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sunquick
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Homework Statement
A plank of length 2L leans against a wall. It starts to slip downward without friction. Show that the top of the plank loses contact with the wall when it is at two-thirds of its initial height.
(from Kleppner&Kolenkow,An Introduction to Mechanics 1st ed., problem 6.41)
Homework Equations
Equation of motion for translation of center of mass:
The coordinates of the center of mass are
[tex] x = L \cos(\theta) [/tex]
[tex] y= L \sin(\theta) [/tex]
Differentiating twice with respect to time, the acceleration of CoM is
[tex] \ddot{x} = - L \ddot{\theta} \sin(\theta) - L \dot{\theta}^2 \cos(\theta) [/tex]
[tex] \ddot{y} = L \ddot{\theta} \cos(\theta) - L \dot{\theta}^2 \sin(\theta) [/tex]
So the equations of motion for the center of mass are
[tex] m \ddot{x} = - m L \ddot{\theta} \sin(\theta) - m L \dot{\theta}^2 \cos(\theta) = N_x [/tex]
[tex] m \ddot{y} = m L \ddot{\theta} \cos(\theta) - m L \dot{\theta}^2 \sin(\theta) = - m g +N_y[/tex]
Conservation of energy:
Since there is no friction, energy is conserved and so
[tex]m g y_0=\frac{1}{2} m v^2 + \frac{1}{2} I_0 \dot{\theta}^2 + m g y[/tex]
We can make the following substitutions
[tex] I_0 = \frac{1}{3} m L^2 [/tex]
[tex] v^2 = \dot{x}^2 + \dot{y}^2 = L^2 \dot{\theta}^2 [/tex]
So we get
[tex]m g y_0=\frac{1}{2} m L^2 \dot{\theta}^2 + \frac{1}{6} m L^2 \dot{\theta}^2 + m g L \sin(\theta)[/tex]
The final equation we need is the angular equation of motion, which we can get making use of the fact that energy is conserved. We can differentiate both sides of the conservation of energy equation and get
[tex] 0 = m L^2 \dot{\theta} \ddot{\theta} + \frac{1}{3} m L^2 \dot{\theta} \ddot{\theta} + m g L \dot{\theta} \cos(\theta) [/tex]
Which can be further simplified into
[tex] \ddot{\theta} = - \frac{3}{4} \frac{g}{L} \cos(\theta) [/tex]
One other way to get the angular equation of motion is to balance the torques around the axis of the center of mass
[tex] I_0 \ddot{\theta} = N_y L \sin(\theta) - N_x L cos(\theta) [/tex]
The Attempt at a Solution
I've been able to solve this problem and show what was asked, but I'd like to show you my first atttempt at solving it and understand what I did wrong and why was wrong.
The plank loses contact with the wall when [tex] N_x = 0 [/tex] so from the equation of motion for the x coordinate of the center of mass we get
[tex] - m L \ddot{\theta} \sin(\theta) - m L \dot{\theta}^2 \cos(\theta) = 0 [/tex]
And substituting into it [tex] \ddot{\theta} = - \frac{3}{4} \frac{g}{L} \cos(\theta) [/tex] we get the following equation
[tex] \dot{\theta}^2 \cos(\theta) - \left( \frac{3}{4} \frac{g}{L} \cos(\theta) \right) \sin(\theta) = 0 [/tex]
which can be solved to
[tex] \dot{\theta}^2 = \frac{3 g}{4 L} \sin(\theta) [/tex]
Knowing [tex] \ddot{\theta} [/tex] and [tex] \dot{\theta}^2 [/tex] I can make use of the equation of motion for the y coordinate of the center of mass and find the force [tex] N_y [/tex] at the instant the plank loses contact with the wall, at which time [tex] N_x = 0 [/tex].
[tex] N_y = m g + m L \cos(\theta) \left(- \frac{3 g}{4 L} \cos(\theta) \right) - m L \sin(\theta) \left(\frac{3 g}{4 L} \sin(\theta) \right) [/tex]
[tex] N_y = \frac{1}{4} m g [/tex]
The torque balance equation then reads
[tex] I_0 \ddot{\theta} = N_y L \sin(\theta) = \frac{1}{4} m g L \sin(\theta) [/tex]
[tex]\frac{1}{3} m L^3 (-\frac{3g}{4L} \cos(\theta) ) = \frac{1}{4} m g L \sin(\theta) [/tex]
Which I then tried to solve to find [tex] \theta [/tex] but ended up with something weird,
[tex] - \cos(\theta) = \sin(\theta) [/tex]
from which I can tell something went wrong along the way, and I suspect it is from finding the [tex] N_y [/tex] force, hence the title of the post.
What did I do wrong? Why is it wrong?