OII na Sperimentale

Experiment Italy E2 page 1 of 13

Experiment Italy E2 page 2 of 13

x y $y_0$

Experiment Italy E2 page 3 of 13 Il potenziale a gobba di cammello della trappola magnetica funziona come un oscillatore unidimensionale. Ciò vuol dire che se si allontana di poco la grafite dalla posizione di equilibrio lungo l’asse

z M $\chi$ $\chi$ Q $\mu \text{A}$

Experiment Italy E2 page 4 of 13

HB/0.3, HB/0.5, HB/0.7, e HB/0.9. Prendi nota delle costanti e dei dati per il calcolo degli esatti diametri. Puoi aver bisogno di rompere le sbarrette di grafite riducendole alla lunghezza richiesta dal problema. 8. 9.

Experiment Italy E2 page 5 of 13 10. 11. 12. 13. 14. 15. 1. 2. 3. 4. R L ${\epsilon }_M$ $\rho$ 3 d d d d T ${\mu }_0$ $\times$ $-6$ $k_B$ $\times$ $-23$ g 2

Experiment Italy E2 page 6 of 13

M m M M = m $\pi$ L $R^2$ R L M M x B x $\le$ (x) = BI m ${\mu }_0$ $2\pi$ L $x_P$ x x

Experiment Italy E2 page 7 of 13

$B_0$ B x $\le x\le$ x p M

x R

Experiment Italy E2 page 8 of 13

$x_{\text{OFFSET}}$ x x $\chi$ $y_0$ $y_0$ ( ) $F_M$ $y_0$ $\chi$ $y_0$ $\mu =(1+\chi ){\mu }_0$ $\mu$ ( ) = $-$ f ( ) $F_M$ $y_0$ $\chi$ ${\mu }_0$ $M^2$ $V_r$ 2 $R^4$ $a^5$ $y_0$ a ( ) $F_M$ $y_0$ $V_r$ M a a = R + /2 ${\epsilon }_M$ ${\epsilon }_M$ ${\epsilon }_M$ f (u) f(u) = $4u(3-$ $)(1-$ ) $u^2$ $u^2$ $(1+u^2{)}^5$ $y_0$

Experiment Italy E2 page 9 of 13 $\chi$

Experiment Italy E2 page 10 of 13

$\chi$ y0 $\chi$ z

$U_M$ 1 $2k_z{z}^2$ $k_z$ z $k_z$ M $\chi$ = $-$ $\chi$ $k_z$ $C_1{\mu }_0$ $M^2{V}_r$ ${\mu }_0$ $V_r$ $C_1$ 2 l A

Experiment Italy E2 page 11 of 13 $\chi$

Experiment Italy E2 page 12 of 13

Si osserva che l’oscillazione della barra di grafite è frenata dall’attrito con l’aria. Si vuole capire qual è la dipendenza dell’attrito dalla forma della sbarretta di grafite (diametro e lunghezza) per stimare la viscosità dell’aria .

$\mu \text{A}$ z(t) = A $\sin (\omega t)$ $e^{-t/\tau }$ A $\omega =2\pi f$ t $e^{-t/\tau }$ $\tau$ $Q=\omega \tau /2$ Q Q Q $\tau$ $\tau$ = ln(0.607 $\times$ ) 2 3 $\rho r^2$ $\mu \text{A}$ l r $\rho$ r l $\mu \text{A}$ $\tau$ $\tau$ $\tau$ $\mu \text{A}$ 2 ) 0.5

$a_n$ 4 $T\omega$ $k_B$ QmR ​ $k_B$ T mR $\tau$

Experiment Italy E2 page 13 of 13

$a_n$ S S l $\Delta z$ S N $\Delta z$ N S Q

Topic: Magnetism, Oscillations & Waves, Fluid Mechanics Metodi: Simple Harmonic Motion Analysis, Experimental Data Analysis, Physical Modeling Competenze: Experimental Data Analysis, Mathematical Modeling, Graph Linearization Objects: Magnet, Rod Fonte: Testo (PDF) — p.1 Soluzione: Soluzioni (PDF)

Experiments Italy E2 page 1 of 13

Experiments Italy E2 Page 2 of 13

x y $y_0$

Experiments Italy E2 page 3 of 13 The camel-shaped potential of the magnetic trap works like an oscillator. It’s one-dimensional. This means that if you move graphite away from the equilibrium position a little bit, along the axis

z M $\chi$ $\chi$ Q $\mu \text{A}$

Experiments Italy E2 Page 4 of 13

HB/0.3, HB/0.5, HB/0.7, e HB/0.9. Note the constants and data for calculating the exact diameters. You may need it. The problem is to break down the graphite bars and reduce them to the length required by the problem. 8. 9.

Experiments Italy E2 page 5 of 13 10. 11. 12. 13. 14. 15. 1. 2. 3. 4. R L ${\epsilon }_M$ $\rho$ 3 d d d d T ${\mu }_0$ $\times$ $-6$ $k_B$ $\times$ $-23$ g 2

Experiments Italy E2 page 6 of 13

M m M M = m $\pi$ L $R^2$ R L M M x B x $\le$ (x) = BI m ${\mu }_0$ $2\pi$ L $x_P$ x x

Experiments Italy E2 page 7 of 13

$B_0$ B x $\le x\le$ x p M

x R

Experiments Italy E2 page 8 of 13

$x_{\text{OFFSET}}$ x x $\chi$ $y_0$ $y_0$ ( ) $F_M$ $y_0$ $\chi$ $y_0$ $\mu =(1+\chi ){\mu }_0$ $\mu$ ( ) = $-$ f ( ) $F_M$ $y_0$ $\chi$ ${\mu }_0$ $M^2$ $V_r$ 2 $R^4$ $a^5$ $y_0$ a ( ) $F_M$ $y_0$ $V_r$ M a a = R + /2 ${\epsilon }_M$ ${\epsilon }_M$ ${\epsilon }_M$ f (u) f(u) = $4u(3-$ $)(1-$ ) $u^2$ $u^2$ $(1+u^2{)}^5$ $y_0$

Experiments Italy E2 page 9 of 13 $\chi$

Experiments Italy E2 page 10 of 13

$\chi$ y0 $\chi$ z

$U_M$ 1 $2k_z{z}^2$ $k_z$ z $k_z$ M $\chi$ = $-$ $\chi$ $k_z$ $C_1{\mu }_0$ $M^2{V}_r$ ${\mu }_0$ $V_r$ $C_1$ 2 l A

Experiments Italy E2 page 11 of 13 $\chi$

Experiments Italy E2 page 12 of 13

It is noted that the oscillation of the graphite bar is slowed by friction with the air. You want to understand What is the dependence of friction on the shape of the graphite bar (diameter and length) for estimate the viscosity of the air .

$\mu \text{A}$ z(t) = A $\sin (\omega t)$ $e^{-t/\tau }$ A $\omega =2\pi f$ t $e^{-t/\tau }$ $\tau$ $Q=\omega \tau /2$ Q Q Q $\tau$ $\tau$ = ln(0.607 $\times$ ) 2 3 $\rho r^2$ $\mu \text{A}$ l r $\rho$ r l $\mu \text{A}$ $\tau$ $\tau$ $\tau$ $\mu \text{A}$ 2 ) 0.5

$a_n$ 4 $T\omega$ $k_B$ The following is the list of the countries: ​ $k_B$ T mR $\tau$

Experiments Italy E2 page 13 of 13

$a_n$ S S l $\Delta z$ S N $\Delta z$ N S Q

Topic: Magnetism, Oscillations & Waves, Fluid Mechanics Metodi: Simple Harmonic Motion Analysis, Experimental Data Analysis, Physical Modeling Competenze: Experimental Data Analysis, Mathematical Modeling, Graph Linearization Objects: Magnet, Rod Fonte: Testo (PDF) — p.1 Soluzione: Soluzioni (PDF)