rotational physics question?

Question

A light string 4.00m long is wrapped around a solid cylindrical spool with a radius of .075m and a mass of .500kg. A 5.00 kg mass is then attached to the free end of the string, causing the string to unwind from the spool.

a. What is the angular acceleration of the spool

b. How fast will the spool be rotating after all of the string has unwound?

the answers are:
2600rad/s^2 (a)
530r/s (b)

If anyone can explain to me how to get those two answers i will greatly appreciate it, thanks.

Answer 1

That's too hard for us dummies out here..

Answer 2

A)

it is a problem that has to treated in two parts...What happens to the cylinder and what happens to the falling mass. It easy to resolve this problem with newton's 2nd law.

First let's write the second law for the falling mass

T-mg=-ma........eq 1
here my reference system is + upward and - downward

On the cylinder we have
To=I alpha....eq 2
where To is the torque = TR

I is the inertia of the cylinder=1/2(MR^2)

we are searching alpha..the angular acceleration

from eq 1 we have T=mg-ma

so the torque becomes

(mg-ma)R=1/2(MR^2)alpha

mg-ma=1/2(MR)alpha

we have two acceleration the linear acceleration and the angular acceleration wich are related together with alpha= a/R
then a=alpha R
if we mix that
mg-malphaR=1/2MR alpha
mg=1/2MRalpha+malphaR
isolating alpha

mg=alpha(1/2MR+mR)

alpha=mg/(1/2MR+mR)

here we have m=5 kg, R=.075m, M=0.5kg,g=9.8

B) we will find the solution by asking first how many radians does the cylinder turn

we know tha teta=L/R by definition of an angle soo here L= 4m and R=0.075m
then teta=4/.075 radians

to find the angular speed we will use the equation

Wf^2=Wo^2 +2 alpha (teta f-tetao)
knowing that Wo=0 and( teta f-tetao)=4/.075

then

Wf=( 2 alpha 4/.075)^0.5

alpha is the value found in question A


I am pretty sure of my solution so you will have to verify it and do the calculus

note.

This problem could be treated also by conservation of energy, that would be a good exercice for you...
good luck
P.A.