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.