How Does Acceleration Affect Pendulum Deflection?

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The discussion centers on how a pendulum deflects when subjected to a sideways force (Fh) while accelerating. It explores the relationship between the deflection angle (alpha) and the forces acting on the pendulum, particularly the deflection force (Fd) that opposes Fh. Participants clarify that while Fd is not equal to Fh, it exists in a non-inertial frame as a pseudoforce, but in an inertial frame, only Fh is relevant. The conversation also touches on the analogy of a crane's hook deflecting when the trolley accelerates, confirming that deflection occurs regardless of the frame of reference. Ultimately, the pendulum does deflect under constant acceleration, as the net force acting on it results in a stable angle from the vertical.
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I have a strange question and have internet-wide not found answers:

An ideal pendulum hanging straight down gets a sidewards force Fh attacking at the pivot point.
The whole pendulum starts accelerating along Fh.

Fh: sidewarts force at the pivot point
Fd: horizontal force attacking at the pendulum C/G that causes the pendulum to deflect
alpha: deflection angle
m: pendulum mass

Fd = m*g / tan (alpha)

Now, how much does the pendulum deflect?

The feelings says the pendulum remains deflected as long as Fh causes a constant acceleration.

But, when the pendulum is deflected from where comes the force Fd (deflection force, opposite direction of Fh)? Fd can not be equal to Fh (because that would cause the acceleration to stop).
Is Fd smaller than Fh causing a slower acceleration (compared to Fh attacking directly a body with mass m)?
Is there a Fd at all? If not, remains the pendulum really hanging straight down?

Only a constant Fh shall be considered (no dynamic deflections because of the initial Fh step).
 
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mrueedi said:
I have a strange question and have internet-wide not found answers:

An ideal pendulum hanging straight down gets a sidewards force Fh attacking at the pivot point.
The whole pendulum starts accelerating along Fh.

Fh: sidewarts force at the pivot point
Fd: horizontal force attacking at the pendulum C/G that causes the pendulum to deflect
alpha: deflection angle
m: pendulum mass

Fd = m*g / tan (alpha)

Now, how much does the pendulum deflect?

The feelings says the pendulum remains deflected as long as Fh causes a constant acceleration.
I imagine a massless support free to slide along a frictionless table, from which the pendulum hangs. A horizontal force acts on the support.
But, when the pendulum is deflected from where comes the force Fd (deflection force, opposite direction of Fh)?
Viewed from an inertial frame, there's no need for a "deflection force"--the pendulum is being accelerated.
Fd can not be equal to Fh (because that would cause the acceleration to stop).
Is Fd smaller than Fh causing a slower acceleration (compared to Fh attacking directly a body with mass m)?
Viewed from the non-inertial frame of the pendulum, there will be a "pseudoforce" exactly equal and opposite to Fh acting on the pendulum bob. Note that Fh does not act on the bob, it acts on the support, so those two forces don't "cancel out".
 
Thank you very much for the answer!
Doc Al said:
Viewed from the non-inertial frame of the pendulum, there will be a "pseudoforce" exactly equal and opposite to Fh acting on the pendulum bob. Note that Fh does not act on the bob, it acts on the support, so those two forces don't "cancel out".

Ok, but once the deflection becomes stable that "pseudoforce" Fd works against Fh (if I observe from an inertial frame). But because the whole pendulum is accelerating, a net force to the side must exist.

I would expect that any Fd reduces the achievable acceleration. At the same time Fd can not have the equal and opposite size as Fh otherwise the whole pendulum would stop accelerating.
 
mrueedi said:
Ok, but once the deflection becomes stable that "pseudoforce" Fd works against Fh (if I observe from an inertial frame). But because the whole pendulum is accelerating, a net force to the side must exist.
If you observe from an inertial frame, there is no pseudoforce. (The pseudoforce is only an artifact of observing things in a non-inertial frame.) There is certainly a net force to the side: Fh.
I would expect that any Fd reduces the achievable acceleration. At the same time Fd can not have the equal and opposite size as Fh otherwise the whole pendulum would stop accelerating.
Again, the pseudoforce (what you call Fd) only exists within the non-inertial frame of the pendulum. But with respect to that frame, the pendulum is not accelerating: the net force (including both real and pseudo forces) is zero.
 
Thanks for your patience. I still am unsure. My focus is on the deflection.

Doc Al said:
If you observe from an inertial frame, there is no pseudoforce. (The pseudoforce is only an artifact of observing things in a non-inertial frame.) There is certainly a net force to the side: Fh.

If there is no pseudoforce (looking from inertial frame): -> no deflection? Is that right?

Imagine a crane with a trolley (like on this link
http://www.towercranetraining.co.uk/towercranetypes_files/image004.jpg ):

M is loaded on the hook and the trolley starts accelerating steadily to the right:
Is there a steady deflection alpha from the vertical for the hook & m?
 
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mrueedi said:
If there is no pseudoforce (looking from inertial frame): -> no deflection? Is that right?
I don't understand what you are trying to say.

There is never a pseudoforce when viewing from an inertial frame. Pseudoforces only arise in non-inertial frames.

No matter which frame you use to describe the situation, everyone agrees (of course!) that the pendulum bob is deflected when the apparatus is accelerated.

Try this: Describe the forces that are exerted on the pendulum bob. (Use an inertial frame.)

Imagine a crane with a trolley (like on this link
http://www.towercranetraining.co.uk/towercranetypes_files/image004.jpg ):

M is loaded on the hook and the trolley starts accelerating steadily to the right:
Is there a steady deflection alpha from the vertical for the hook & m?
Sure. (Ignoring oscillations, of course.)
 
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