hey guys, I have a question regarding how to get the potential energy for this. I can get the correct answer, but the solutions do a step that makes absolutely no sense to me, so hopefully someone leads me in the right direction(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Using energy method,

derive the equation of motion for the pendulum which consists of the slender uniform rod of mass m and the bob of mass M. assume small oscillations, and neglect the radius of the bob.

***see attachment for diagram***

2. Relevant equations

[tex] I_o = \frac{1}{3}mL^2 + ML^2 [/tex] is the moment of inertia

[tex] V = mgh [/tex]

is the gravitational potential energy

[tex] T = \frac{1}{2}I_o \omega^2 = \frac{1}{2}I_o \dot{\theta}^2 [/tex]

is the kinetic energy

[tex] \frac{d}{dt}(T+V)=0 [/tex]

to get equation of motion

3. The attempt at a solution

My expression for V is this;

[tex] V = -mg\frac{L}{2}cos(\theta)-MgLcos(\theta) [/tex]

where I have taken the datum as being at the pivot acting postive up. This is different to the solution, which has written

[tex] V_{solution} = mg\frac{L}{2}(1-cos(\theta))+ MgL(1-cos(\theta)) [/tex]

I don't understand why they have 1 - cos(theta). Wouldn't you just want the cosine of the angle?

Anyway, if I continue to the derivtive step (doing the small angle approximation), this doesn't seem to have an effect. So...

[tex] \frac{d}{dt}(T+V)=\frac{d}{dt}\left ( \frac{1}{2}(M+\frac{m}{3})L^2\dot{\theta}^2 -mg\frac{L}{2}cos(\theta)-MgLcos(\theta)\right ) =0 [/tex]

which yields, after alot of manipulation,

[tex] \ddot{\theta}+\frac{3g(m+2M)}{2L(m+3M)}\theta = 0 [/tex]

which is what the textbook says is correct.

is the textbooks expressionfor the potential energywrong in this case? or did I get lucky in getting the right answer? It doesn't make sense to me why they have 1 - cosine.

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# Homework Help: Equation of motion for pendulum with slender rod (energy method)

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