How Does a Pendulum Behave When Its Support Accelerates Horizontally?

[Dazed&Confused]

Homework Statement

A pendulum is at rest with its bob painting toward the center of the earth. The support of the pendulum is moved horizontally with uniform acceleration $a$, and the pendulum starts to swing. Neglect the rotation of the earth. Consider the motion of the pendulum as the pivot moves over a small distance $d$ subtending at angle $\theta_0 ≈ d/R_e << 1$ at the center of the earth. Show that if the period of the pendulum is $2\pi \sqrt{R_e/g}$, the pendulum will continue to point toward the center of the earth, if effects of order ${\theta_0}^2$ and higher are neglected.

The Attempt at a Solution

I am not clear on how to tackle this. First of all, if the period of the pendulum for small angles is approximatly $2\pi \sqrt{R_e/g}$, then the moment of inertia is $m{R_e}^2$. This cannot be realistic for a simple pendulum. The question is part of an accelerated reference frame chapter so the forces on the pendulum in that frame are $-ma$ and $mg$. Any further advice would help. Thank you.

 

[BiGyElLoWhAt]

Why can that be realistic? Take a marble, treat it as a point mass. What's it's moment of inertia about pluto?

 

[Dazed&Confused]

Why can that be realistic? Take a marble, treat it as a point mass. What's it's moment of inertia about pluto?

Ok but isn't the pendulum close to the Earth?

 

[BiGyElLoWhAt]

Yes, it is. But what's the moment of inertia for a point mass? and where is it rotating about?

 

[Dazed&Confused]

Yes, it is. But what's the moment of inertia for a point mass? and where is it rotating about?

For a point mass it is $md^2$, with $d$ being the distance to the mass. I would say it is rotating about the support of the pendulum.

 

[BiGyElLoWhAt]

Look at the diagram you posted again.

 

[BiGyElLoWhAt]

Also note that in the problem statement the base of the pendulum is moving with a constant acceleration a.

 

[Dazed&Confused]

Look at the diagram you posted again.

I assumed that the diagram was drawn for clarity and not accuracy and since the angle is approximately $d/R_e$, the length of the pendulum is negligible compared to the Earth's radius. In short, the pendulum seems to have a base slightly above the Earth, and is accelerated with rate $a$.

Thank you for your patience by the way.

 

[BiGyElLoWhAt]

Yes the base is slightly above the earth, but basically what's being done here is this: the point of rotation is the center of the earth, the whole base of the pendulum is moving around the Earth's surface, and it is being assumed that the height of the pendulum is negligable wrt the radius of the earth.

 

[Dazed&Confused]

Yes the base is slightly above the earth, but basically what's being done here is this: the point of rotation is the center of the earth, the whole base of the pendulum is moving around the Earth's surface, and it is being assumed that the height of the pendulum is negligable wrt the radius of the earth.

Just to clarify, is the base the part of the pendulum that the string is attached to above the mass or is it essentially the mass? Again, I would say the height of the pendulum is what determines the moment of inertia as we are measuring the moment of inertia from the base (the former in the previous sentence).

Really sorry to keep responding like this.

From the illustration, I think the solid line represents the pendulum; the asterisks is the base and the circle is the mass.
Ok I think I see what you are saying but it doesn't seem to make sense to measure the moment of inertia from the centre of the Earth

 

[BiGyElLoWhAt]

Hey man not to leave you hanging like this, but I got to go for a few hours. Reread the problem statement where it talks about the subtending angle.