 #1
SilverSoldier
 26
 3
 Homework Statement:
 Suppose we have a simple pendulum, let's say with mass ##m##, attached to a massless string of length ##l##, on an ##xy## coordinate plane as indicated in the diagram below. I want to model the motion of this pendulum over time.
 Relevant Equations:

##\tau=I\alpha##
Centripetal force = ##\dfrac{mv^2}{r}##
Potential energy = ##mgh##
Kinetic energy = ##\dfrac{1}{2}mv^2##
Conservation of energy
Suppose we displace the pendulum bob ##A## an angle ##\theta_0## initially, and let go.
This is equivalent to giving it an initial horizontal displacement of ##X## and an initial vertical displacement of ##Y##. Let ##Y## initially be a negative number, and ##X## initially be positive.
I observe that in this initial state, it has some potential energy ##mg(l+Y)## with respect to point ##B##. This energy must be conserved.
Next, we look at some such general position as ##A'## of the pendulum bob, where its horizontal and vertical displacements are ##x## and ##y## respectively (##y<0##), as shown below. Let us suppose it has a net velocity ##v## at this instant. Let ##\theta## be the angle it forms with the vertical, and consider this angle to be negative.
Here, the bob would have lesser potential energy, and the difference, will have converted into its kinetic energy. So we have
$$
\begin{align*}
mg(l+Y)&=mg(l+y)+\dfrac{1}{2}mv^2\\
v^2&=2g(Yy).
\end{align*}
$$
Now this pendulum is also undergoing nonuniform circular motion, so there must be a centripetal force directed towards the center to keep it going in the circle. This force must be provided by the tension force ##T## and the component of the weight force along the string. Therefore,
$$
\begin{align*}
Tmg\cos\theta&=\dfrac{mv^2}{l}\\
T&=mg\cos\theta+\dfrac{mv^2}{l}.
\end{align*}
$$
We know ##v^2## from the Principle of Conservation of Energy, so
$$
T=mg\cos\theta+\dfrac{2mg(Yy)}{l}.
$$
Now, we resolve the forces on the pendulum bob vertically and horizontally, and use Newton's Second Law to find the accelerations in these directions. If the net horizontal acceleration in the right direction is ##a_x##, and net vertical force in the upward direction is ##a_y##, we have
$$
\begin{align*}
a_x&=\dfrac{T\sin\theta}{m}\\
a_x&=g\cos\theta\sin\theta+\dfrac{2g\sin\theta (Yy)}{l},
\end{align*}
$$
and
$$
\begin{align*}
a_y&=\dfrac{T\cos\thetamg}{m}\\
a_y&=g\cos^2\theta+\dfrac{2g\cos\theta (Yy)}{l}g\\
&=\dfrac{2g\cos\theta (Yy)}{l}g\sin^2\theta.
\end{align*}
$$
We can also say that ##\sin\theta=\dfrac{x}{l}## and ##\cos\theta=\dfrac{y}{l}##.
The net horizontal and vertical accelerations ##a_x## and ##a_y## that the object experiences, in terms of its position ##(x,y)## are therefore
$$
a_x=\dfrac{gx}{l^2}\left(2Y3y\right)
$$
and
$$
a_y=\dfrac{g}{l^2}\left(x^2+2y(Yy)\right).
$$
Is this correct?