Find Pendulum Length from Kinetic Energy & Angle Graph

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The discussion focuses on calculating the length of a simple pendulum using kinetic energy and angle data from a graph. The pendulum bob has a mass of 0.260 kg, and the maximum kinetic energy is given as 0.015 J. The approach involves relating gravitational potential energy to kinetic energy and using the equations of motion for simple harmonic motion (SHM). The user successfully computes the pendulum length to be 1.177 meters, confirming the methodology used in the calculations. This solution effectively demonstrates the relationship between energy, angle, and pendulum length.
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Homework Statement


The figure below shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.260 kg. What is the length of the pendulum?
W0353-N.jpg


Homework Equations


K=(1/2)mv2
U=mgh
x(t)=xm*cos(w*t+phi)

*where w is the angular frequency, and the wt + phi being the phase

The Attempt at a Solution



Basically I have gone through a series of approaches to this problem, but I have come to settle with this one:

I started by getting phi from the graph, .100 radians. Then, I attempted to relate gravitational potential energy to the max kinetic energy that was given from the graph (.015 J). This way, I could get h, and relate h to the maximum angle and the pendulum length. To do that, I would use x(t)=xm*cos(w*t+phi). I'm not sure if I should use the simple pendulum equations, but I don't think I need to know the inertia.
 
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x(t)=xm*cos(w*t+phi)
From the above equation find the maximum velocity x(t)=xm*cos(w*t+phi)m.
KE is given. Mass is given. Find the value of Vm.
Since angular displacement is only 0.1 rad, it is SHM.
In SHM f = -kx, ...(1)
In simple pendulum, restoring force = -mg(theta) = -mgx/L...(2)
From eq. 1 and 2. find the value of k = mg/L ...(3)
From 3 find the value of omega. And proceed.
 
Thanks for the help. : ) I computed the answer to be 1.177 m. This checked out. I understand the methodology behind it too, thanks again.
 
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