Equation of motion for pendulum with slender rod (energy method)

[NEGATIVE_40]

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

Homework Statement

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***

Homework Equations

$$I_o = \frac{1}{3}mL^2 + ML^2$$ is the moment of inertia

$$V = mgh$$
is the gravitational potential energy

$$T = \frac{1}{2}I_o \omega^2 = \frac{1}{2}I_o \dot{\theta}^2$$
is the kinetic energy

$$\frac{d}{dt}(T+V)=0$$
to get equation of motion

The Attempt at a Solution

My expression for V is this;

$$V = -mg\frac{L}{2}cos(\theta)-MgLcos(\theta)$$
where I have taken the datum as being at the pivot acting postive up. This is different to the solution, which has written$$V_{solution} = mg\frac{L}{2}(1-cos(\theta))+ MgL(1-cos(\theta))$$
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...
$$\frac{d}{dt}(T+V)=\frac{d}{dt}\left ( \frac{1}{2}(M+\frac{m}{3})L^2\dot{\theta}^2 -mg\frac{L}{2}cos(\theta)-MgLcos(\theta)\right ) =0$$
which yields, after a lot of manipulation,
$$\ddot{\theta}+\frac{3g(m+2M)}{2L(m+3M)}\theta = 0$$
which is what the textbook says is correct.

is the textbooks expression for the potential energy wrong 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.

 

[rock.freak667]

The Attempt at a Solution

My expression for V is this;

$$V = -mg\frac{L}{2}cos(\theta)-MgLcos(\theta)$$
where I have taken the datum as being at the pivot acting postive up. This is different to the solution, which has written


$$V_{solution} = mg\frac{L}{2}(1-cos(\theta))+ MgL(1-cos(\theta))$$
I don't understand why they have 1 - cos(theta). Wouldn't you just want the cosine of the angle?

The change in height from when you displace it at the angle is L-Lcosθ or L(1-cosθ). Remember at the angle the length of the adjacent side is Lcosθ, but the original length was L, so change = L-Lcosθ

Anyway, if I continue to the derivtive step (doing the small angle approximation), this doesn't seem to have an effect. So...
$$\frac{d}{dt}(T+V)=\frac{d}{dt}\left ( \frac{1}{2}(M+\frac{m}{3})L^2\dot{\theta}^2 -mg\frac{L}{2}cos(\theta)-MgLcos(\theta)\right ) =0$$
which yields, after a lot of manipulation,
$$\ddot{\theta}+\frac{3g(m+2M)}{2L(m+3M)}\theta = 0$$
which is what the textbook says is correct.

is the textbooks expression for the potential energy wrong 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.

Why it worked without the 1-cosθ was that you would end up differentiating 1-cosθ, which would be like this

d/dt(1-cosθ) = d/dt(1) -d/dt(cosθ) = -d/dt(cosθ)

so you'd still end up just differentiating cosθ wrt to t alone.

 

[NEGATIVE_40]

so I need to be doing the CHANGE in height from the equilibrium position to some displaced position?
I thought you only needed the potential AT the displaced position, ie you don't care about where it ends up only where its at when you do the energy stuff?

I think why I was confused with this is because the textbook always has the datum at some convenient point so the potential for one position is zero.

so in the formula
$$\frac{d}{dt}(T+V)=0$$
should it really be
$$\frac{d}{dt}(\Delta T+\Delta V)=0$$
?

 

[vela]

You simply used a different zero point than what the solutions did. Your zero of potential energy is at the level of the pivot. In the solutions, there are two zeros, one for the rod and one for the mass. The zeros are where the mass and the center of mass are located when the pendulum is vertical. Neither method is incorrect.

 

[DeeNos]

First of all, great job on deriving the equation of motion for the pendulum using the energy method! Your approach and final result are both correct.

Now, regarding the potential energy expression, both yours and the textbook's expressions are actually correct, just written in slightly different ways. The reason for the 1-cosine term in the textbook's expression is because they have taken the datum as being at the lowest point of the pendulum, where the potential energy is zero. In this case, the height of the bob is equal to the length of the rod, so the potential energy expression becomes V = mgh + Mgh = (m+M)gh = (m+M)g(L-Lcos(theta)) = (m+M)gL(1-cos(theta)).

In your expression, you have taken the datum as being at the pivot point, where the potential energy is not zero. In this case, the height of the bob is equal to L/2, so the potential energy expression becomes V = mgh + Mgh = (m+M)g(L/2-Lcos(theta)) = (m+M)gL(1/2-cos(theta)).

Both expressions are equivalent, it just depends on where you take the datum to be. So to answer your question, the textbook's expression is not wrong, it's just written in a different way.

Keep up the good work in your studies of physics!