# Linear momemtum principle problem

1. Jan 11, 2009

### pentazoid

1. The problem statement, all variables and given/known data

A uniform rope of mass M and length a is held at rest with its two ends close together and the rope hanging symmetrically below.((in this position, the rope has two long vertical segments connected by a small curved segment at the bottom.) One of the ends is then released. It can be shown by energy conservation that the velocity of the free end when it has descended by a distance x is given by

v^2= (x(2a-x))g/(a-x)

Find the reaction R exerted by the support at the fixed end when the free end has descended a distance x. The support will collapse if R exceeds 3/2Mg. Find how far the free end will fall before this happens.

2. Relevant equations

dp/dt= Mg-R, dp/dt is change in momemtum and R is the reaction force.
3. The attempt at a solution

Mass of moving rope is changing. At rest, T=0 . When rope is release T=1/2*Mv^2 and V= Mghr+Mghl.

hl=x, hr=1/2(a-x)

R=Mg-dp/dt

should I differentiate v^2 in order to obtain dv/dt and then should I proceed to multiply dv/dt by M in order to get dp/dt. Now that I have dp/dt I can calculate the R now right?

2. Jan 12, 2009

### pentazoid

Any body not understand my solution or the problem ?

3. Jan 12, 2009

### chrisk

The mass of the moving portion of the rope is not constant. Therefore,

p = mv

dp/dt = mdv/dt +vdm/dt

m = density*length = M/L*x

dm/dt = M/L*dx/dt = M/L*v

vdm/dt = v*M/L*v = M/L*v2

Your were given v2 in terms of x so dv/dt can be found. Therefore, dp/dt can be found as a function of x.

4. Jan 12, 2009

### pentazoid

For the second part of the problem , should I plug in R= 3/2Mg and solved for x?

5. Jan 13, 2009

### chrisk

Yes, your approach to the second part of the problem is correct.

6. Jan 14, 2009

### pentazoid

Not sure how to solve for x:

Last part of my second solution: 6*a=(2a^2-2ax-3x^2)/(a-x)

7. Jan 15, 2009

### chrisk

Multilply both sides of the equation by (a - x), distribute this factor on the left side of the equation with 6a. Then move all terms to one side thus setting one side equal to zero. This will give a quadratic equation in x equal to zero. Solve for x using the quadratic formula or completing the square.