- #1
Adel A
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Homework Statement
(Sorry for my bad English)
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!
Homework Equations
* T = m*(v^2)*1/2
* -0.5⋅k⋅r^2
* An integration
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!