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# Horizontal Circular Motion-m2 watch

1. 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
2. (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.
3. (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
4. (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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