Energy Lost to Friction in Spool Experiment

[sullyjared]

[SOLVED] help! where can i find mass?

Homework Statement

A spool of thin wire (with inner radius R=0.50m outer radius R=o.65m, and a moment of inertia I_cm=0.8957 kg*m^2) pivots on a shaft. The wire is pulled down by a mass M=1.5 kg. After falling a distance D=0.51m, starting from rest, the mass has a speed of v=64.5 cm/s. Calculate the energy lost to friction during that time.

Homework Equations

KE=.5Iw^2
KE=.5mv^2
v=rw
I=.5M(r^2+R^2)

The Attempt at a Solution

I have made two attempted solutions for this problem. The first was simple. I neglected the spool all together. Then I solved for the potential energy of the mass pulling down and the kinetic energy of this mass, and found the difference... that didn't work.
My second attempt was to calculate the kinetic energy of the spool with KE=.5Iw^2 and compare it to the Kinetic energy of the mass pulling down but to solve for the spools kinetic energy i need the moment of inertia about the axis of rotation, which means i need the mass of the spool, which i don't have... any suggestions?

 

[hage567]

You were given the moment of inertia in the question.

 

[sullyjared]

yeah, i was given the moment of inertia about the center of mass, not the moment of inertia about the axis of rotation. which is I= I_cm + mr^2.

 

[physixguru]

If you are given the moment of inertia about the axis of rotation ..then you can calculate it about any axis.

 

[sullyjared]

that is true, but i am not given the MoI about the axis of rotation so then i am stuck.

 

[hage567]

Where do you think the axis of rotation is if it's not about the centre?

 

[sullyjared]

sorry... you all were right about the moment of inertia. when the axis of rotation is located on the center of mass then the moments of inertia are the same. i did finally figure this problem out and will post the method for others learning purposes.

D=given
M=given
g=given

First, you must calculate potential energy of the mass using PE=MgD. This should equal the kinetic energy of the falling mass (KE_m=1/2(M)(v^2) + the rotational kinetic energy of the spool (KE_s=1/2(I)(w^2) + the energy lost (to friction-x). Use w=v/r for angular velocity in the rotational kinetic energy.
finally, PE= KE_m + KE_s + x (where x is the only unknown and the variable asked for in the problem.

Thanks everyone for your help.

 

[boymk]

which r did you use?