Lagrangian: Pendulum down a slope

[WWCY]

Homework Statement

I have the answer for part a, which is:
$$\theta '' + \frac{a}{r} \cos \theta + \frac{g}{r} \sin \theta$$

My issue lies with getting the following equation of motion for part b,
$$\theta '' + \frac{g}{r} \cos \alpha \sin \theta = 0$$

Homework Equations

The Attempt at a Solution

Since the acceleration downhill is a constant, I took $a$ from part a's answer and replaced it with $g\sin \alpha$. I also replaced $g$ with $g\cos \alpha$ due to deal with the new component of gravity.

I get the following equation,
$$\theta '' + \frac{g\sin \alpha}{r} \cos \theta + \frac{g\cos \alpha}{r} \sin \theta = 0$$
which doesn't look like it's going to reduce to the target EOM.

Could anyone help point out what I'm missing? Thanks.

 

[Orodruin]

Please show your work.

 

[WWCY]

Please show your work.

Referencing the leftmost figure

Take $a = g\sin \alpha$

I defined coordinates and energy are defined as such (Origin at the fixed end of the rope),
$$X = x_0 + ut + \frac{1}{2} at^2 + r\sin \theta$$
$$Y = r\cos \theta$$
$$T = \frac{1}{2}m(\dot{X^2} + \dot{Y^2}) = \frac{1}{2}m [(u + at)^2 + 2r\dot{\theta} (u+at)\cos \theta + r^2\dot{\theta} ^2 \cos ^2 \theta]$$
$$U = -m \tilde{g} r\cos \theta$$
with $\tilde{g} = g\cos \alpha$, (this felt somewhat wonky to me)

Lagrangian:
$$L = T - U = \frac{1}{2}m(\dot{X^2} + \dot{Y^2}) = \frac{1}{2}m[(u + at)^2 + 2r\dot{\theta} (u+at)\cos \theta + r^2\dot{\theta} ^2 \cos ^2 \theta] + m \tilde{g} r\cos \theta$$

Applying Euler-Lagrange equation,
$$\frac{\partial L}{\partial \dot{\theta}} = (u - at)mr\cos \theta + mr^2 \dot{\theta}$$
$$ \frac{dP_{\theta}}{dt} = -\dot{\theta} (u + at)mr\sin \theta + amr\sin \theta + mr^2\ddot{\theta} $$
$$\frac{\partial L}{\partial \theta} = -(u + at)mr\dot{\theta}\sin \theta - m\tilde{g} r\sin \theta$$
Equating both parts of the equation and substituting $a$ and $\tilde{g}$ gives
$$\theta '' + \frac{a}{r} \cos \theta + \frac{\tilde{g}}{r} \sin \theta$$
$$\theta '' + \frac{g\sin \alpha}{r} \cos \theta + \frac{g\cos \alpha}{r} \sin \theta = 0$$

Thanks for your assistance.

 

[Orodruin]

You need to think more about the form of the potential.

Edit: For example, in the case of $a = 0$, e.g., the system being at rest, where will the minimum of the potential be? Your potential has a minimum at $\theta = 0$.

 

[WWCY]

For example, in the case of $a = 0$, e.g., the system being at rest, where will the minimum of the potential be? Your potential has a minimum at $\theta = 0$.

May I know why would potential be a minimum at $\theta = 0$ rather than $\theta = \alpha$? Is potential not defined with respect to "flat" ground?

Thanks for your assistance.

 

[Orodruin]

May I know why would potential be a minimum at $\theta = 0$ rather than $\theta = \alpha$? Is potential not defined with respect to "flat" ground?

Exactly, but your potential is not. The angle $\theta$ is an angle relative to the normal of the inclined plane.

Your potential would be correct if you considered everything in the accelerated frame, but you are not giving the kinetic energy in the accelerated frame, you are giving it in the frame of the ramp.

Edit: Of course, the elegant way of solving the problem is just to go to the accelerated frame and consider the known solution for the period of oscillations of a pendulum in a gravitational field with the gravitational field replaced by the gravitational acceleration plus the acceleration due to the additional inertial force.

 

[WWCY]

Exactly, but your potential is not. The angle $\theta$ is an angle relative to the normal of the inclined plane.

Your potential would be correct if you considered everything in the accelerated frame, but you are not giving the kinetic energy in the accelerated frame, you are giving it in the frame of the ramp.

I believe I was mixing up frames, apologies. So would I be right in saying that to do this calculation the ramp frame, I need to take potential energy relative to some imaginary line parallel to flat ground?

Thanks for your assistance.

 

[Orodruin]

I believe I was mixing up frames, apologies. So would I be right in saying that to do this calculation the ramp frame, I need to take potential energy relative to some imaginary line parallel to flat ground?

Yes.

Also note that if you want to compare the Lagrangians you write down in the accelerated and ramp frames, they will not be exactly equal. They will differ by a total derivative. Of course, this will not affect the equation of motion as total derivatives only affect the boundary terms and not the EL equations.

 

[WWCY]

Yes.

Also note that if you want to compare the Lagrangians you write down in the accelerated and ramp frames, they will not be exactly equal. They will differ by a total derivative. Of course, this will not affect the equation of motion as total derivatives only affect the boundary terms and not the EL equations.

I'm having some difficulty regarding defining the potential in the ramp frame, could you guide me through it?

Also, I'm not exactly familiar with non-inertial frames, what would I need to consider in coming up with the kinetic energy terms in such a frame?

Thank you for your continued patience.

 

[Orodruin]

I'm having some difficulty regarding defining the potential in the ramp frame, could you guide me through it?

Take a reference point fixed in the carriage. This reference point will have a known height $h(t)$ relative to your fixed reference point on the ramp as a function of time only and the potential of the mass is the potential relative to the reference point in the carriage plus $mgh(t)$. Since $mgh(t)$ is only a function of time, it is a total derivative (of its anti-derivative wrt $t$) and does not affect the equation of motion and you can simply drop it from your expression for the potential without screwing up your EoM. There remains only the potential relative to the fixed point in the carriage.

 

[WWCY]

There remains only the potential relative to the fixed point in the carriage.

Wouldn't this then be simply $mg\cos \theta$?

If so, wouldn't this mean that I'd only need to substitute $a = g\sin \alpha$, giving
$$\theta '' + \frac{g \sin \alpha}{r} \cos \theta + \frac{g}{r} \sin \theta$$
which still doesn't look too right.

Thanks!

 

[Orodruin]

Wouldn't this then be simply $mg\cos \theta$?

No. First of all, you are missing an $r$. More importantly, that has a maximal value when $\theta = 0$ and a minimal value when $\theta = \pi$. Where should your potential have its minimal value?

Edit: Note that when $\theta = 0$, the pendulum is hanging at an angle $\alpha$ relative to the vertical.

 

[WWCY]

No. First of all, you are missing an $r$. More importantly, that has a maximal value when $\theta = 0$ and a minimal value when $\theta = \pi$. Where should your potential have its minimal value?

Edit: Note that when $\theta = 0$, the pendulum is hanging at an angle $\alpha$ relative to the vertical.

Oops, I meant $-mgr\cos \theta$ if it makes it slightly more correct.

But since we are taking the flat ground as a reference, am i right to say that the minimal potential energy should be "centered" about $\alpha$ and therefore $U = -mgr\cos (\theta - \alpha)$?

Thanks for your assistance. It is greatly appreciated

 

[Orodruin]

But since we are taking the flat ground as a reference, am i right to say that the minimal potential energy should be "centered" about $\alpha$ and therefore $U = -mgr\cos (\theta - \alpha)$?

Yes, the angle to the vertical is $\theta - \alpha$ so this makes sense.