1. ## [SOLVED] Physics Problem

A rope of mass M and length L lies on a frictionless table, with a short portion Lo hanging through a hole. Initially the rope is at rest. Evaluate A and B so that the initial conditions are satisfied.

Ok, so I understand the problem and have gotten the final 2 equations (according to my book):
yAe^(yt)-yBe^(-yt)=0
Lo=Ae^(yt)+Be^(-yt)

It looks like, according to the first equation, Be^(-yt)=Ae^(yt). Plugging this into the second equation, Lo=2Ae^(yt) and Lo=2Be^(-yt). Is this all that is meant by "evaluate A and B"? If not, where do I go from here?

2. Originally Posted by davesface
A rope of mass M and length L lies on a frictionless table, with a short portion $L0$ hanging through a hole. Initially the rope is at rest. Evaluate A and B so that the initial conditions are satisfied.

Ok, so I understand the problem and have gotten the final 2 equations (according to my book):
yAe^(yt)-yBe^(-yt)=0
Lo=Ae^(yt)+Be^(-yt)

It looks like, according to the first equation, Be^(-yt)=Ae^(yt). Plugging this into the second equation, Lo=2Ae^(yt) and Lo=2Be^(-yt). Is this all that is meant by "evaluate A and B"? If not, where do I go from here?
What are A and B? What is your initial condition? Perhaps if you told us the whole problem we could help you better.

-Dan

3. A and B are just constants that the book gave as part of the answer. Actually, the problem devolves into a differential equation, and integrating A and B as constants was the only way to confirm that the equation was correct (according to the TA) since I have not taken a differential equations class.

Sorry about the initial conditions thing. I thought the equations made it clear, but initially t=0, and therefore v=0 (first equation). Also, Lo=x in the position equation, which I've already substituted in (second equation).

Also, I tried that physics forum earlier, and my post is already buried on page 2.

4. Originally Posted by davesface
A rope of mass M and length L lies on a frictionless table, with a short portion Lo hanging through a hole. Initially the rope is at rest. Evaluate A and B so that the initial conditions are satisfied.

Ok, so I understand the problem and have gotten the final 2 equations (according to my book):
yAe^(yt)-yBe^(-yt)=0
Lo=Ae^(yt)+Be^(-yt)

It looks like, according to the first equation, Be^(-yt)=Ae^(yt). Plugging this into the second equation, Lo=2Ae^(yt) and Lo=2Be^(-yt). Is this all that is meant by "evaluate A and B"? If not, where do I go from here?
Your initial conditions pertain to a time when t=0, which will simplify things.

RonL

5. The first equation would (obviously) simplify to A=B. The second equation would simplify to Lo=A+B, therefore Lo=2a or 2B. Again, the problem wasn't that I can't get to those equations. It's clear to me that this is where I'll end up using the equations I mentioned in the first post. My point was simply that the final answers seems (only intuitively) to be wrong. Is that just me being paranoid, or is there something else to this that I'm missing?

6. Originally Posted by davesface
The first equation would (obviously) simplify to A=B. The second equation would simplify to Lo=A+B, therefore Lo=2a or 2B. Again, the problem wasn't that I can't get to those equations. It's clear to me that this is where I'll end up using the equations I mentioned in the first post. My point was simply that the final answers seems (only intuitively) to be wrong. Is that just me being paranoid, or is there something else to this that I'm missing?
$A=B=L_0/2$

since $A$ and $B$ are parapeters that are determined by the problem and not vice versa.

RonL

7. Good point. I guess it just doesn't simplify any more than that. Thanks.