- #1

ourheroine

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**1.**Homework Statement

An 100g bead is free to slide along an 80cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40cm apart. When the pole is rotated, AB becomes horizontal.

a) find the tension in the string

b) find the speed of the bead at B.

in the diagram, A is the point the string is attached at the bottom of the pole, B is where the bead is (horizontally in line with A), and C is where the string is attached at the top of the pole (40cm up from A).

**2.**Relevant equations

F=ma, a=v^2/r

**3.**The attempt at a solution

For a free body diagram I have drawn the forces acting on the bead; the weight of the bead (mg) acting vertically downwards, the y component of Ftens acting upwards on the bead, and the x component of Ftens acting horizontally towards the pole.

I have figured that since the bead rotates in a horizontal plane that mg = Ftensy

therefore: (0.1kg)(9.8N) = 0.98N = Ftensy

I have also figured by the ole 3-4-5 rule that the horizontal part of the string (aka the radius) is 30cm and that the hypotenuse of the triangle is 50cm. Via trig, the angle at point C is 36.9o and the angle at point B (aka the bead) is 53.1o.

I am stuck now, as I think the next step should be to figure out the radial acceleration via the net force, but I am unsure as to how to figure out the value for Ftensx...which, if Ftensy = mg (but in opposition), would be Fnet...?

Note: I did find a couple of posts on this in the archive but was still unclear after reading through them, so any help is much appreciated!