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

In summary, the physics lab involved a hanging mass on a string that was spooled around a rotating spool on a table. The radius of the spool was measured and the mass needed to counteract friction was determined. The work done by friction was calculated and the final velocity of the falling mass was found. The goal was to calculate the experimental value for the moment of inertia and compare it to the theoretical value for a disk. To do this, the work-energy equation was used to derive the expression for the experimentally determined moment of inertia, taking into account the frictional losses and using kinematics equations to express final velocity and angular velocity in terms of time and height.
  • #1
mattpd1
13
0

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:

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

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

[tex]W_{f}=\Delta E=E_{f}-E_{i}[/tex]

I know that:

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

So how in the world do I go from work/energy -> moment of inertia?
 
Physics news on Phys.org
  • #2
mattpd1 said:
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 mf 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 = mf gh
The formula for I was given to us as:

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

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

[tex]W_{f}=\Delta E=E_{f}-E_{i}[/tex]

I know that:

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

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. :smile: 'Pretty simple really.

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

There is still one little complication however. Your work energy equations have the variable vf 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 vf 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 vf as a function of t and h. Then substitute that into vf in your other equation where you solved for I. And that should do it. :wink:
 
  • #3
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?
 
Last edited:
  • #4
mattpd1 said:
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! :approve:
 
  • #5
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:

[tex]mgh=\frac{1}{2}m(\frac{2h}{t})^2+\frac{1}{2}I(\frac{2h}{rt})^2[/tex]

and is this right?:

[tex](\frac{2h}{t})^2=(\frac{4h^2}{t^2})[/tex]
[tex](\frac{2h}{rt})^2=(\frac{4h^2}{r^2t^2})[/tex]

Now subtract linear part?:

[tex]mgh-\frac{1}{2}m(\frac{4h^2}{t^2})=\frac{1}{2}I(\frac{4h^2}{r^2t^2})[/tex]

This does not look right, what should I be doing?
 
Last edited:
  • #6
mattpd1 said:
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. :approve:
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:

[tex]mgh=\frac{1}{2}m(\frac{2h}{t})^2+\frac{1}{2}I(\frac{2h}{rt})^2[/tex]
You forgot about the mf part. But other than that, it looks right.

Start with

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

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

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

1. What is the work-energy theorem?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. This means that the work done by all forces acting on an object will result in a change in its speed or direction, or both.

2. How does the work-energy theorem relate to the moment of inertia?

The work-energy theorem can be used to derive an expression for the moment of inertia, which is a measure of an object's resistance to rotational motion. This is because work done on an object can also result in a change in its rotational kinetic energy, which is affected by the object's moment of inertia.

3. What is the equation for the moment of inertia?

The equation for moment of inertia is I = mr^2, where I is the moment of inertia, m is the mass of the object, and r is the distance of the object from the axis of rotation. This is a simplified equation that assumes the object is a point mass.

4. How can the work-energy theorem be used to derive the moment of inertia for a rigid body?

The work-energy theorem can be used to derive the moment of inertia for a rigid body by considering the work done by all forces acting on the object, and the resulting change in rotational kinetic energy. This results in the equation I = Σmiri^2, where Σm is the sum of all the masses of the individual particles in the body, and ri is the distance of each particle from the axis of rotation.

5. Can the work-energy theorem be applied to non-rigid bodies?

Yes, the work-energy theorem can be applied to non-rigid bodies by considering the rotational kinetic energy of each individual particle in the body. This results in an integral equation for the moment of inertia, which takes into account the changing distances of the particles from the axis of rotation as the object moves.

Similar threads

  • Introductory Physics Homework Help
2
Replies
40
Views
3K
Replies
25
Views
542
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
678
  • Introductory Physics Homework Help
Replies
28
Views
576
  • Introductory Physics Homework Help
Replies
3
Views
908
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top