Is there an instantaneous angular acceleration for a conical pendulm?

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Discussion Overview

The discussion revolves around the concept of instantaneous angular acceleration in the context of a conical pendulum. Participants explore the relationship between centripetal acceleration, angular velocity, and the radius vector, examining whether an instantaneous angular acceleration exists as the pendulum swings in a conical motion.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that there is an instantaneous centripetal acceleration for a conical pendulum, questioning if this implies an instantaneous angular acceleration towards the center.
  • Others seek clarification on the definitions of angle and center, with one participant defining the angle as the one between the string and the axis of symmetry.
  • One participant argues that since the angle remains constant as the pendulum swings, there is no instantaneous angular acceleration.
  • Another participant counters that while the radius of circular motion is constant, there is still centripetal acceleration directed towards the center, suggesting a nuanced view on the relationship between radius and acceleration.
  • Some participants agree that angular velocity is constant when defined around the axis of symmetry, while others discuss the implications of changing radius vectors and their relationship to velocity and acceleration.
  • A later reply emphasizes that a changing radius vector indicates a velocity, but questions whether this alone implies acceleration.

Areas of Agreement / Disagreement

Participants express disagreement regarding the existence of instantaneous angular acceleration in a conical pendulum, with multiple competing views on the relationship between centripetal acceleration, angular velocity, and the radius vector.

Contextual Notes

Limitations in definitions and interpretations of angular acceleration and centripetal acceleration are present, as well as the dependence on the specific conditions of the pendulum's motion.

jason12345
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For a conical pendulum, there is an instantaneous centripetal acceleration. Does this mean there is an instantaneous angular acceleration of the pendulum towards the center?
 
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Can you define what your angle and center refer to?
 
olivermsun said:
Can you define what your angle and center refer to?

The angle is between the string and the axis of symmetry the pendulum rotates around.
 
I see, you're talking about a pendulum which swings about the center axis in a cone.

Your angle, as defined, rotates with the pendulum string and remains constant, so I would say "no."
 
olivermsun said:
I see, you're talking about a pendulum which swings about the center axis in a cone.

Your angle, as defined, rotates with the pendulum string and remains constant, so I would say "no."

Thanks for your reply, although I disagree with it :) I could also argue that the radius of the circular motion is constant and so there isn't an acceleration towards the centre - but there is: v^2/r
 
There is an acceleration (which happens to be toward the center) because the radius vector is not constant. Only the radius magnitude is constant.

As far as I can tell, the angular velocity is constant if defined around the axis of symmetry.
 
olivermsun said:
There is an acceleration (which happens to be toward the center) because the radius vector is not constant. Only the radius magnitude is constant.

I think you mean velocity where you state radius.

As far as I can tell, the angular velocity is constant if defined around the axis of symmetry.

I agree that angular velocity is constant.
 
jason12345 said:
I think you mean velocity where you state radius.

You're right. Change in radius vector per time (velocity) changes.
 
Last edited:
How does the radius vector not change? Doesn't it's magnitude stay the same, however the direction is changing?
 
  • #10
The radius does change (dr/dt is nonzero), so that there is a velocity, but he was talking about whether or not there is an acceleration. There is, since d^2/dt^2 = dv/dt is nonzero. A changing radius vector isn't enough to imply an acceleration, although it is enough that the magnitude stays the same while the direction is changing (as you say).
 

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