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    ------I'm using * to indicate ^2 since it's less confusing to read.-------

    Say you have an object on a string and are spinning it in the vertical plane.
    To calculate the centripetal force when the object is at the top, you use F=T+mg, since the tension and weight are acting in the same direction. When the object is at the bottom, they act in opposite directions, so you use F=T-m. That's fairly simple and I get it.

    However, things stop making sense when the tension is at a right angle to the weight (i.e. the object is exactly on the left or right). Should the resultant force not be sqrt(T*+(mg)*)? I can't work out why the resultant force is just T. (I always say this: my teacher never explains anything, just flicks through a slideshow and doesn't get why none of us know anything.)

    Leading on from that, what happens when the object is at a random point on the circular path? I'm not sure if I will need to know how to calculate the resultant force in that situation, but I need to know just in case.

    Sorry if this is a stupid question but I have a test on Monday and I can't learn the concept if I don't understand it.

    Thanks!
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    Remember, centripetal force is always directed to the centre of the circle.

    Centripetal force is calculated by resolving the forces acting away or towards the centre in a parallel line. Just like how you said when it's at the top, the centripetal force is T+mg. You were able to resolve because weight and tension both act in the same parallel line.

    When the mass is on the left or right, the weight isn't parallel to the tension. Rather, it is perpendicular. Therefore the only force causing the circular motion is the tension.

    Sorry if that was a bad explanation. Also i'm not sure how to do it at any random point.
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    (Original post by JustJusty)

    However, things stop making sense when the tension is at a right angle to the weight (i.e. the object is exactly on the left or right). Should the resultant force not be sqrt(T*+(mg)*)? I can't work out why the resultant force is just T.

    Thanks!
    When the weight of the object is perpendicular to the tension as explained by The_Big_E, tension is the "resultant force" for the centripetal acceleration  a_c while weight is the "resultant force" for the tangential acceleration  a_t . The magnitude of the "the total resultant force" is indeed what you have said.

     |\vec{F}_{net}| = \sqrt{T^2 + (mg)^2}

    (Original post by JustJusty)
    Leading on from that, what happens when the object is at a random point on the circular path? I'm not sure if I will need to know how to calculate the resultant force in that situation, but I need to know just in case.
    Thanks!
    Consider the diagram below and assume it is moving in a clockwise directionName:  vertical_circular_motion.JPG
Views: 50
Size:  19.0 KB

    The forces that are responsible for centripetal acceleration are

     T - mg \cos(\theta) = ma_c

    while the force responsible for tangential acceleration is
     mg \sin(\theta) = ma_t

    The magnitude of the resultant acceleration at any point is

     |\vec{a}| = \sqrt{a^2_c + a^2_t}

    Hope it is clear now.
 
 
 
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