OII 2016 Nazionale Teorica

P1 Lo stucco nel secchiello

1 Equilibrio – Dimostrazione: La forza necessaria per garantire l’equilibrio è minore di quella massima (di distacco). Forza d’attrito: $A=m_{1}g=8.32\le 8.34\le 8.35\text{N}$

2 Valore minimo dell’intervallo $\Delta t$: $(\Delta t{)}_{\text{min}}=\frac{m_{2}v}{(\alpha {\mu }_{\text{s}}M-\alpha m_1-m_2)g}=0.79\le 0.82\le 0.86\text{s}$

3 Velocità del secchiello $\vec{V}$: $V=\frac{m_{2}v}{m_1+m_2+M}=0.551\le 0.554\le 0.557{\text{m s}}^{-1}$ verso il basso.

4 Accelerazione del secchiello $\vec{a}$: $a=\frac{(m_1+m_2-{\mu }_{\text{d}}M)g}{m_1+m_2+M}=-1.02\le -0.99\le -0.96{\text{m s}}^{-2}$ verso l’alto.

5 Velocità del secchiello ${\vec{V}}^{\prime }$: $V^{\prime }=\frac{m_{2}v}{m_1+m_2+m_3/2+M}=0.507\le 0.510\le 0.513{\text{m s}}^{-1}$ verso il basso.

6 Accelerazione del secchiello ${\vec{a}}^{\prime }$: $a^{\prime }=\frac{(m_1+m_2-{\mu }_{\text{d}}M)g}{m_1+m_2+m_3/2+M}=-0.93\le -0.91\le -0.89{\text{m s}}^{-2}$ verso l’alto.

OLIMPIADI DI FISICA – Gara Nazionale Teorica: 15.04.2016 Riservato alla COMMISSIONE

Topic: Newtonian Mechanics, Conservation of Momentum Metodi: Free-Body Diagram, Impulse-Momentum Theorem, Kinematic Equations Competenze: Diagrammatic Reasoning, Mathematical Modeling, Physical Reasoning Objects: Block, Pulley Fonte: Testo (PDF) — p.1 Soluzione: Soluzioni (PDF)

P1 Stok in the bucket

1 Balance Demonstration: The force required to ensure balance is less than the maximum (out of detachment). The force of friction: $A=m_{1}g=8.32\le 8.34\le 8.35\text{N}$

2 Minimum value of the range $\Delta t$: $(\Delta t{)}_{\text{min}}=\frac{m_{2}v}{(\alpha {\mu }_{\text{s}}M-\alpha m_1-m_2)g}=0.79\le 0.82\le 0.86\text{s}$

3 Bottle speed $\vec{V}$: $V=\frac{m_{2}v}{m_1+m_2+M}=0.551\le 0.554\le 0.557{\text{m s}}^{-1}$ Down the hallway.

4 Bottle speed $\vec{a}$: $a=\frac{(m_1+m_2-{\mu }_{\text{d}}M)g}{m_1+m_2+M}=-1.02\le -0.99\le -0.96{\text{m s}}^{-2}$ Up the hill.

5 Bottle speed ${\vec{V}}^{\prime }$: $V^{\prime }=\frac{m_{2}v}{m_1+m_2+m_3/2+M}=0.507\le 0.510\le 0.513{\text{m s}}^{-1}$ Down the hallway.

6 The speed of the bucket ${\vec{a}}^{\prime }$: $a^{\prime }=\frac{(m_1+m_2-{\mu }_{\text{d}}M)g}{m_1+m_2+m_3/2+M}=-0.93\le -0.91\le -0.89{\text{m s}}^{-2}$ Up the hill.

Theoretical national competition: 15.04.2016 Reserved to the Commission

Topic: Newtonian Mechanics, Conservation of Momentum Metodi: Free-Body Diagram, Impulse-Momentum Theorem, Kinematic Equations Competenze: Diagrammatic Reasoning, Mathematical Modeling, Physical Reasoning Objects: Block, Pulley Fonte: Testo (PDF) — p.1 Soluzione: Soluzioni (PDF)

P2 Da lontano verso il solenoide

1 Campo e flusso interno: $B_{\text{int}}={\mu }_{0}nI$ ${\Phi }_{\text{int}}=\pi {\mu }_0{R}^{2}nI$

2 Flusso del campo esterno: Dimostrazione: Uso del teorema di Gauss o valutazione del numero di linee di campo. ${\Phi }_{\text{est}}=-\pi {\mu }_0{R}^{2}nI$

3 Forza sull’elettrone e traiettoria: $\vec{F}=-e\vec{v}\times \vec{B}$ Traiettoria $\Rightarrow$ Dimostrazione: Velocità e accelerazione sempre nello stesso piano.

4 Momento della forza: Dimostrazione: $\mathcal{M}=erv{B}_{\text{est}}(r)\text{sen}\alpha =-er{B}_{\text{est}}(r)v_r$.

5 Relazione tra $dL$ e $d\Phi$: $dL=\frac{e}{2\pi }d\Phi$

6 Espressione di $L$ in funzione di ${\Phi }_{\text{est}}$: $L(P)=\frac{e}{2\pi }{\Phi }_{\text{est}}$

7 Espressione della corrente minima: $I_{\text{min}}=\frac{2m{v}_0}{e{\mu }_{0}nR}$

OLIMPIADI DI FISICA – Gara Nazionale Teorica: 15.04.2016 Riservato alla COMMISSIONE

Topic: Magnetism, Rotational Dynamics, Electromagnetism Metodi: Gauss’s Law, Lorentz Force Analysis, Torque & Angular Momentum Analysis Competenze: Mathematical Modeling, Diagrammatic Reasoning, Physical Reasoning Objects: Solenoid, Electron Fonte: Testo (PDF) — p.2 Soluzione: Soluzioni (PDF)

P2 From far to the solenoid

1 Field and internal flow: $B_{\text{int}}={\mu }_{0}nI$ ${\Phi }_{\text{int}}=\pi {\mu }_0{R}^{2}nI$

2 Outward field flow: The following is the list of the samples: Use of Gauss’s theorem or evaluation of the number of field lines. ${\Phi }_{\text{est}}=-\pi {\mu }_0{R}^{2}nI$

3 Electron force and trajectory: $\vec{F}=-e\vec{v}\times \vec{B}$ Traiettoria $\Rightarrow$ The following is the list of the samples: Speed and acceleration always on the same plane.

4 Moment of force: The following is the list of the samples: $\mathcal{M}=erv{B}_{\text{est}}(r)\text{sen}\alpha =-er{B}_{\text{est}}(r)v_r$.

5 Relationship between $dL$ and $d\Phi$: $dL=\frac{e}{2\pi }d\Phi$

6 Expression of $L$ by ${\Phi }_{\text{est}}$: $L(P)=\frac{e}{2\pi }{\Phi }_{\text{est}}$

7 Minimum current expression: $I_{\text{min}}=\frac{2m{v}_0}{e{\mu }_{0}nR}$

Theoretical national competition: 15.04.2016 Reserved to the Commission

Topic: Magnetism, Rotational Dynamics, Electromagnetism Metodi: Gauss’s Law, Lorentz Force Analysis, Torque & Angular Momentum Analysis Competenze: Mathematical Modeling, Diagrammatic Reasoning, Physical Reasoning Objects: Solenoid, Electron Fonte: Testo (PDF) — p.2 Soluzione: Soluzioni (PDF)

P3 Scalda e raffredda

1 Potenza erogata e temperatura finale: $W_{\text{r}}={\mathcal{E}}^2/r$ $T_1=T_0+\frac{2{\mathcal{E}}^2}{3Rnr}\Delta t$

2 Potenza massima dispersa: $W_{\text{d}}=k\frac{{\ell }^2}{d}\frac{2{\mathcal{E}}^2}{3Rnr}\Delta t$

3 Condizione su $k$: $k\ll \frac{3dRn}{2{\ell }^2\Delta t}$

4 Forza sul blocco: $F_{\text{b}}=\frac{{\mathcal{E}}^2}{3hr}\Delta t$

5 Temperatura di equilibrio $T_{\text{eq}}$: $T_{\text{eq}}=(T_0+T_1)/2$

6 Aumento di entropia $\Delta S$: $\Delta S=\frac{3Rn}{2}\ln \frac{(T_0+T_1{)}^2}{4T_0{T}_1}$

7 Differenza delle temperature di equilibrio: Spiegazione: Lavoro prodotto sull’esterno

8 Temperatura di equilibrio $T_{\text{rev}}$: $T_{\text{rev}}=\sqrt{T_0{T}_1}$

Topic: Thermodynamics, Circuits, Conservation of Energy Metodi: First Law of Thermodynamics, Energy Conservation Method, Thermodynamic Cycle Analysis Competenze: Mathematical Modeling, Physical Reasoning, Diagrammatic Reasoning Objects: Block, Resistor Fonte: Testo (PDF) — p.3 Soluzione: Soluzioni (PDF)

P3 Heat and cooling

1 Power delivered and final temperature: $W_{\text{r}}={\mathcal{E}}^2/r$ $T_1=T_0+\frac{2{\mathcal{E}}^2}{3Rnr}\Delta t$

2 Maximum power dispersed: $W_{\text{d}}=k\frac{{\ell }^2}{d}\frac{2{\mathcal{E}}^2}{3Rnr}\Delta t$

3 Condition on $k$: $k\ll \frac{3dRn}{2{\ell }^2\Delta t}$

4 Force on the block: $F_{\text{b}}=\frac{{\mathcal{E}}^2}{3hr}\Delta t$

5 Balance temperature $T_{\text{eq}}$: $T_{\text{eq}}=(T_0+T_1)/2$

6 Increased entropy $\Delta S$: $\Delta S=\frac{3Rn}{2}\ln \frac{(T_0+T_1{)}^2}{4T_0{T}_1}$

7 Difference in equilibrium temperatures: Explanation: Work produced abroad

8 Balance temperature $T_{\text{rev}}$: $T_{\text{rev}}=\sqrt{T_0{T}_1}$

Topic: Thermodynamics, Circuits, Conservation of Energy Metodi: First Law of Thermodynamics, Energy Conservation Method, Thermodynamic Cycle Analysis Competenze: Mathematical Modeling, Physical Reasoning, Diagrammatic Reasoning Objects: Block, Resistor Fonte: Testo (PDF) — p.3 Soluzione: Soluzioni (PDF)