# Speed of a bead turned through X degrees

1. Sep 15, 2016

### moenste

1. The problem statement, all variables and given/known data
A small bead is threaded on a smooth circular wire of radius r which is fixed in a vertical plane. The bead is projected from the lowest point of the wire with speed √6gr. FInd the speed of the bead when it has turned through: (a) 60°, (b) 90°, (c) 180°, (d) 300°.

Answers: (a) √5gr, (b) √4gr, (c) √2gr, (d) √5gr

2. The attempt at a solution
While this question is explained here, I do not understand one part. Why PE = m * g * h = m * g * (r - r * cos(θ)). Specifically h = r - r * cos(θ)? Why isn't it equal to just h = r cos 60°?

Taking this image (as I see the problem):

We have φ = 60°. R is h and equals to r cos 60°, where r is the adjacent line to the angle. Why the correct height h is r - r cos 60° and not just r cos 60°?

2. Sep 15, 2016

### BvU

The potential energy is taken as zero when $\phi=0$ So you want h to be the height above the reference height zero. $h - h_0 = ...$

3. Sep 15, 2016

### TSny

h is the vertical height of the bead above the bottom of the wire loop, as explained by BvU.

Your picture indicates that the loop is rotating about the vertical axis with angular velocity $\Omega$. But, apparently the problem wants you to assume the loop is not rotating.

4. Sep 15, 2016

### moenste

What would be the correct graphical representation of the problem? Could you please give an example?

And with your image I think I got it. If we take two positions h = 0 at α = 0 and r = 10 cm (for example). When the bead moves to α = 60° the height is not 0, since the bead is in the air (if we take h = 0 as the moment when the bead is on the ground). In order to find h we need to subtract 10 cos 60° from the original radius, which is equal to 10. So h = 10 - 10 cos 60 = 5 cm. Is this the right logic?

5. Sep 15, 2016

### TSny

Yes, that's right.