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Horizontal Circular Motion-m2 Watch

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    Hello, can you help me figure out the triangle being described in this question for horizontal circular motion. As far as I can see, its describing a right angled triangle of hypotenuse 4a,base 3a and height 5a o_O

    15) The point A is vertically above the point B and a distance 5a from it. The particle, P, of a mass m, is attact to A by a light inextensible string of length 4a. The particle is also attached to B by a light inextensible string of length 3a. P movies in a horizontal circle with both strings taut. Find the tension in the strings and show that omega square is greater or equal to 5g/16a
    * P moves not movies
    i hate questions without diagrams

    Thanks
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    (Original post by jarasta)
    Hello, can you help me figure out the triangle being described in this question for horizontal circular motion. As far as I can see, its describing a right angled triangle of hypotenuse 4a,base 3a and height 5a o_O
    You have a 3,4,5 triangle, which is right-angled. But the hypotenuse (the longest side in a right-angled triangle) is 5a (the vertical), and both 3a, and 4a are inclined with the right-angle between them.
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    (Original post by ghostwalker)
    You have a 3,4,5 triangle, which is right-angled. But the hypotenuse (the longest side in a right-angled triangle) is 5a (the vertical), and both 3a, and 4a are inclined with the right-angle between them.
    aahh thank you. May I ask how you got to that conclusion though
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    (Original post by jarasta)
    aahh thank you. May I ask how you got to that conclusion though
    Which bit causes the problem?

    It's a 3,4,5 right-angled triangle with the 5 side vertical (AB=5a).
 
 
 
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