Uniform Circular Motion: Bead on a String Problem Solution

[aal0315]

Homework Statement

A 100 g bead is free to slide along an 80 cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40 cm apart. When the pole is rotated
about its axis, AB becomes horizontal.
(a) Find the tension in the string.
(b) Find the speed of the bead at B.

Homework Equations

The Attempt at a Solution

I am just wondering how to find the length of AB and BC without knowing any of the angles of the triangle besides the right angle one. can anyone help me figure that out?

 

[LowlyPion]

It says that AB is horizontal and apparently AC is vertical.

So BC would be the hypotenuse apparently without seeing your drawing.

You know what AC and you know what the others add to.

Think Pythagoras.

 

[aal0315]

does this make sense:
BC = 80-AB
then
80-AB = 40 + AB
40 = 2AB
AB = 20

 

[LowlyPion]

does this make sense:
BC = 80-AB
then
80-AB = 40 + AB
40 = 2AB
AB = 20

Does that satisfy the Pythagorean relationship?

 

[aal0315]

if i cancel out the squares than yes but i don't know if i can do that or not.

 

[LowlyPion]

if i cancel out the squares than yes but i don't know if i can do that or not.

It is a right triangle.

40² + AB² = BC²

AB + BC = 80

Just solve.

 

[aal0315]

ok so i got AB = 30 cm
so i know that i need to break the tension into x and y components.
Fy = mg = 0.1 kg(9.8) = 0.98 N
Fx = m (v^2/r), but we don't have the velocity. i know that v = 2(pi)r/T, but i don't know the T. what can i do?

 

[LowlyPion]

What must the vertical component of the Tension be?

 

[aal0315]

the vertical component is equal to the mass of the bead x gravity
Ty = mg = 0.1 (9.8)
= 0.98 N

 

[LowlyPion]

the vertical component is equal to the mass of the bead x gravity
Ty = mg = 0.1 (9.8)
= 0.98 N

And you know the cosθ because you know the ratio directly from 40 cm/50 cm.

So what's holding you back from the answer for T?

 

[aal0315]

so i can do cos 36.9 = 0.98/Ft
Ft = 1.23 N ?
but don't i have to account for the centripetal acceleration at all for finding the tension in the string?
also, is there a second string force acting out from the horizontal because it is one piece of string?

 

[aal0315]

can someone please tell me if this is correct or not?
is there two tensions to the string since the string is attached at two points on the pole and there are two parts of the string divded by the bead.

 

[LowlyPion]

The string is free to move in the bead, so the string isn't accelerating relative to the bead. The Tension in the string then from the top = the tension horizontally. And sure the centripetal acceleration is translated outward and is part of the tension. But they don't give you ω, so I dare say you would have to figure it already from the Tension from the m*g anyway.

As to the numbers, if you are satisfied you have the right value for Cos, then ... I'm not doing your math for you.