Determine the moment of inertia of the wheel

[skysunsand]

Homework Statement

A cord 3.0 m long is coiled around the axle of a wheel. The cord is pulled with a constant force of 40N. When the cord leaves the axle, the wheel is rotating at 2.0 rev/s. Determine the moment of inertia of the wheel and axle. Neglect friction. (Hint: The easiest solution is by the energy method)

Homework Equations

KE= 1/2 mv^2 + 1/2 I w^2
I= Mr^2

The Attempt at a Solution

I'm completely lost. I don't have a mass to work with, so I can't use these equations. But then I have no clue what equation I WOULD use. I know the answer is 1.5kg*m^2, but I have no clue where that comes from.
Ideally, I think I'd like the radius of the wheel, but I'm not told how many times that cord is wrapped around the wheel, so I can't use the circumference to find the radius...

 

[cepheid]

Homework Statement

A cord 3.0 m long is coiled around the axle of a wheel. The cord is pulled with a constant force of 40N. When the cord leaves the axle, the wheel is rotating at 2.0 rev/s. Determine the moment of inertia of the wheel and axle. Neglect friction. (Hint: The easiest solution is by the energy method)

Homework Equations

KE= 1/2 mv^2 + 1/2 I w^2
I= Mr^2

The Attempt at a Solution

I'm completely lost. I don't have a mass to work with, so I can't use these equations. But then I have no clue what equation I WOULD use. I know the answer is 1.5kg*m^2, but I have no clue where that comes from.
Ideally, I think I'd like the radius of the wheel, but I'm not told how many times that cord is wrapped around the wheel, so I can't use the circumference to find the radius...

The work-energy theorem says that the work done is equal to the change in kinetic energy. For rotation under constant torque, the work done is W = τΔθ, where Δθ is the total angular displacement.

The distance Δs, traveled by any point on the circumference of the wheel must be equal to rΔθ. However, it must also be equal to the length of rope that was unwound, and hence Δs = rΔθ = 3.0 m. So anyway, the work energy-theorem says that:$$\tau \Delta \theta = \frac{1}{2}I\omega^2$$I think you'll find that if you plug in τ = rF and Δθ = Δs/r, then you'll find that it is not necessary to know r. You should be able to solve for I in terms of the other known quantities.

 

[wwj]

use intergration!

I=mr^2
and m= m/A * 2 pai r * dr (m/A= mass/area)
Intergrate mr^2 which is m/A * 2 pai r * dr * r^2
you will get 1/2 mr^2