- #1

freutel

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## Homework Statement

Two identical disks with mass

*m*and radius

*r*are connected via a massless wire of length

*L*which is winded up around both disks. Disk B is connected to the ceiling and is free to rotate around its axis. Disk A is besides disk B and will fall due to the gravitational force with acceleration of

*g*.

**Visualization**

http://i.imgur.com/LZAk170.png

While disk A falls it will unwind. Question 1 is to find the acceleration of disk A - assuming it moves straight down. Question 2 is to find the speed of disk A when the massless wire has reached its maximum extension. Question 3 is to explain what happens when the wire has reached its maximum extension.

## Homework Equations

- F
_{g}=mg, ΣF=ma - Torque=Fr, Torque=Iα (I is moment of inertia and α is angular acceleration)
- I=½mr
^{2} - α=a
_{tan}/r (rolling without slipping) - Conservation of energy: K
_{1}+U_{1}=K_{2}+U_{2} - Kinetic energy(translational + rotational): K=½mv
_{cm}^{2}+ ½Iω^{2} - v
_{cm}=rω (rolling withouth slipping) - Potential energy: U=mgh

## The Attempt at a Solution

I'm stuck at question 2 but to know if I'm correct I have also put my answer for question 1 here:

For question one I divided the situation in two parts. First part is to determine the tangular acceleration of disk B. The force applied on disk B is F

_{g}from disk A which is

*mg*. By equalising both torque terms you get

*mgr = ½mr*. Which gives

^{2}*a_{tan}/r*a*

_{tan}=2g.The second part is to find the downwards acceleration of disk A. There are two forces applying on disk A and that is the tension of the rope

*T*and the

*F*Force

_{g}.*T*is the only force applying torque on disk A. So we can write

*ΣF=mg-T=ma*To solve this first we equalise the torque equations again:

_{cm}.*½mr**

^{2}*a*/

_{cm}*r = Tr.*This gives

*T=½ma*. Subbing this in the force equation you get:

_{cm}*g-½a*=

_{cm}*a*-->

_{cm}*a*.

_{cm}=2/3gThe total downwards acceleration of disk A is

*a*

_{tan}+a_{cm}=8/3g.For question 2 I started with the conservation of energy. The starting position is when both disks are at rest at the same height. There is no kinetic energy and only potential energy with height

*L.*When disk A falls the potential energy will turn into kinetic energy of both translation and rotation. Disk B stays at the same height but will also have kinetic energy of only rotation. Because disk B stays at the same height the potential energy of disk B will cancel out. So we get:

*U*=

_{A1}*K*.

_{A2}+ K_{B2}*mgL = ½mv*The speed at which disk A moves down due to the rotation of disk B is

_{cm}^{2}+ ½Iω_{A}^{2}+*½Iω*_{B}^{2}.*v*=rω

_{tan}_{B}. The speed at which disk A moves down due to its own rotation is

*v*. After subbing everything in you get:

_{cm}=rω_{A}*gL=¾v*

_{cm}^{2}

*+¼v*

_{tan}^{2}.

I know the total speed of disk A going down is

*v*but because of those fractions it's not that easy to sub it in. I have a feeling the solution is really easy but I cannot see it. Any help would be much appreciated.

_{tan}+v_{cm}For question 3 i have a feeling disk B will start the oscillate from right to left and disk A will move along with it as a pendulum.