can assume that the mass falls to the ground with uniform acceleration (although at the moment, you don't yet know what that acceleration is -- but you could find it if you had to). You already know what h and t are. Use your kinematics equations for uniform acceleration to express v_f as a function of t and h. Then substitute that into v_f in your other equation where you solved for I. And that should do it.

 

[mattpd1]

Hmm... vf as a function of t and h? Is it just vf=h/t?

Also, what about angular velocity? I assume it needs to be written in other terms as well. Does it equal v/r?

 

[collinsmark]

Hmm... vf as a function of t and h? Is it just vf=h/t?

No. That would be true if the block was falling at a constant velocity. But it's not. It's falling at a constant acceleration.

Given that the initial position and initial velocity are zero, use h = (1/2)at² to find the acceleration. Then use another one of your kinematics equations to find the final velocity. (You should be able to express the final velocity in terms of h and t).

Also, what about angular velocity? I assume it needs to be written in other terms as well. Does it equal v/r?

Yes, that's right!

 

[mattpd1]

So, 2h/t^2=a?
vf=2h/t^2*t=2h/t?

And when rewriting omega, should I use 2h/tr?

I still can't get it exactly right. I guess my algebra skills are lacking, because I went about it a few different ways, getting a different answer each time.

I start with:

$$mgh=\frac{1}{2}m(\frac{2h}{t})^2+\frac{1}{2}I(\frac{2h}{rt})^2$$

and is this right?:

$$(\frac{2h}{t})^2=(\frac{4h^2}{t^2})$$
$$(\frac{2h}{rt})^2=(\frac{4h^2}{r^2t^2})$$

Now subtract linear part?:

$$mgh-\frac{1}{2}m(\frac{4h^2}{t^2})=\frac{1}{2}I(\frac{4h^2}{r^2t^2})$$

This does not look right, what should I be doing?

 

[collinsmark]

So, 2h/t^2=a?
vf=2h/t^2*t=2h/t?

And when rewriting omega, should I use 2h/tr?

'Looks okay to me.

I still can't get it exactly right. I guess my algebra skills are lacking, because I went about it a few different ways, getting a different answer each time.

I start with:

$$mgh=\frac{1}{2}m(\frac{2h}{t})^2+\frac{1}{2}I(\frac{2h}{rt})^2$$

You forgot about the m_f part. But other than that, it looks right.

Start with

$$mgh=\frac{1}{2}(m+m_f) \left( \frac{2h}{t} \right)^2 +\frac{1}{2}I \left( \frac{2h}{rt} \right)^2$$

Now continue with the method you were doing before (algebra), and solve for I!