Why is mg=mv^2/r for centripetal motion instead of mg+T=mv^2/r?

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In centripetal motion, the equation mg = mv^2/r is used because, at the top of the circular path, the only force acting on the pendulum is gravity, as tension becomes zero. The tension in the rope does not contribute to the centripetal force at this point, as it acts perpendicular to the motion. The discussion clarifies that while both gravitational force and tension act on the pendulum, only gravity is relevant for maintaining circular motion at the top of the swing. The misconception arises from misunderstanding the role of tension in relation to the forces acting on the pendulum. Understanding these dynamics is crucial for accurately applying the principles of centripetal motion.
jack1234
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Hi, for this question
http://tinyurl.com/2qgco5
this is the solution for (b)
http://tinyurl.com/yrl64q

May I know why for centripetal motion, it is mg=mv^2/r, but not mg+T=mv^2/r (T is the tension of the rope)?
 
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Because the rope is acting perpendicular to the motion and is prescisely enough to offset the force of gravity in that direction.
 
Sorry, I am not very clear for the explanation.
I assume "the rope is acting perpendicular to the motion" is the downward force for the tension of rope, visually, it is two forces acting on the object, one is the tension of rope, one is the gravitational force, I am not sure what is the incorrectness of my thinking...
 
jack1234 said:
Sorry, I am not very clear for the explanation.
I assume "the rope is acting perpendicular to the motion" is the downward force for the tension of rope, visually, it is two forces acting on the object, one is the tension of rope, one is the gravitational force, I am not sure what is the incorrectness of my thinking...
Don't forget that you are looking for just enough energy to have the pendulum complete a full circle around the peg...in which case there is no tension at the top of its circle, and the only force acting on the pendulum is gravity at that particular point.
 

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