Angular momentum conservation and energy considerations

[rcgldr]

I suggest thus that the OP was measuring the radial speed.

The OP mentions an increase in speed when the pole is replaced by a very small hole and someone pulls the string up through the hole, in this case there is work done, (math already done in this case, see earlier posts). The OP was wondering why a pole doesn't have the same effect as a hole, and it's because the curve for the pole case is an involute of circle, whilethe hole case could be anything, even an ellipse if the tension is varied to be 1/r^2 (like gravity). I did the string through the hole case first, then the pole case.

 

[rcgldr]

Cutting to the chase I don't know how to solve analytically.

Use cartesian coordiantes instead.

Equation of involute of circle:
x\ =\ R\ (\ cos(\phi) \ +\ \phi \ sin(\phi)\ )
y\ =\ R\ (\ sin(\phi) \ -\ \phi \ cos(\phi)\ )

Derivative of involute of circle:
d(y(\phi))/d(x(\phi)=(R(cos(\phi)\ + \phi \ sin(\phi) \ - cos(\phi)))\ / \ (R(-sin(\phi)\ + \phi \ cos(\phi) \ + sin(\phi)))
d(y(\phi))/d(x(\phi)=(R\ \phi \ sin(\phi))\ /\ (R\ \phi \ cos(\phi)) \ = \ tan(\phi)

Equation of normal line of involute of circle:
-(dy_0/dx_0)(y - y_0)=(x - x_0)
-tan(\phi _0)(y - y_0)=(x - x_0)

Equation of circle:
x\ = \ R\ cos(\phi)
y\ = \ R\ sin(\phi)

Derivative of circle:
d(y(\phi))/d(x(\phi)=-R sin(\phi) \ /\ R cos(\phi) = -cot(\phi)

Equation of tangent line of circle:
(y - y_0)=(dy_0/dx_0)(x - x_0)
(y - y_0)=(-cot(\phi _0))(x - x_0)
-tan(\phi _0)(y - y_0)=(x - x_0)

For both equations:
-tan(\phi _0)(y - y_0)=(x - x_0)
y - y_0=-cot(\phi _0)(x - x_0)
y + cot(\phi _0)x = y_0 + cot(\phi _0)x_0

For involute of circle:
y + cot(\phi _0)x = R(sin(\phi_0)-\phi_0 cos(\phi_0) + cot(\phi _0)R(cos(\phi_0)+\phi_0 sin(\phi_0))
y + cot(\phi _0)x = R(sin(\phi_0)-\phi_0 cos(\phi_0) + cos(\phi_0)^2/sin(\phi_0) + \phi_0 cos(\phi_0))
y + cot(\phi _0)x = R(sin(\phi_0) + cos(\phi_0)^2/sin(\phi_0))

For circle:
y + cot(\phi _0)x = R(sin(\phi_0)) + cot(\phi _0)R(cos(\phi_0))
y + cot(\phi _0)x = R(sin(\phi_0)) + cos(\phi_0)^2/sin(\phi_0))

So the the equations for the normal line of the involute of circle and tangent line of the circle are the same with respect to \phi , and no work is done since the tangent line of the circle, which is the string and the line of force, is normal to the path of the puck. (and now my head hurts). Examples of the normal+tangent line equations for specific values of \phi .

For \phi = \pi / 2, y = R
For \phi = \pi / 4, y = -x + \sqrt{2} \ R

 

[fantispug]

Then, for the second part, we replaced the bracket with the eye in it with post that was lowered right to the table top. We got the puck moving in a circle and had pencil marks on the table for R and R/2. When the puck reached the mark for R/2 the measured velocity also changed by a factor of 2 just like it did in the original experiment. I have never found anyone who could explain where the extra energy came from and I'm just trying again.

This seems to imply to me that LeeB is claiming that when the string wraps around the pole the puck speeds up in an inversely proportional manner. I argue with this claim on theoretical and physical grounds (I wish I had the apparatus to carry out a reasonable experiment).

Anyway Jeff I checked and my equations are consistent with the path being involute of a circle. I don't think that 2nd order DE I presented is right.

Anyway I ended up trying again, getting kinetic energy
T=\frac{m\dot{r}^2r^2}{2a^2}
Which leads to a DE in terms of R:
\ddot{r}{r}+\dot{r}^2
Which has solution
r=\sqrt{2r_0\dor{r}_0t+r_0^2}
where the puck has initial position r_o and velocity \dot{r}_0. The precise motion can be worked out using the relation between r and theta
\frac{dr}{d\theta}=\frac{ar}{\sqrt{r^2-a^2}}
Which is the differential form of the involute of a circle, which has solutions
r=R\sqrt{\phi^2+1}
where phi is given implicitly by
\theta=\phi-\arctan(\phi)
where theta is the polar coordinate.

The explicit solution isn't too helpful though. Conservation of energy, since there are no sources of potential energy, implies the speed is constant throughout the motion. As Jeff pointed out this is very different to the case where someone is pulling on the string supplying energy.

 

[rcgldr]

The string through the hole case, assuming that angular momentum is conserved when the string is pulled through a hole. For example, angular momentum is m x r x s, so if r is halfed, then to conserve momentum, it becomes m x (r/2) x (2s). Based on this assumption, then the relation ship between tension and r is:

t(r)\ =\ -m\ (s_0)^2\ (r_0)^2\ /\ r^3

So if string is pulled from r_0 to r_0/2, then work done is:

\int _{r_0} ^{r_0/2} (-m\ (s_0)^2\ (r_0)^2\ /\ r^3)\ dr

\left[ \frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ r^2 \right]_{r_0}^{r_0/2}

\frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ (\frac{r_0}{2})^2 - \frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2

\frac{4}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2 - \frac{1}{2}\ m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2

\frac{3}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2

\frac{3}{2} m\ (s_0)^2

Original KE:

{KE}_0 = \frac{1}{2} m (s_0)^2

KE after work done:

{KE}_1= \frac{1}{2} m(s_0)^2 + \frac{3}{2} m\ (s_0)^2 = \frac{4}{2} m\ (s_0)^2

{KE}_1= \frac{1}{2} m (2\ s_0)^2

So the speed is doubled if the radius is decreased by 1/2. In the case of the post, relative to a circular path, the post applies a backwards and inwards tension, which may account for the reason that angular momentum is not conserved in that case. Also in the case of the post, the tension is normal to the actual path of the puck, so no work is done and the puck's speed remains constant.

 

[tabchouri]

Use cartesian coordiantes instead.

Equation of involute of circle:
x\ =\ R\ (\ cos(\phi) \ +\ \phi \ sin(\phi)\ )
y\ =\ R\ (\ sin(\phi) \ -\ \phi \ cos(\phi)\ )

Derivative of involute of circle:
d(y(\phi))/d(x(\phi)=(R(cos(\phi)\ + \phi \ sin(\phi) \ - cos(\phi)))\ / \ (R(-sin(\phi)\ + \phi \ cos(\phi) \ + sin(\phi)))
d(y(\phi))/d(x(\phi)=(R\ \phi \ sin(\phi))\ /\ (R\ \phi \ cos(\phi)) \ = \ tan(\phi)

Equation of normal line of involute of circle:
-(dy_0/dx_0)(y - y_0)=(x - x_0)
-tan(\phi _0)(y - y_0)=(x - x_0)

You are right, now i see the error of ly ways :)
Error in my algebra led to my erronous conclusions.
I opologies for any trouble.

-----------------------------------------------------
http://ghazi.bousselmi.googlepages.com/présentation2

 

[rcgldr]

So if there is no work done, then you are suggesting that the new velocity WON'T be twice the old velocity under these circumstances? When we actually did this on an air table, the results were virtually identical to the first case.

Are you sure about this? After going through the math, it seems that the puck shouldn't have an increase in energy, but this is based on the assumption that the puck does follow an involute of circle path, and this sure looks like the proper model for anything winding around a post. Assuming the puck is attached at the side, then it's angular momentum (rotation wise) should slow things down a bit.

I do realize that the string is moving inwards with tension, but it's tension is perpendicular to the path of the puck so I'm not sure where the increase in energy would originate from. Relative to the string (which isn't an inertial frame since it's rotating), the string is being wrapped around the pole similar to the pole rotating and pulling in the string. Using this analogy, the only work done is to increase the potential energy (relative to the pole), but not the kinetic energy of the puck, similar to rasing a weight with a string (increase in potential energy, no increase in kinetic energy).

 

[TVP45]

Measurement error

Straightforward thermodynamics says that the linear velocity cannot increase when you do this. If this were not so, David could have whacked an entire army of Goliaths by simply shortening up his sling. If you have measured otherwise, the problem is in the measurement which is difficult to make for small circles. Measure omega and calculate v.

 

[rcgldr]

Regarding the post case (involute of circle), and taking into account the post experiences a torque and applies this torque to the earth, increasing the Earth's angular momentum, then angular momentum for the entire system, earth, post, and puck are conserved. In the hole case, no there no torque applied at the hole or to the earth, so there's no change in angular momentum to the earth, so the puck retains it's angular momentum.

 

[Prologue]

Wow Jeff, I've got to dig this one up to thank you big time for your explanation. Pretty much every post enlightened me at least a little bit on angular momentum, major kudos to you.