# Motion on 2D surface - particle hanging from string

• Zatman
In summary, the conversation discusses the derivation of the expression \dot{r}^2 = A - \frac{B}{r^2} - gr, where r is the distance of one particle from a hole on a horizontal table and A and B are constants. The equations of motion for each particle in polar coordinates are used to obtain a system of equations. The conversation also mentions using conservation laws to derive the expression, specifically conservation of angular momentum.
Zatman
1. Homework Statement
Two particles of mass m are connected by a light inextensible string of length l. One of the particles moves on a smooth horizontal table in which there is a small hole. The string passes through the hole so that the second particle hangs vertically below the hole.

Derive the expression

$\dot{r}^2 = A - \frac{B}{r^2} - gr$

where r is the distance of the first particle from the hole, and A and B are constants. Use the equations of motion for each particle in polar coordinates.

## Homework Equations

General acceleration in polar coordinates:

$\mathbf{\ddot{r}}=(\ddot{r}-r\dot{\theta}^2)\mathbf{\hat{r}}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \mathbf{ \hat{\theta} }$

## The Attempt at a Solution

I'm using a plane-polar coordinate system in the plane of the table with the origin at the hole. The forces acting on the first particle (on the table) (in the plane) are just the tension, T. So:

$-T\mathbf{\hat{r}} = (\ddot{r}-r\dot{\theta}^2)\mathbf{\hat{r}}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \mathbf{ \hat{\theta} }$

And for the hanging particle, the forces are the tension and weight, so (aligning the z axis perpendicular to the table):

$(T-mg)\mathbf{\hat{z}} = -m\ddot{r}\mathbf{\hat{z}}$

From these equations, I get the following system:

$\ddot{r} - r\dot{\theta}^2 = -\frac{T}{m}$

$r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0$

$\ddot{r} = g - \frac{T}{m}$

I've tried manipulating these equations for a while now but don't get anything close to the required form. Are these equations incorrect? I would be very grateful for any hints! :)

Zatman said:
And for the hanging particle, the forces are the tension and weight, so (aligning the z axis perpendicular to the table):

$(T-mg)\mathbf{\hat{z}} = -m\ddot{r}\mathbf{\hat{z}}$

I believe there is a sign error here. Let ##h## be the distance from the hole to the hanging particle, pointing downward. Clearly, ##h = l - r##. Then ##\ddot h = mg - T ##. ##mg## is positive and ##T## negative because ##h## is chosen to point downward. So ## - \ddot r = mg - T ## or ## \ddot r = T - mg ##.

From these equations, I get the following system:

$\ddot{r} - r\dot{\theta}^2 = -\frac{T}{m}$

$r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0$

$\ddot{r} = g - \frac{T}{m}$

The second equation can be integrated. That can also be obtained by recalling conservation laws in a central force field.

1 person
voko said:
The second equation can be integrated. That can also be obtained by recalling conservation laws in a central force field.

There was actually a previous part of the question asking for the derivation of the same equation using conservation laws, so I can't use that here. I don't understand how that equation can be integrated.

Multiplied by ##r##, it becomes...?

1 person
voko said:
Multiplied by ##r##, it becomes...?

...exact. So this gives:

$r^2\dot{\theta} = k$

where k is a constant.

Rearranging for $\dot{\theta}$ and substituting into the first equation:

$\ddot{r} - r\frac{k^2}{r^4} = \ddot{r} - g$

where I have also used the other eqn. to eliminate T. EDIT: just realized this part is wrong, now correcting... This gives

$\frac{k^2}{r^3} = g$

$\frac{k^2}{r^2} = gr$

Which is close, but no constant term and no mention of $\dot{r}^2$

Last edited:
So I actually get:

$\ddot{r} - r\dot{\theta}^2 = -\ddot{r} - g$

$2\ddot{r} - r\dot{\theta}^2 + g = 0$

Now substituting $\dot{\theta}$, from above post:

$2\ddot{r} - \frac{k^2}{r^3} + g = 0$

$2r\ddot{r} - \frac{k^2}{r^2} + gr = 0$

So still not correct.

That is correct, but not finished. Use the same trick again - but I will let you find the integrating factor this time. The equation you are supposed to derive in the end is a good hint.

1 person
Oh, okay, I've got it now (multiply by $\frac{\dot{r}}{r}$).

Thank you so much for your help, voko.

Have you figured out the part with conservation laws? What law does the integral you found in #5 correspond to?

Yes, that was conservation of angular momentum, so that $k=\frac{L}{m}$

## 1. How does the force of gravity affect the motion of a particle hanging from a string on a 2D surface?

The force of gravity acts in a downward direction, causing the particle to accelerate towards the ground. This acceleration is known as the acceleration due to gravity, and it is constant for all objects near the Earth's surface.

## 2. What other forces are involved in the motion of a particle hanging from a string on a 2D surface?

In addition to gravity, there is also the tension force from the string that is holding the particle in place. This force acts in the opposite direction of gravity and keeps the particle from falling to the ground.

## 3. How does the length of the string affect the motion of the particle?

The length of the string does not affect the acceleration of the particle, but it does affect the speed at which the particle moves in a circular motion. A longer string will result in a slower speed, while a shorter string will result in a faster speed.

## 4. Can the particle ever stop moving while hanging from a string on a 2D surface?

No, the particle will continue to move in a circular motion as long as the string remains intact and there is a force of gravity acting on it. However, the speed and direction of the motion may change depending on the forces involved.

## 5. How can we calculate the acceleration and velocity of the particle in this motion?

To calculate the acceleration, we can use the formula a = F/m, where F is the resultant force (in this case, the force of gravity) and m is the mass of the particle. To calculate the velocity, we can use the formula v = ωr, where ω is the angular velocity (which can be found using the equation ω = √(g/r)) and r is the radius of the circular motion (equal to the length of the string).

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