Unwinding a Light String from a Spool

[artsim2011]

Homework Statement

A light string 4.00m long is wrapped around a solid cylindrical spool with a radius of 0.075m and a mass of 0.500kg. A 5.00kg mass is then attached to the free end of the string, causing the string to unwind from the spool.

Homework Equations

I know I have to use the equations that are the transitions from acceleration force and other equations.

The Attempt at a Solution

I'm not sure how to identify the variables I have to masses and two distances. The line of thought I've had so far is that the 4.00m and the 5.00kg are going to be used to find the speed with which the string unwinds and the other two variables, 0.500kg and the radius 0.075m are going to be used to find the angular acceleration.

The only thing I can see that I could find from this would be the moment of inertia, but I don't need that as far as I can see. The only other thing that I think I might be missing is that the omega final will be 0.0rad/s. If you want more information or have more questions please post and I'll try and post what ever else I can.

 

[cepheid]

Welcome to PF artsim2011

You haven't posed any sort of question here. What is the problem asking for?

 

[artsim2011]

I try and make it more clear sorry about that.
The Question from the homework:
A light string 4.00m long is wrapped around a solid cylindrical spool with a radius of 0.075m and a mass of 0.500kg. A 5.00kg mass is then attached to the free end of the string, causing the string to unwind from the spool.
Find the angular acceleration.
Find the speed at which the string unwound from the spool.

I just got into rotational physics and I'm in a high school class. I know that I'm using the same equations that are used to find acceleration, force and the other stuff, but with different variable symbols and different units of measurements. I know the equations that I have to use, but I don't see which ones I have to use with the information given. I'm looking for someone to help me put the proper label on each variable given. If I'm still not being specific enough please let me know and thanks for looking.

 

[cepheid]

Because the string is unwinding from the spool, if the string moves by 1 cm, then a point on the circumference of the spool moves by 1 cm around as well. Therefore, it stand to reason that the speed of the mass (and the string) is the same as the speed of any point on the outside of the spool, and their accelerations are the same.

You can figure out the acceleration of the mass (and string) using Newton's laws. Therefore, you can figure out the acceleration of the spool. What is the relationship between acceleration and angular acceleration?

 

[artsim2011]

Okay I get what your saying about the relationship between the movement of the string and the movement of the spool. What I still don't understand is using Newtons laws or the angular acceleration equation. The equation for Newton's law is F=ma and changes to T=I X alpha. The acceleration equation is a = delta v / delta t. I only have the two masses and the two distances. I know I can find the moment of inertia with 1/2 mr^2. I also know that I can find alpha with a/d. I hope you can help me further because I'm really lost on this problem.

 

[cepheid]

Maybe I'm missing something, but I think that if you ignore friction, the mass at the end of the string is in free-fall, so you know its acceleration. Therefore, you know the acceleration of any point on the surface of the spool. From this, you can compute the spool's angular acceleration (using alpha = a/d as you stated).

I think the second part is asking what the speed is at the time when the string has become *completely* unwound, which you can just figure out since you know alpha, and you know the distance that has to be traveled (and therefore the total angular displacement of the spool). If this is not what the second part is asking, then I don't know how to interpret it, since the speed is not constant. It changes continuously.

This is entirely a kinematical problem. You shoudn't have to deal with torque.

 

[artsim2011]

Thanks for the help that you gave me, but we ended up going over the problem in class.
The solution:
We took the equation for torque for rotational and translational motion and put it together. I would explain the math, but I really don't completely understand it all. Starting equation is torque = Fdsin(theta) = I x alpha, I didn't copy all the steps for the equation I apologize so I will give the final result, alpha = (2)(g)(msub2)/(msub1)(r). msub1 is the 0.500 and 5.00 is the msub2.
Next we us S = (r)(delta theta), S is the arc length then we moved the equation around to get delta theta = S/r. then we used the rest of the given information to find omega final with this equation. omega final^2 = omega initial^2 + 2 x alpha x delta theta. We moved it around to get the square root of omega initial^2 + 2 x alpha x delta theta. If the variables that I listed don't sound right then I can try and throw together something more readable in an excel file and post it again. Hope someone gets some use out of this.