# Rotating and translating spool across a table

In summary: That's a recipe for wasted effort.I think you can solve this problem by considering the conservation of energy for the system, that is, the spool and the hanging mass. The tricky part is figuring out what the potential energy of the spool is for a given point in time because the spool is rotating as it unwinds. You can figure that out by noting that the spool is effectively a pulley of radius d/2 and multiplying by the circumference of the spool at any given time.

## 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?

## 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 vf2 = vi2+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 ωf= √(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?

## 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?

## 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 vf2 = vi2+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 ωf= √(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.

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.

Hajytel

## 1. How does the rotation of a spool affect its movement across a table?

When a spool is rotated, it creates a force known as torque. This torque causes the spool to move in a circular motion, which in turn affects its movement across a table. The direction and speed of rotation will determine the direction and speed of movement across the table.

## 2. What factors influence the translation of a spool across a table?

The translation of a spool across a table is influenced by several factors, including the force applied to the spool, the mass of the spool, and the surface properties of the table. Friction between the spool and the table can also affect the spool's translation.

## 3. How does the shape of the spool affect its rotation and translation across a table?

The shape of the spool can greatly impact its rotation and translation across a table. A spool with a larger diameter will have a greater torque and will rotate and translate faster than a spool with a smaller diameter. Additionally, the shape of the spool can also affect its stability and balance, which can impact its movement across the table.

## 4. Can the surface properties of the table affect the rotation and translation of a spool?

Yes, the surface properties of the table can have a significant impact on the rotation and translation of a spool. A surface with high friction will slow down the movement of the spool, while a surface with low friction will allow the spool to move more freely. The texture and material of the table can also play a role in the spool's movement.

## 5. How can we calculate the speed and direction of a spool's rotation and translation across a table?

The speed and direction of a spool's rotation and translation can be calculated using various mathematical equations, such as the equations for torque and rotational motion. The specific calculations will depend on the variables involved, such as the mass of the spool, the force applied, and the surface properties of the table. These calculations can help us understand and predict the movement of a spool across a table.

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