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Is there an instantaneous angular acceleration for a conical pendulm? 
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#1
May312, 10:13 AM

P: 111

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?



#2
May312, 10:45 AM

P: 865

Can you define what your angle and center refer to?



#3
May312, 11:13 AM

P: 111




#4
May312, 11:28 AM

P: 865

Is there an instantaneous angular acceleration for a conical pendulm?
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." 


#5
May312, 01:20 PM

P: 111




#6
May312, 01:25 PM

P: 865

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. 


#7
May312, 05:46 PM

P: 111




#8
May312, 06:23 PM

P: 865




#9
May312, 08:46 PM

P: 29

How does the radius vector not change? Doesn't it's magnitude stay the same, however the direction is changing?



#10
May412, 06:42 AM

P: 865

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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