# Mass on a string struck by an impulsive force

1. Oct 7, 2009

### nothingislost

A mass hangs on a string and is struck by a horizontal impulsive force that introduces an angular velocity w, using a lagrangian multiplier what are the conditions that causes the string to go slack.

the constraint is simply that the string is of constant length and therefore the radius of motion is of constant distance, r=a. I write the lagrangian equations for r and theta and then solve for the constraint force/multiplier. for the string to go slack the constraint force in r goes to zero and therefore lambda goes to zero.

solving for theta dot or omega i get that the condition for a slack string is that w=sqrt (g/a)

This doesnt make sense however as this is the condition when theta is at 180 and by logical reasoning i know that the string starts to go slack when theta max is just over the horizontal or 90.

What am i doing wrong here and how do i account for the fact that this is a string and not a solid rod?

thanks :)

Last edited: Oct 7, 2009
2. Oct 8, 2009

### tiny-tim

Hi nothingislost!

(have a square-root: √ and an omega: ω and a theta: θ )
This is the same problem as a mass moving on the inside of a sphere … the string going slack corresponds to the object losing contact with the sphere.

But if you want us to check what went wrong, you'll need to show us your full calculations.

3. Oct 9, 2009

### nothingislost

Ok. So my lagrangian is

$$L = \frac{1}{2}m(\dot r^2 + r^2 \dot \theta ^2 ) + mgr\cos (\theta )$$

with the constraint that r=a or f=r-a

from this i can get the equations of motion for $$\theta$$ and $$\omega$$

next i solve for the lagrangian multiplier and get $$\lambda$$ for the r equation.

but i feel that i have not accounted for the fact that this is a string and not a solid rod with my constraint. how can i account for this?

I end up solving for the slack condition where the constraint for on r goes to zero and get

$$\omega=\sqrt (g/a)$$

and i know this ofcourse cant be true for a string. what am i missing?

4. Oct 9, 2009

### Pengwuino

You solved for $$\omega$$? It is given in the problem. Remember, $$\omega$$ could be so high that it results in the mass fully rotating around the center without going slack.

5. Oct 9, 2009

### nothingislost

ok so these are my two equations of motion

$$mg\cos\theta + ma\dot\theta^2=\lambda$$

and

$$ma^2\ddot\theta + mga\sin\theta = 0$$

plus the equations of constraint that r=a.

then what i was doing was setting lambda to zero to find where the tensions force goes to zero for slack situation. is this the right direction?

Last edited: Oct 9, 2009
6. Oct 12, 2009

### nothingislost

No responses :( Help.

When i go further i get that $$\omega^2=(g/a)(\cos\theta - 2)$$