Difference between revisions of "Angular Momentum Conservation"

From wwwelab
Jump to navigation Jump to search
Line 104: Line 104:
 
$V_{rms}=\frac{V_{medida}}{\sqrt{3}\sqrt{2}}$
 
$V_{rms}=\frac{V_{medida}}{\sqrt{3}\sqrt{2}}$
  
 
+
In the setup used the energy dissipates in 3 branches so the power comes multiplied by 3. Also, each winding is in parallel with a resistor with the same resistence value $4,7\Omega$, what means that $R=4,7\Omega$ can be used in the equation multiplying the power by 2.
$4,7\Omega$ $R=4,7\Omega$
 
  
 
$P=3\times2\times\frac{V_{rms}^2}{R}=3\times2\times\left(\frac{V_{medida}}{\sqrt{3}\sqrt{2}}\right)^2\frac{1}{R}$
 
$P=3\times2\times\frac{V_{rms}^2}{R}=3\times2\times\left(\frac{V_{medida}}{\sqrt{3}\sqrt{2}}\right)^2\frac{1}{R}$

Revision as of 20:57, 12 January 2013

UNDER CONSTRUCTION

Description of the Experiment

This control room allows the confirmation of angular momentum conservation by colliding a spinning disk with another. Moreover, the disk inertia momentum can be extrapolated from energy conservation principles.

Experimental Apparatus

The experimental apparatus is based in a PC hard disk drive motor and its spinning disk. A second disk is held on top of it and can be dropped by a servo motor actuator.

The apparatus´ motor can be used as a generator equipped with a switchable resistor acting as an electromagnetic brake. The braking current&voltage characteristic is measured allowing a rigorous energy dissipation calculation.

Protocol - Angular Momentum Conservation

Figura 1: Angular velocity (rpm) as a function of time for a collision at 1000 rpm.

A disc with a total mass of 115g is accelerated by the hard-drive motor until they reach a selected angular velocity. In this instant the motor is disconnected from supply and the disc allowed to rotate freely being their speed monitored. When a certain pre-determined speed is reached, the servo lets the suspended disc with a total mass of 69g initially at rest fall on top of the rotating disc.

Data taken from the experiment is given and plotted with the disc speed in function of time.

Figure1 is a plot of the results of an experience in which a servo lets the suspended disc fall when the disc below reach 1000 rpm.

Doing a linear regression between the deceleration and fall of the disc, it's possible to obtain the predicted rotational speed at any time. This give us the thumb rule for the friction deceleration related to angular speed.

Physics

Using the following quantities:

L - angular momentum

I - moment of inertia

ω - angular velocity

m - mass in rotation.

For the angular momentum conservation:

$L_i=L_f$

$I_i \omega_i=I_f \omega_f$

$\frac{I_i}{I_f}=\frac{\omega_f}{\omega_i}$

$\frac{\frac{m_i\left (r_1^2+r_2^2 \right )}{2}}{\frac{m_f\left (r_1^2+r_2^2 \right )}{2}}=\frac{\omega_f}{\omega_i}$

$\frac{m_i}{m_f}=\frac{\omega_f}{\omega_i}$

The experimental results give:

$\frac{\omega_f}{\omega_i}=\frac{623}{950}=0,656$

while the predicted mass ratio is

$\frac{m_i}{m_f}=\frac{115}{115+69}=0,625$

Evaluating the accuracy:

$\frac{\left|0,656-0,625\right|}{\left|0,625\right|}\times 100=4,9\%$

The speed ratio is different from the mass ratio by 4,9% which gives a good approximation for the angular momentum conservation.

Knowing the exact dimensions of the disks ($r_1=13mm, r_2=47mm$) and adding an error momentum on the equations one can infer an approximated value for the motor rotor momentum of inertia (or its mass knowing its average radius).

$I_i \omega_i=I_f \omega_f$

$\left (I_m + I_{Di}\right ) \omega_i=\left (I_m + I_{Df}\right ) \omega_f$

Solving in order to $I_m$

$I_m = \frac{I_{Df} \omega_f - I_{Di} \omega_i}{\omega_i-\omega_f}$

Advanced Protocol - Moment of Inertia Evaluation

Figura2: rotational speed
Figura3: voltage measurement
Figura4: voltage between two phases

A disc with a total mass of 115g are accelerated by the hard-drive motor until they reach a pre-defined angular velocity. At that time the motor is disconnected and the disc allowed to rotate freely. When a certain speed that the user defines previously is reached, a relay puts each motor winding in parallel with a resistor which resistance is the same as the motor's windings (Figure 3). These resistors will dissipate energy acting as an electromagnetic brake. Voltage and speed in funtion of time are given in a table of results in the end of the session.

Figures 2 and 4 are plots created in Microsoft Excel using the table of results of an experiment in which the relay turns on when the rotating discs reach 1400 rpm.

Using the first data to do a linear regression it is possible to get the speed of the motor over time if the relay didn't turn on.

Between each speed acquisition it is done an energy balance. The loss of total mechanical energy must be equal to the sum of losses by friction and electromagnetic breaking.

$\Delta E_{mec} = \Delta E_{atrito} + \Delta E_{ele}$

The energy of a rotating body is $E_{rot}=\frac{Iw^2}{2}$ I being the moment of inertia, so the variation of mechanical energy between each acquisition will be:

$\Delta E_{mec}=\frac{I(w_{2exp}^2-w_{1exp}^2)}{2}$

$w_{2exp}$ and $w_{1exp}$ being the angular velocity in that acquisition and in the acquisition that preceded respectively.

The loss of energy by friction will be:

$\Delta E_{atrito}=Iw_{exp}\left(w_{2wo/fr}-w_{1wo/fr}\right)$

$w_{exp}$ is the angular velocity of the disc in that acquisition and $w_{2wo/fr}$ $w_{1wo/fr}$ are the extrapolated velocity of the disc in a no friction situation at the time of the acquition and the previous one respectively.

The dissipated power is:

$P=VI=\frac{V^2}{R}$

The rms voltage across one winding is:

$V_{rms}=\frac{V_{medida}}{\sqrt{3}\sqrt{2}}$

In the setup used the energy dissipates in 3 branches so the power comes multiplied by 3. Also, each winding is in parallel with a resistor with the same resistence value $4,7\Omega$, what means that $R=4,7\Omega$ can be used in the equation multiplying the power by 2.

$P=3\times2\times\frac{V_{rms}^2}{R}=3\times2\times\left(\frac{V_{medida}}{\sqrt{3}\sqrt{2}}\right)^2\frac{1}{R}$

$P=\frac{V^2}{R}$

$\Delta E_{ele}=P*\Delta t$

$\Delta t$

$Balance = \Delta E_{mec} - \Delta E_{friction} - \Delta E_{ele}$

$1,274\times10^{-4}$ $kg$ $m^2$

$I=\frac{m\left(r_1^2+r_2^2\right)}{2}=\frac{0,115\left(0,013^2+0,047^2\right)}{2}=1,367\times 10^{-4}kg \; m^2$

$\frac{\left|1,274\times 10^{-4}-1,367\times 10^{-4}\right|}{\left|1,367\times 10^{-4}\right|}\times 100=6,8\%$