# Homework Help: Finding spring constant

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1. Mar 24, 2015

1. The problem statement, all variables and given/known data
A slippery (frictionless), light horizontal bar rotates about a vertical axis with a constant angular velocity ω. A cylinder with mass m, is initially attatched to a thread with length a and to a spring, which from the beginning has its "natural" length.
Suddenly the thread breaks. Now the maximum distance between the cylinder and the axis is b.
Determine the spring constant k for the spring.

I have uploaded a picture of the problem.

Thanks!

2. Relevant equations
* T = m*(v^2)*1/2
* -0.5⋅k⋅r^2
* An integration

3. The attempt at a solution
I got the correct answer by the following equation, but my teacher says that the "procedure" is wrong. The solution involves an integral.

Here is my solution:

In the beginning (when the thread is not cut), the cylinder only has the kinetic energy: T1 = 0.5⋅m(ω⋅a)^2.

When the thread is cut, the cylinder is rotating with a new radius, b, thus the new kinetic energy is T2= 0.5⋅m(ω⋅b)^2. However, since the thread is cut, the spring makes the energy: -0.5⋅k⋅(b-a)^2.
At this stage, the energy is: T2 + the energy from the spring = 0.5⋅m(ω⋅b)^2 - 0.5⋅k⋅(b-a)^2.

Then I set the energy from the first stage equal to the energy from the second stage:

0.5⋅m(ω⋅a)^2 = 0.5⋅m(ω⋅b)^2 -0.5⋅k⋅(b-a)^2.

I solved for k, and got the correct answer:
k = (m⋅(ω^2)⋅(b + a))/(b - a).

My teacher said that the procedure is wrong. He said:
⋅The energy is not preserved (as I have assumed).
⋅There should be an integration involved.
⋅The work from the spring is negative.
⋅The Kinetic energy increases.

Another teacher said:
⋅We have an external force which rotates the axis. The energy is not preserved.
He showed me this integral, which is not complete:

∫m⋅r⋅(ω^2) - k⋅(...) dr = 0

I don't know how to solve this problem, and I really need some help.
Thank you!

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2. Mar 24, 2015

### Orodruin

Staff Emeritus
I am sorry for your teacher, but he is wrong. Your solution does make sense, but perhaps not in exactly the way you imagine it.

The first thing you need to do is to decide whether to consider the system from the fixed coordinate system or from a coordinate system that rotates with the bar. This is going to determine what "energy" means (note that energy is not independent of the reference frame). Your choice essentially corresponds to choosing a coordinate system that rotates with the bar. In this system there is a fictitious centrifugal force with a corresponding potential energy, which decreases as the cylinder travels outwards. This energy is essentially what you have called the "kinetic energy" of the cylinder (the cylinder also has a kinetic energy due to its radial movement, but you are looking for a point where this is zero). The total effective potential in this rotating system (which only has one degree of freedom) is the sum of the potential stored in the spring and the effective potential from the centrifugal force. Setting the potential at $b$ equal to the original potential at $a$ will give you the turning point as in any one-dimensional potential.

What your teacher's integral is doing is simply computing the potentials as the integral of the forces.