# Acceleration of the ring

1. Mar 23, 2014

### nil1996

1. The problem statement, all variables and given/known data

A ring is moving on a vertical rod attached to a spring. Find the velocity of the ring when spring becomes horizontal. The spring is ideal with spring constant 400N/m. Mass of ring is 10kg. Natural length of spring is 4m. Initially the ring is at rest as shown in figure.(attachment)

2. Relevant equations

3. The attempt at a solution
I want to solve the problem using Newtons laws and not by conservation of energy method. I have no idea on where should I start. I have made the following equation:

Fcosθ + mg = ma ........where F is the force by the spring
........ θ is the angle between spring and rod
........ a is the acceleration of the ring

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2. Mar 23, 2014

### haruspex

Next you need some geometry to relate theta to the length of the spring, and an equation relating the length of the spring to F.

3. Mar 23, 2014

### nil1996

From hooks law:
F=k*x ......where x is the length of spring.

from geometry:

cosθ=h/x

so x=h/cosθ

so F=k*h/cosθ

4. Mar 23, 2014

### haruspex

No, that would be where x is the extension of the spring.
This way is going to get messy because h and x are both variable. Let the horizontal distance be D (=4m) and write the height and spring extension in terms of D and theta.

5. Mar 23, 2014

### nil1996

So i should write it as:

F=k(D/sinθ -3) ..... D =horizontal distance

so putting this value in first equation:

ma=Fcosθ + mg

ma=k(D/sinθ-3)cosθ + mg

dv/dt = k/m(D/sinθ-3)cosθ + g

$\frac{dv}{dt}$=$\frac{k}{m}$($\frac{D}{sinθ}$-3)+g

Last edited: Mar 23, 2014
6. Mar 25, 2014

### haruspex

Not 3. The relaxed length is also D.
You dropped the cos.

7. Mar 25, 2014

### nil1996

$\frac{dv}{dt}$=$\frac{k}{m}$($\frac{D}{sinθ}$-D)cosθ + g

now what to do after this?Should we consider a small change in θ and integrate it over the "θ" range to find the velocity??

8. Mar 25, 2014

### haruspex

You need to express the vertical height as a function of theta, and thereby express dv/dt in terms of theta.

9. Mar 25, 2014

### nil1996

sorry but i am not getting what you saying?Can you please explain a little.