Measuring the moment of inertia

[HclGuy]

Homework Statement

You are asked to measure the moment of inertia of a large wheel for rotation about an axis through its center. You measure the diameter of the wheel to be 0.740 m and find that it weighs 280 N. You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang a 8.00 kg mass from the free end of the rope. You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has a speed 5.00 m/s after it has descended 2.00m. What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center?

Homework Equations

E initial = mgh
E final = 1/2mv^2 + 1/2$$I$$$$\omega^2$$

The Attempt at a Solution

I tried using conservation of energy and then solving for $$I$$
But I don't think that is the way to go.

Any help is appreciated, Thanks!

 

[alphysicist]

Hi HclGuy,

What did you get from using conservation of energy?

 

[tiny-tim]

Hi HclGuy!

I tried using conservation of energy and then solving for $$I$$
But I don't think that is the way to go.

Well it should work.

Show us your working, so that we can see what's going wrong.

 

[HclGuy]

So what I did was
mgh = 1/2mv^2 + 1/2$$I$$$$\omega^2$$
Relate angular velocity back to linear velocity by using v = r * angular velocity
mgh = 1/2mv^2 +1/2$$I$$(v/r)^2
mgh - 1/2mv^2 = 1/2$$I$$(v/r)^2
2mgh - mv^2 = $$I$$(v/r)^2
$$I$$ = (2mgh - mv^2)(r/v)^2
m is 8.00 kg
h is distance the mass traveled = 2.00m
r= radius of wheel = 0.0370m
I get a really small number.. for my answer.. so that leads me to believe it is incorrect.

 

[alphysicist]

The radius of the wheel is 0.37 m, not 0.037 m. This would make the answer you're getting 100 times smaller than the real answer. Once you correct that, is the answer more reasonable?

 

[HclGuy]

Well , I ended up with .622 kgm^2, still not sure if that is right but it seems more reasonable..