Pendulum: Calculating Work

[babaliaris]

I know that I'm rushing too much but I wanted to see if I can calculate the work of a pendulum that does oscillation with a similar way you calculate the work of a spring. Consider the following free body diagram:

Things i noticed:
1) The position is changing both in x and y-axis while the ball is doing oscillation.
2) The force from the rope to the ball (T) is changing only in direction while the ball changes position.

Things we know:
1) The height where the rope is tied on the ceiling which is the same with the rope's length.
2) The magnitude of the force from the rope to the ball (T)

First thing i did was to create a formula that describes the force T while the position of the ball changes.

$$\vec{T} = |\vec{T}| \cdot \vec{direction}$$

In order to find the direction vector all i need to do is subtract the position of the ball from the position of the wall where the rope is tied and normalize the outcome vector.

$$
\vec{diff} = ( 0\vec{i} + H\vec{j} ) - ( x\vec{i} + y\vec{j} )
= -x\vec{i} + (H - y)\vec{j}
$$

This gives me a vector that points from the ball's position towards the wall where the rope is tied and also the magnitude of this vector is the distance from the ball to the wall.

$$
\vec{direction} = \frac{\vec{diff}}{|\vec{diff}|}
= \frac{-x\vec{i} + (H - y)\vec{j}}{\sqrt{x^{2} + (H - y)^{2}}}
$$

But:
$$
|\vec{diff}| = \sqrt{x^{2} + (H - y)^{2}} = H
$$

because the distance from the ball to the wall is always the length of the rope.

So:
$$
\vec{direction} = \frac{\vec{diff}}{|\vec{diff}|}
= \frac{-x\vec{i} + (H - y)\vec{j}}{H}
$$

Finally:
$$
\vec{T} = |\vec{T}| \cdot \vec{direction}
= |\vec{T}| \cdot \frac{-x\vec{i} + (H - y)\vec{j}}{H}
$$

Now we can calculate the net force:
$$
\vec{F_{net}} = \vec{T} + \vec{F_{g}}
= |\vec{T}| \cdot \frac{-x\vec{i} + (H - y)\vec{j}}{H} - mg\vec{j}
$$

The work when the forces are constant is:
$$
work = \vec{F_{net}} \cdot \vec{d}
$$

Where
$$
\vec{d} = ( x_{2} - x_{1} ) \vec{i} + ( y_{2} - y_{1} ) \vec{j}
$$

Since T is changing while the position of the ball changes, we can't use that formula, but we can divide each displacement to small fractions with length close to zero and take the sum of all the works to calculate the final work. In other words we can integrate the formula of Fnet to the displacement d.

$$
work = \int_{y_{1}}^{y_{2}}\int_{x_{1}}^{x_{2}} [ T \cdot \frac{-x + (H - y)}{H} - mg] \cdot dxdy = ...
$$

I don't really believe that I'm correct, probably i missed something but am I close?

I'm just trying to take my mathematics skills to another level (from just solving exercise solving real problems!)

Thanks

 

[Doc Al]

For a force to do work, it must have a component parallel to the velocity of the object on which it acts. Is that the case here for the tension in the string?

 

[tnich]

$$
work = \int_{y_{1}}^{y_{2}}\int_{x_{1}}^{x_{2}} [ T \cdot \frac{-x + (H - y)}{H} - mg] \cdot dxdy = ...
$$

Not bad, so far. A couple of things for you to consider
1) Can you find a way to determine $|\vec T|$? Let me suggest that you look at the components of the force due to gravity in the $\vec {direction}$ direction and in the direction tangent to the path of the ball.
2) You are integrating over x and y. Do you really have two independent variables here? Given the value of x, what is the value of y?
3) You might be able to solve this problem more directly by considering conservation of energy.

 

[tnich]

Not bad, so far. A couple of things for you to consider
1) Can you find a way to determine $|\vec T|$? Let me suggest that you look at the components of the force due to gravity in the $\vec {direction}$ direction and in the direction tangent to the path of the ball.
2) You are integrating over x and y. Do you really have two independent variables here? Given the value of x, what is the value of y?
3) You might be able to solve this problem more directly by considering conservation of energy.

4) Also, force $\vec T$ is not very useful in this problem since what you want to find is the work against the force of gravity. Once again, the components of the force of gravity tangent and perpendicular to the ball's path would be useful.

 

[Chandra Prayaga]

$\vec T . d \vec r$ is always zero since $\vec T$ is always perpendicular to $d \vec r$

 

[babaliaris]

I understand what you told me so far. But let's say, if T was doing work, then the final integral which i showed you is correct? Or i must have something like:

$$
\int_{x_{1}}^{x_{2}} Fnet_{x} \cdot dx + \int_{y_{1}}^{y_{2}} Fnet_{y} \cdot dy
$$

?

Or maybe this is the same with the double integral? I'm not good with integrals this is why I'm asking.
I'm not really sure if i have understand correctly what is a double integral.

 

[BvU]

I understand what you told me so far.

No. re-read #3 and translate it for us to see what blocks you.

But let's say, if T was doing work, then the final integral which i showed you is correct?

No. See #3. Work is only done by the component parallel to the path: $W = \vec F \cdot\vec s$ for a straight path. For a curved path we subdivide in infinitesimally small steps $d\vec s$ and intergate along the path: $$dW = \vec F \cdot d\vec s \Rightarrow W = \int_{\rm path}\vec F \cdot d\vec s $$

The path is a one-dimensional beast, so no worrying about double integrals