 #1
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Main Question or Discussion Point
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
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
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