Rotating and translating spool across a table

[hadsox]

Homework Statement

A uniform spool of mass M and diameter d rests on end on a frictionless table. A massless string wrapped around the spool is attached to a weight m which hangs over the edge of the table. If the spool is released from rest when its center of mass is a distance l from the edge of the table, what is the velocity of the weight m when the center of mass of the spool reaches the edge of the table?

Homework Equations

The Attempt at a Solution

My attempt:
I thought of breaking up the problem into two cases and the combining them at the end.

case1: Pretend no rotation:

With no rotation the spool has forces Tension acting on it. T = Ma
The mass attached to the string has forces Tension and gravity. solved for T' = mg - ma
Since the acceleration for both we can get to [a = (mg)/(M+m)
So, we can get a final velocity of v = √(2*(mg)/(M+m)*l). where I started with v_f² = v_i²+2*a*l, l being the displacement of the spool on the table.

Case2: Pretend no translation:

With no translation, I believe then the Tension and Torque are equal to each other. Then we can get α = (τ/I). and we can get θ = l/(π*d),
What I end up using is the angular kinematics to get ω_f= √(2*(τ/I)*(l/(π*d))

So this is my work...am I on the rigth track or completely wrong? And how can I relate these two to get a uniform equation?

 

[vela]

Homework Statement

A uniform spool of mass M and diameter d rests on end on a frictionless table. A massless string wrapped around the spool is attached to a weight m which hangs over the edge of the table. If the spool is released from rest when its center of mass is a distance l from the edge of the table, what is the velocity of the weight m when the center of mass of the spool reaches the edge of the table?

Homework Equations

The Attempt at a Solution

My attempt:
I thought of breaking up the problem into two cases and the combining them at the end.

case1: Pretend no rotation:

With no rotation the spool has forces Tension acting on it. T = Ma
The mass attached to the string has forces Tension and gravity. solved for T' = mg - ma
Since the acceleration for both we can get to [a = (mg)/(M+m)
So, we can get a final velocity of v = √(2*(mg)/(M+m)*l). where I started with v_f² = v_i²+2*a*l, l being the displacement of the spool on the table.

Case2: Pretend no translation:

With no translation, I believe then the Tension and Torque are equal to each other.

Torque has units of force x length whereas tension has units of force. How can they possibly be equal to each other?

Then we can get α = (τ/I). and we can get θ = l/(π*d),

The table's frictionless, right? So the spool is probably going to slip, and your expression for ##\theta## won't hold.

What I end up using is the angular kinematics to get ω_f= √(2*(τ/I)*(l/(π*d))

So this is my work...am I on the rigth track or completely wrong? And how can I relate these two to get a uniform equation?

You can't look at the translational and rotational motion separately.

Note that the acceleration of the falling mass isn't going to be the same as the acceleration of the spool because the spool will unwind as the mass falls.

 

[haruspex]

I thought of breaking up the problem into two cases and the combining them at the end.

As vela notes, you cannot do that. For future reference, it was probably not a good strategy to do the work for the separate cases before having any idea how you would combine them.