# Work-Energy for Bead on Rotating Stick

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1. Jan 4, 2017

### GL_Black_Hole

1. The problem statement, all variables and given/known data
Verify the Work-Energy Theorem W=ΔK for a bead of masd m constrained to lie on a frictionless stick rotating with angular velocity ω in a plane.

2. Relevant equations
W =∫ F⋅dr, K =m/2 v^2

3. The attempt at a solution
Adopting polar coordinates the velocity is v = r' +r*Θ'. Finding the radial and tangential accelerations we may argue that since the stick is frictionless on the tangential force of the stick on the bead may do work. As the angular velocity was assumed constant this works out to just be the Coriolis force, 2mωr'. The line integral of this from the initial position, r_0, of the bead to a final r gives the work done as mr^2 *ω^2 -mr_0^2 *ω^2.
But the Kinetic energy is K= m/2 (r' ^2 +r^2 *ω^2), justified by the use of polar coordinates.
The 'non-conservation' of energy makes sense since work must be done to keep the rod spinning at a constant rate but if I take ΔK I don't recover W. Instead I get m/2 r'^2 +m/2 r^2 * ω^2 - m/2 r_0 ^2 *ω^2.
Where do I have an error?

2. Jan 4, 2017

### TSny

OK. I assume you mean this to be a vector equation in which the first term on the right is in the radial direction and the second term is in the tangential direction (i.e., direction of increasing θ).
Yes, the stick exerts a force on the bead that is perpendicular to the stick (i.e., in the tangential direction).
Most people would not call this the Coriolis force. The Coriolis force is a fictitious force that occurs in the frame rotating with the stick. In the lab frame, there is only the real force that the stick exerts on the bead. But, you are right that the magnitude of this force is 2mωr'.
Yes, that looks right.
This also looks correct.
(Looks like you're assuming the initial radial velocity is zero. But, you don't need to make this assumption.)
You need to show that there is no contradiction between your two results for ΔK. They only appear to be different.

3. Jan 9, 2017

### GL_Black_Hole

I found my error when I went back and explicitly computed the line integral for the work done by the constraining force. The integrand is proportional to 2*sinh(ωt)*cosh(ωt), which equals sinh(2*ωt), which integrates to cosh(2*ωt)/2ω. So I recover the factor of 1/2 I was missing. For the kinetic energy an explicit expression is (m*ω^2 *r_0^2)/2 *(cosh(2*ωt) so the change in the kinetic energy is indeed equal to the work done by the constraint.

4. Jan 9, 2017

### TSny

OK, good. You don't actually need to solve for r(t) explicitly in terms of exponential (or hyperbolic) functions to get the result. But, it's certainly OK if you want to.