Throwing baton modeled as inverted pendulum

[helemekoko]

Homework Statement

The juggler twirls a baton as high as she can into the air. To do this she applies a fast and powerful wrist flexion torque of T=70Nm. The baton will fly as high as possible when it leaves the girl's hand with its max take-off velocity. In order to maximize take-off velocity she must apply maximum flexion torque. The baton will leave the girl's hand when the vertical component of force exerted on the baton by the hand becomes zero.

While she is holding the baton, the system can be modeled by a single inverted pendulum (shown in the figure). Assuming that the baton begins from rest and is initially oriented parallel to the ground and 1.2 m above the ground, determine how high above the ground will the baton fly.

Given:
T_W = 70 Nm
Mass of baton, m = 0.75kg
Length of baton, l = 0.2m
Moment of inertia, I = 0.05 kgm²

Homework Equations

Kinematics

$x=lcosθ$
$y=lsinθ$

Force Balance

$F_{x}=m\ddot{x}$
$F_{y}=mg+m\ddot{y}$

Energy Equations

$mgh=\frac{1}{2}Iω^{2}+\frac{1}{2}mv_{takeoff}^{2}$
where
$ω=\dot{θ}$

Torque Equilibrium

$T_{W}=I\ddot{θ}-F_{x}lsinθ+F_{y}lcosθ$

The Attempt at a Solution

Found the Accelerations

$\ddot{x}=-l\ddot{θ}sinθ-l\dot{θ}^{2}cosθ$
$\ddot{y}=l\ddot{θ}cosθ-l\dot{θ}^{2}sinθ$

Multiplying by mass then subbing them into the Torque balance gave me

$T_{W}=I\ddot{θ}+ml^{2}\ddot{θ}+mglcosθ$

Rearranging

$\ddot{θ}(t)=\frac{T_{W}-mglcosθ(t)}{I+ml^{2}}$

Numerically Integrating to find θ(t) using initial conditions θ(0)=0 and ω(0)=0
After sketching graph, I see that the graph always increases, similar to a exponential graph.
How do I use the graph to help me find the height of the baton. Using energy equations I can get ω by integrating once via analytically then using the initial condition ω(0)=0. But i get stuck at finding V(take-off). Also once the baton leaves the hand I can't quite understand how the equations of the pendulum are useful

 

[Greg Bernhardt]

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