# Dynamics Problem involving ball-tension-mass HELP need ASAP

• Jun 12th 2007, 03:42 AM
hrshah
Dynamics Problem involving ball-tension-mass HELP need ASAP
I pretty much need an answer in 2 hours; plz n ty

http://img.photobucket.com/albums/v3...sQuestion2.jpg

I can get some parts but I generally think I don't know how to do it properly

i will post my attempted solution in a couple minutes
• Jun 12th 2007, 03:50 AM
topsquark
A question, though I can't guarentee I can get back to you before your time limit: Do you know Lagrangian methods?

If so then my suggestion is to set up your generalized coordinates as the coordinate of the CM of the block and the angle the rod makes.

I'll get back to you on this later today, when I can.

-Dan
• Jun 12th 2007, 03:53 AM
hrshah
Langranian method... I honestly have never heard of that; I know that you can solve it using components taking the ball as 1 body and the block as another, only thing common would b the tension in between and you could sum up forces in x and y direction .

Also, the prof said that there is no friction acting on the block so it "could" move, and when I was summing up forces I gave it I made it equal to =ma assuming its accelerating to the right when the swing moves right as well
• Jun 12th 2007, 05:26 AM
topsquark
Quote:

Originally Posted by hrshah
Langranian method... I honestly have never heard of that; I know that you can solve it using components taking the ball as 1 body and the block as another, only thing common would b the tension in between and you could sum up forces in x and y direction .

Also, the prof said that there is no friction acting on the block so it "could" move, and when I was summing up forces I gave it I made it equal to =ma assuming its accelerating to the right when the swing moves right as well

Ouch! (Winces in pain.) No Lagrangian method. There's probably a simple way to look at this problem, then, that will provide you with the answers. At the moment this kind of viewpoint is escaping me. (I've never tried to do one like this without using the Lagrangian method, which provides a short-cut out of the Physics logic in the more complicated problems.) I'll work at it, but no promises. Sorry!

-Dan
• Jun 12th 2007, 05:36 AM
hrshah
honestly, the more i look at these forums i get more excited about math.. alot is new to me thats one of the reasons its so cool.

I haven't heard of the method but if you could solve it that way and show me how maybe I can understand it and check solution with the way I do it. I really in the end just need solution no matter how I do it.

Thank again
Warmest Regards
• Jun 12th 2007, 10:49 AM
topsquark
Quote:

Originally Posted by hrshah

Well, I ran into a bit of a problem, but here's the general idea. We can (presumably) find the equations of motion of the block and the pendulum bob. From these we can derive all else.

I am going to write down the Lagrangian of the system, which is simply L = T - V, where T is the kinetic energy of the system and V is the potential energy of the system. Using a principle called "Conservation of Action" we may derive the following set of equations:
$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot{q_k}} \right ) - \frac{\partial L}{\partial q_k} = 0$
where the $q_k$ is a set of generalized coordinates of the system, $\dot{q_k}$ are defined as $\frac{dq_k}{dt}$ (and there are no nonconservative forces in the system.)

I'm not going to explain "generalized coordinates" to you. You can look those up if you like. I'll simply mention that we have two of them here: The position coordinate x of the block, and the angle $\theta$ the rod makes with the vertical. (I'm calling +x to the right and + $\theta$ counterclockwise.) I'm going to call the length of the pendulum b.

So we need to find L. First find T, the kinetic energy. The block has only a linear motion in the x direction so: $\frac{1}{2}M \dot{x}^2$ is the kinetic energy of the block.

The pendulum bob has the same linear velocity as the block, but also has rotational motion. If the pendulum were oscillating by itself the speed of the bob would be: $v_m = b \dot{ \theta}$. The overall velocity of the bob is the sum of the linear motion and the rotational motion, so:
$\bold{v} = \bold{v_M} + \bold{v_m}$

So $v^2 = v_M^2 + v_m^2 + 2 \bold{v_M} \cdot \bold{v_m} = v_M^2 + v_m^2 + 2v_Mv_m \, cos( \theta )$

So the overall kinetic energy of the system will be:
$T = \frac{1}{2}M \dot{x}^2 + \frac{1}{2} m(\dot{x}^2 + (b \dot{ \theta })^2 + 2 \dot{x} (b \dot{ \theta }) \, cos( \theta ) )$

Now to find V. I'm going to define the 0 point for the gravitational potential energy to be at the floor, with positive upward. The only potential energy in the system is due to the pendulum, so
$V = -mgb \, cos( \theta )$

Thus
$L = T - V = \frac{1}{2}(M + m) \dot{x}^2 + mb \dot{x} \dot{ \theta } \, cos( \theta ) + \frac{1}{2}mb^2 \dot{ \theta }^2 + mgb \, cos( \theta )$
(after some rearranging.)

Now we apply the equations of motion for x and $\theta$:
$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot{x}} \right ) - \frac{\partial L}{\partial x} = 0$

This gives
$\frac{d}{dt}((M + m)\dot{x} + mb \dot{\theta} \, cos( \theta ) ) = 0$

Or
$(M + m)\ddot{x} + mb\ddot{\theta} \, cos(\theta) - mb\dot{\theta}^2 \, sin( \theta ) = 0$

The other equation of motion gives:
$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot{\theta}} \right ) - \frac{\partial L}{\partial \theta} = 0$

This gives:
$mb \ddot{x} + mb^2 \ddot{ \theta } - mgb \, sin( \theta ) = 0$

So we need to solve the system of coupled (nonlinear) equations:
$(M + m)\ddot{x} + mb\ddot{\theta} \, cos(\theta) - mb\dot{\theta}^2 \, sin( \theta ) = 0$
$mb \ddot{x} + mb^2 \ddot{ \theta } - mgb \, sin( \theta ) = 0$

The problem is that after working on it for a bit I can't find a way to solve it! (Which is irritating. I distinctly remember doing this problem in grad school.) Perhaps you or someone else can make something of it.

-Dan
• Jun 13th 2007, 11:04 PM
hrshah
does anyone know how to work it out n actually get an answer?
• Jun 14th 2007, 06:44 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
"Conservation of Action" we may derive the following set of equations:
$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot{q_k}} \right ) - \frac{\partial L}{\partial q_k} = 0$

Just curious are those derived from Euler-Lagrange equations? Because it looks like it to me.
• Jun 14th 2007, 06:46 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Just curious are those derived from Euler-Lagrange equations? Because it looks like it to me.

(chuckles) They are the Euler-Lagrange equations. Nothing more, nothing less.

-Dan
• Aug 5th 2007, 05:54 AM
Rebesques
Topsq, you are a little distracted lately, is it woman-related? :p

Quote:

The other equation of motion gives:

$
\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot{\theta}} \right ) - \frac{\partial L}{\partial \theta} = 0
$

This gives:
$
mb \ddot{x}cos(\theta) + mb^2 \ddot{ \theta } - mgb \, sin( \theta ) = 0
$

and also the first equation is straightforwardly integrated: $
(M + m)\dot{x} + mb \dot{\theta} \, cos( \theta ) = c
$
. I think this where we use the IC (which I failed miserably to determine :o)
• Aug 5th 2007, 08:44 AM
topsquark
Quote:

Originally Posted by Rebesques
Topsq, you are a little distracted lately, is it woman-related? :p

$
mb \ddot{x}cos(\theta) + mb^2 \ddot{ \theta } - mgb \, sin( \theta ) = 0
$

and also the first equation is straightforwardly integrated: $
(M + m)\dot{x} + mb \dot{\theta} \, cos( \theta ) = c
$
. I think this where we use the IC (which I failed miserably to determine :o)

Let me just take the opportunity to say:

:o

Guilty as charged!

-Dan