Difference between revisions of "Angular Momentum Conservation"
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[[File:Discos_balanco_energetico.png|thumb|alt=|Figura5: Final energy balance showing electrical and mechanic component making it possible to get the total moment of inertia]] | [[File:Discos_balanco_energetico.png|thumb|alt=|Figura5: Final energy balance showing electrical and mechanic component making it possible to get the total moment of inertia]] | ||
− | The bottom disc is accelerated by the motor to a selected angular velocity. At that time the motor supply is disconnected and the disc is allowed to rotate freely. When a certain pre-determined velocity is reached, a relay puts each motor winding in parallel with a resistor | + | The bottom disc is accelerated by the motor to a selected angular velocity. At that time the motor supply is disconnected and the disc is allowed to rotate freely. When a certain pre-determined velocity is reached, a relay puts each motor winding in parallel with a resistor resistance is the same as the motor's windings ('''Figure 3'''). These resistors will dissipate energy acting as an electromagnetic brake. Voltage and velocity as functions of time are given in a table of results in the end of the session. |
'''Figures 2''' and '''4''' are plots obtained through the table of results of an experiment in which the relay turns on when the rotating discs reach 1400 rpm. | '''Figures 2''' and '''4''' are plots obtained through the table of results of an experiment in which the relay turns on when the rotating discs reach 1400 rpm. | ||
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This experiment gave results that differ by '''6,8%''' from the ones calculated theoreticaly. | This experiment gave results that differ by '''6,8%''' from the ones calculated theoreticaly. | ||
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=Physics= | =Physics= |
Revision as of 12:07, 15 October 2013
Contents
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.
Links
- Video: rtsp://elabmc.ist.utl.pt/inertiadisks.sdp
- Laboratory: Intermediate in e-lab.ist.eu
- Control room: [unavaliable]
- Level: ***
Experimental Apparatus
The experimental apparatus is based in a PC hard disk drive motor and its spinning disc with a mass of 115g, 12.5mm internal radius and 47.5mm external. A second disc with a mass of 69g and the same dimensions 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 and voltage characteristic is measured allowing a rigorous calculation of energy dissipation.
Protocol - Angular Momentum Conservation
The bottom disc is accelerated by the motor until it reaches a selected angular velocity. At this instant the motor is disconnected from supply and the disc allowed to rotate freely. When a certain pre-determined velocity is reached, the servo lets the suspended disc initially at rest fall on top of the rotating disc.
Data taken from the experiment is given and plotted with the disc velocity in function of time.
Figure1 is a plot of the results of an experient in which a servo lets the suspended disc fall when the disc below reaches 1000 rpm.
Doing a linear regression between the deceleration and fall of the disc, it is possible to obtain the predicted velocity at any time. This gives us the rule of thumb for the friction deceleration related to angular velocity.
Advanced Protocol - Moment of Inertia Evaluation
The bottom disc is accelerated by the motor to a selected angular velocity. At that time the motor supply is disconnected and the disc is allowed to rotate freely. When a certain pre-determined velocity is reached, a relay puts each motor winding in parallel with a resistor resistance is the same as the motor's windings (Figure 3). These resistors will dissipate energy acting as an electromagnetic brake. Voltage and velocity as functions of time are given in a table of results in the end of the session.
Figures 2 and 4 are plots obtained through the table of results of an experiment in which the relay turns on when the rotating discs reach 1400 rpm.
Using the first data from Figure 2, before the braking, to do a linear regression, one can get the angular deceleration caused by friction, assumed constant at all time, from the slope of the rect. From the deceleration it will be possible to calculate differentially the instantaneous loss of angular momentum.
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.
$\Delta E_{mec} = \Delta E_{fr} + \Delta E_{ele}$
The energy of a rotating body is $E_{rot}=\frac{I w^2}{2}$ I being the moment of inertia. Then, 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 two consecutive acquisitions.
The loss of energy by friction will be:
$\Delta E_{fr}=Iw_{exp}\left(w_{2wo/fr}-w_{1wo/fr}\right)$
$w_{exp}$ 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:
$V_{rms}=\frac{V_{measured}}{\sqrt{3}\sqrt{2}}$
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 $4,7\Omega$, which means the power equation will come multiplied by 2 and $R=4,7\Omega$ will be used.
$P=3\times2\times\frac{V_{rms}^2}{R}=3\times2\times\left(\frac{V_{measured}}{\sqrt{3}\sqrt{2}}\right)^2\frac{1}{R}$
$P=\frac{V^2}{R}$
The energy dissipated will be:
$\Delta E_{ele}=P*\Delta t$
Where $\Delta t$ is the time between acquisitions.
In the end the ballance between each consecutive acquitition is summed.
$Balance = \Delta E_{mec} - \Delta E_{friction} - \Delta E_{ele}$
Finaly the goal-seek function of Microsoft Excel is used to force the sum of balances to be 0 (zero) changing the value of I.
Using this method, it was reached an experimental value of $1,274\times10^{-4}$ $kg$ $m^2$ for the moment of inertia.
Figure 5 shows the energy of the disc in function of time, the energy lost by friction and electromagnetic breaking and the sum of all energies that allows the verification of the conservation of energy through all the experiment.
The disc is in fact a ring with interior radius 13mm and exterior 47mm so it's theoretical moment of inertia is:
$I=\frac{m\left(r_1^2+r_2^2\right)}{2}=\frac{0,115\left(0,013^2+0,047^2\right)}{2}=1,367\times 10^{-4}kg \; m^2$
Evaluating the accuracy:
$\frac{\left|1,274\times 10^{-4}-1,367\times 10^{-4}\right|}{\left|1,367\times 10^{-4}\right|}\times 100=6,8\%$
This experiment gave results that differ by 6,8% from the ones calculated theoreticaly.
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 ($r_1=12,5mm, r_2=47,5mm$) 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 $I_m$
$I_m = \frac{I_{Df} \omega_f - I_{Di} \omega_i}{\omega_i-\omega_f}$