Engineering Why wasn’t gravity included in the potential energy for this problem?

[SafiBTA]

Homework Statement

Derive the equation of motion using the principle of conservation of energy

Relevant Equations

T = ½ m ẋ² + ½ J₀ θ̇²

x = 4 r θ

U_spring = ½ k (4 r θ)²

(What’s confusing me is: shouldn’t there also be a gravitational potential term U_g = m g x ?)

I’m looking at the attached vibration problem. The solution in the manual includes the spring potential energy but does NOT include the gravitational potential energy of the hanging mass.

Can someone explain why gravitational potential energy is not included when deriving the equation of motion? I tried asking ChatGPT but kept going in circles and couldn't figure out.

Thanks!

 

[PeroK]

Why not add GPE to the analysis and see what equations you get?

 

[SafiBTA]

Why not add GPE to the analysis and see what equations you get?

As I understand, there will be an mgx term that will turn into mg after differentiation

 

[PeroK]

Have you ever studied a mass oscillating vertically on a spring? It might be easier to look at that problem first and figure out why gravity cancels out of the equations. That is to say, the period of the SHM depends only on the spring and not on gravity. See here, for example:

https://phys.libretexts.org/Courses...anical_Phenomena/2.05:_Spring-Mass_Oscillator

The problem you have is more advanced and I suspect that the author expected you to know this already from your previous studies.

 

[SafiBTA]

If I include the GPE, I get:

U_spring = ½ k (4 r θ)²
U_gravity = m g (r θ)

So the total potential energy is:

U = 8 k r² θ² + m g r θ

Now include the kinetic energy:

T = ½ (m r² + J₀) θ̇²

Now apply conservation of energy:

d/dt (T + U) = 0

Therefore the equation of motion is:

(m r² + J₀) θ̈
+ 16 k r² θ
+ m g r
= 0

 

[PeroK]

If I include the GPE, I get:

U_spring = ½ k (4 r θ)²
U_gravity = m g (r θ)

So the total potential energy is:

U = 8 k r² θ² + m g r θ

Now include the kinetic energy:

T = ½ (m r² + J₀) θ̇²

Now apply conservation of energy:

d/dt (T + U) = 0

Therefore the equation of motion is:

(m r² + J₀) θ̈
+ 16 k r² θ
+ m g r
= 0

This doesn't take into account that the equilibrium position depends on gravity. See the analysis I linked to.

 

[SafiBTA]

Thanks! I read through your link on the vertical spring mass.

I understand that:

- When the mass is hanging vertically, the spring is stretched but not "fully stretched".

- Vertical spring mass system can still be assumed as horizontal spring mass system.

- The displacement variable in the potential energy term is zero at this equilibrium position.

I still don't understand why potential energy due to the gravity would not be taken into account.

 

[PeroK]

Thanks! I read through your link on the vertical spring mass.

I understand that:

- When the mass is hanging vertically, the spring is stretched but not "fully stretched".

- Vertical spring mass system can still be assumed as horizontal spring mass system.

- The displacement variable in the potential energy term is zero at this equilibrium position.

I still don't understand why potential energy due to the gravity would not be taken into account.

It is taken into account. See figure 2.5.4 and the subsequent calculations. But, it cancels out of the final equation for SHM.

Once you know this, you don't necessarily have to include gravity in future problems. You can assume it's going to cancel out.

 

[vela]

I still don't understand why potential energy due to the gravity would not be taken into account.

Given the method, I think you should include the potential energy due to gravity (though you should've had $-mgx$). It feels a bit arbitrary to omit it just because you might happen to know from a different analysis it effectively won't matter. It's straightforward to see it does indeed cancel out.

As you found, including the gravitational potential energy makes the differential equation non-homogeneous with a constant forcing term. The particular solution is $\theta_0 = \tfrac{mg}{16kr}$. Note that this is the angle through which the pulley rotates to reach the new equilibrium point. The general solution to the differential equation is $\theta(t) = \theta_0 + \theta_h(t)$ where $\theta_h(t)$ is the solution to the homogeneous differential equation. In other words, the system oscillates about $\theta = \theta_0$.

Alternatively, you can take $\theta=0$ and $x=0$ to be around the new equilibrium point. Then the potential energy of the system should be
\begin{align*}
U &= \tfrac 12 k [4r(\theta + \theta_0)]^2 -mgx \\
&= \tfrac 12 (16kr^2)(\theta + \theta_0)^2 - mgr\theta\\
&= \tfrac 12 (16kr^2)\theta^2 + (16kr^2)\theta_0\theta + \tfrac 12(16kr^2)\theta_0^2 - mgr\theta\\
&= \tfrac 12 (16kr^2)\theta^2 + \tfrac 12(16kr^2)\theta_0^2
\end{align*} Since $16kr^2 \theta_0 = mgr$, the linear term from the spring's potential energy cancels the gravitational potential energy term, and you'll obtain the homogeneous equation after differentiating.