What Forces Act on an Object in Non-Uniform Circular Motion?

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In non-uniform circular motion, the forces acting on an object include weight, tension, and the normal reaction force. The equation of motion can be expressed as mr'' = -|T|er - (mgcos a)er - (mgsin a)eo + |N|er, where er and eo represent radial and tangential components. The discussion emphasizes the need to clarify the forces acting on the object and to resolve the equations of motion separately for radial and tangential directions. It suggests that starting with conservation of energy might provide a clearer understanding of the problem. Overall, identifying and analyzing the forces is crucial for solving the motion of the object effectively.
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I am a bit confused by this question(see attached image). This has been posted before but I am not any further forward. My problems are:


P1 - has this got the following forces acting on it: Weight, Tension, and Natural reaction force?

and thus does the equation of motion end up as:

mr''= -|T|er - (mgcos a)er - (mgsin a)eo + |N|er

where er and eo are the radial and tangential components respectively

Then is it a case of gathering the similar terms and resolving in the er direction to get:

ml0'^2 = mg cos a - |T|


P2 - is this just experiencing the tension of the string and its own weight?
 

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I'm not clear as to what you are asked to solve for.
 
need to identify the forces acting on each particle using appropriate co-ordinates system and then write the equation of motions for each of particle.
 
confusedalot said:
P1 - has this got the following forces acting on it: Weight, Tension, and Natural reaction force?

and thus does the equation of motion end up as:

mr''= -|T|er - (mgcos a)er - (mgsin a)eo + |N|er

where er and eo are the radial and tangential components respectively
(1) Looks like you have tension along the radial direction.
(2) Which way does the tangential component of gravity act?

I think it might be a bit clearer if you wrote the radial and tangential equations separately.

Then is it a case of gathering the similar terms and resolving in the er direction to get:

ml0'^2 = mg cos a - |T|
I don't know what you did here.


P2 - is this just experiencing the tension of the string and its own weight?
Right.
 
Welcome to PF!

Hi confusedalot ! Welcome to PF! :smile:

Hint: forget tension etc … try starting with conservation of energy. :smile:
 

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