# Object on a vertical Ring with spring attached

• Lambdalove
In summary, the conversation discusses the calculation of kinetic and potential energy for a heavy object sliding on a circular ring with a spring attached. The potential energy of the spring is dependent on the square of its elongation, which can be found using the Pythagorean theorem. The potential energy of the spring is (1/2)kL2, where L is the length of the spring. The length of the spring can be determined at different positions, such as the top and leftmost point. The potential energy of the spring will be the same at the bottom and top positions if the spring is stretched the same amount.
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!

Lambdalove said:

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
TSny said:
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
TSny said:
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!

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.

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.

## 1. What is the purpose of the object on a vertical ring with spring attached experiment?

The purpose of this experiment is to study the motion of an object attached to a vertical ring with a spring. It allows scientists to observe how the object moves in response to different forces and to analyze the relationship between the object's motion and the properties of the spring.

## 2. How does the spring affect the motion of the object on the vertical ring?

The spring acts as a restoring force on the object, meaning it pulls the object back towards its equilibrium position when it is displaced. This affects the motion of the object by causing it to oscillate up and down around the equilibrium position.

## 3. What factors influence the motion of the object on the vertical ring?

The motion of the object is influenced by several factors, including the mass of the object, the stiffness of the spring, and the amplitude and frequency of the oscillations. The gravitational force acting on the object and any external forces can also affect its motion.

## 4. How can the motion of the object on the vertical ring be described mathematically?

The motion of the object can be described using equations of motion, such as the simple harmonic motion equation. This equation takes into account the mass of the object, the spring constant, and the amplitude and frequency of the oscillations.

## 5. What real-world applications can be derived from the study of object on a vertical ring with spring attached?

This experiment has many real-world applications, such as understanding the behavior of springs in various mechanical systems, analyzing the motion of objects in simple harmonic motion, and developing mathematical models for oscillatory systems. It can also be used to study the properties of materials and their response to forces.

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