An Atwood's machine consists of two masses, m_{1} = 5 kg and m_{2} = 8 kg, which are connected by a light string that passes over a pulley. The pulley has a mass of 10 kg, a radius of 0.2 m, and a moment of inertia of 0.12 kg m^{2}. If the 5 kg mass is initially on the floor and the 8 kg mass is at a height of 2.0 m, what speed will the heavier mass have just before hitting the floor?

________________________________________________________________________________________________

I first drew my free body diagrams and set PE_{i} = KE_{f}.

All of the PE_{i} of the system comes from the height of the heavier block, so

PE_{i} = (2 m)(8 kg)(9.81 m/s^{2})

I then equated this to the KE_{f} of the system -

(1/2)(m_{1} + m_{2} kg)(v m/s)^{2} + (1/2)(I kg m^{2})(omega rad/s)^{2}

which with the numbers plugged in comes to

(1/2)(13 kg)(v m/s)^{2} + (1/2)(0.12 kg m^{2})(omega rad/s)^{2}

This is where I ran into difficulty. I'm assuming you can't add as is, due to the units being different. I can easily convert omega into v_{t} by

(omega)(radius) = v_{t}).

However, I'm not sure how to convert I into kg's (or do I just use the mass of the entire pulley?) I tried converting I into kg's by (0.12) / (.2)^{2}, but this came out to 3 kg, and the mass of the pulley is 10 kg.

Any advice would be appreciated.

Thanks,

Nick