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1
General guidelines for marking
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0.00 |
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| 2 Rationalizes that the net force on the dipole is zero if the two poles move with equal velocities; Just argument $v= \mathrm{const} \Rightarrow \sum \vec{F}=0$ is $0~$pts. | 0.70 |
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| 3 Concludes that $\omega_{0}=0$. | 0.30 |
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| 4 Using the argument of zero torque, concludes that the velocity should be perpendicular to the dipole; | 0.70 |
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| 5 Just argument $\omega= \mathrm{const} =0 \Rightarrow \vec{\tau}=0$ | 0.40 |
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| 6 States explicitly that $\vec{v}_{0} \| Y$ (or $\perp X$ ). | 0.30 |
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| 1 Derives expression for the magnitude of the net force on the dipole in terms of $\omega$ AND states explicitly that it is parallel to the dipole axis OR derives one single vector expression. | 0.90 |
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| 2 Realizes (drawing or explicit statement) that $\vec{F}$ and the dipole axis point to the center of the orbit, and concludes that $\omega_{0}$ is equal to the orbital angular velocity. | 0.50 |
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| 3 Writes down Newton's second law for the circular motion. | 0.50 |
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| 4 Makes use of the relation $v_{0}=|\omega| R_{c}$. | 0.20 |
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| 5 Derives expression for $v_{0}$ and specifies its direction (drawing or statement) OR derives one single vector expression for $\vec{v}_{0}$ | 0.30 |
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| 6 Direction is wrong or missing | 0.20 |
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| 7 Derives explicitly $R_{c}=q b D /\left(2 m\left|\omega_{0}\right|\right)$. If $|\cdot|$ is omitted, still full points. | 0.30 |
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| 8 Writes down the coordinates of the center of the orbit: | 0.00 |
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| 9 Correct $x_{c}$ (including sign). $x_{c}=q b D /\left(2 m \omega_{0}\right)$ is a correct answer | 0.20 |
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| 10 Correct $y_{c}$ | 0.10 |
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| 1 M1 By integrating the equation(s) of motion derives a "generalized momentum" conservation law - a relationship between the linear momentum $2 m \vec{v}$ and the dipole moment $\vec{p}$ - in vector form OR for the Cartesian components. | 1.50 |
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| 2 M1 States explicitly that the kinetic energy of the dipole conserves. | 0.30 |
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| 3 M1 Writes down explicit expression for the kinetic energy in terms of angular velocity and linear velocity of the center of mass. | 0.50 |
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| 4 M1 Realizes that $\omega_{0}$ is minimal when $\omega_{1}=0$ in the reversed position. | 0.20 |
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| 5 M1 By using the "generalized momentum" conservation, derives explicit expression for the linear velocity $v_{1}$. | 0.50 |
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| 6 M1 Applies the conservation of energy to find relationship between $v_{1}$ and $\omega_{\text {min }}$ | 0.80 |
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| 7 M1 Derives the final expression for $\omega_{\text {min }}$ | 0.20 |
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| M2 Alternative approach: pendulum analogy | ||
| 9 M2 Derives the expression $\tau=-B(\vec{p} \cdot \vec{v})$ for the torque. Even if the derivation has been made in parts (a) or (b), the points should be assigned to Task (c); If term $(\vec{B} \cdot \vec{p})$ is not cancelled, still full points | 0.50 |
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| 10 M2 By integrating the equations of motion, expresses $v_{x}$ and $v_{y}$ in terms of $\theta$. | 1.50 |
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| 11 M2 Writes down the equation of rotational motion in terms of $\sin \theta$. | 0.50 |
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| 12 M2 States that the angular dynamics of the dipole is equivalent to a large-amplitude oscillation of a mathematical pendulum. | 0.30 |
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| 13 M2 Realizes that $\omega_{0}$ is minimal when $\omega_{1}=0$ in the reversed position. | 0.20 |
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| 14 M2 Applies the conservation of energy to the "equivalent pendulum". | 0.80 |
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| 15 M2 Derives the final expression for $\omega_{\text {min }}$ | 0.20 |
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| 1 Rationalizes that the asymptote is parallel to $Y$, i.e. $x= \pm D$. | 0.10 |
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| 2 Rationalizes that asymptotically the motion is linear uniform | 0.20 |
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| 3 Either finds conservation law $\vec{L}_{O}+\vec{B}(\vec{R} \cdot \vec{p})$ OR writes $x_{\infty}$ as integral of $v_{x}$ (with explicit expression for $v_{x}$ ) as a method to find $D$. | 0.30 |
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| 4 Correctly computes generalized angular momentum at 0 and $\infty$ OR uses $\sin \theta \propto \ddot{\theta}$ in integral. | 0.20 |
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| 5 Concludes that $D=d$. | 0.20 |
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