Object on a vertical Ring with spring attached

[Lambdalove]

picture: https://ibb.co/k5P0GG
Two objects slide without friction on a circular ring of radius R, oriented in a vertical plane. The heavier object (of mass 3m) is attached to a spring with an unstretched length of zero (admittedly an unphysical assumption) and spring constant k. The fixed end of the spring is attached to a point a horizontal distance 2R from the center of the circle.

1. What is the kinetic and potential energy of the heavy particle in the initial state, when it passes the left most position, horizontal with respect to the center of the circle, and when it hits the light particle.

Here's what I've come up with (K_f is just before collision)
K_i + U_i + U_spring = K_f + U_f
0 + (3m)g(2R)+k/2(displacement) = 1/2(3m)(v_1)^2+k/2(displacement)

I'm having trouble with finding the displacement of the spring, as it is along a circular path, which I haven't worked with before. Should I only account for the x-axis displacement, add x- and y-axis displacement or am I completely lost?
Any help is much appreciated!

 

[TSny]

K_i + U_i + U_spring = K_f + U_f
0 + (3m)g(2R)+k/2(displacement) = 1/2(3m)(v_1)^2+k/2(displacement)

The potential energy of the spring depends on the square of the elongation of the spring.

I'm having trouble with finding the displacement of the spring, as it is along a circular path, which I haven't worked with before. Should I only account for the x-axis displacement, add x- and y-axis displacement or am I completely lost?
Any help is much appreciated!

Both the x and y components of the displacement of the spring are important. The length of the spring can be found using the Pythagorean theorem.

 

[Lambdalove]

The potential energy of the spring depends on the square of the elongation of the spring.

Both the x and y components of the displacement of the spring are important. The length of the spring can be found using the Pythagorean theorem.

So in this case it would be

The potential energy of the spring depends on the square of the elongation of the spring.

Both the x and y components of the displacement of the spring are important. The length of the spring can be found using the Pythagorean theorem.

Hah... I feel kind of silly now but thank you so much for the help!

 

[Lambdalove]

So apparently I don't really get it after all..
would that be \sqrt((-R)^2+(2R)^2) = 5R^2 or \sqrt((-R)^2+(R)^2 = 2R^2? And is the potential energy of the spring the same at the top, middle-left and bottom point?

I find that spring very confusing.

 

[TSny]

The potential energy of the spring is (1/2)kL²where L is the length of the spring. This is because we assume the unstretched length of the spring is zero.

When the heavier mass is at the top, the spring can be thought of as the hypotenuse of a right triangle. What are the lengths of the legs of this triangle?

When the mass is at the leftmost position, it should be easy to see how long the spring is.

If the spring is stretched the same amount at the bottom as at the top, then the potential energy of the spring will be the same at these two positions.