9/16/2015MasteringPhysics: Ch 21 HW Dipole Motion in a Uniform Field Consider an electric dipole located in a region with an electric field of magnitude E pointing in the positive y direction. The positive and negative ends of the dipole have charges + q and − q , respectively, and the two charges are a distance D apart. The dipole has moment of inertia I about its center of mass. The dipole is released from angle θ = θ 0 , and it is allowed to rotate freely. Part A What is ωmax , the magnitude of the dipole's angular velocity when it is pointing along the y axis? Express your answer in terms of quantities given in the problem introduction. Hint 1. How to approach the problem Because there is no dissipation (friction, air resistance, etc.), you can solve this problem using conservation of energy. When the dipole is released from rest, it has potential energy but no kinetic energy. When the dipole is aligned with the y axis, it is rotating, and therefore has both kinetic and potential energy. The sum of potential and kinetic energy will remain constant. Hint 2. Find the potential energy Find the dipole's potential energy U (θ) due to its interaction with the electric field as a function of the angle θ that the dipole's positive end makes with the positive y axis. Define the potential energy to be zero when the dipole is oriented perpendicular to the field: U (π/2) = 0 . Express your answer in terms of E , q , D, and θ . Hint 1. The formula for the potential energy of a dipole The general formula for the potential energy of an electric dipole with dipole moment P ⃗ in the presence of a uniform electric field E⃗ is U ⃗ ⃗ = −P ⋅ E . Hint 2. The dipole moment The dipole moment of the electric dipole P ⃗ , when it makes an angle θ with the positive y axis can be written as Typesetting math: 52% ⃗ ^ ^ P = qD(sin θ i + cos θ j ) . https://session.masteringphysics.com/myct/itemView?assignmentProblemID=54120380&view=print&offset=next 1/4 Find the total energy when θ = 0 Find an expression for Etotal . q . Express your answer in terms of some or all of the variables E . the total energy (kinetic plus potential) at the moment the dipole is released from rest at angle θ 0 with respect to the y axis. D. Hint 1. Express your answer in terms of quantities given in the problem introduction and ωmax . What is kinetic energy as a function of angular velocity? What is the kinetic energy K of a body rotating with angular velocity ω around an axis about which the moment of inertia is I ? ANSWER: K 1 = 2 Iω 2 ANSWER: Etotal = −EqD + 1 2 I ω max 2 ANSWER: ω max = − −−−−−−−−−−−−−−−−− √ 2 I qED(1 − cos(θ0 )) https://session. ANSWER: Etotal = −EqDcos(θ 0 ) Hint 4.masteringphysics. Find the total energy at the moment of release Find Etotal . and θ 0 .com/myct/itemView?assignmentProblemID=54120380&view=print&offset=next 2/4 . the total energy (kinetic plus potential) at the moment when the dipole is aligned with the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: U (π/2) = 0 .9/16/2015 MasteringPhysics: Ch 21 HW ANSWER: U (θ) = −EqDcos(θ) Hint 3. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: U (π/2) = 0 . This will allow you to read off the expression for \texttip{\omega }{omega}. the dipole will exhibit simple harmonic motion after it is released. (Note: Here. Part B If \texttip{\theta _{\rm 0}}{theta_0} is small. Recall that \texttip{\alpha }{alpha}. as you would expect. How to approach the problem The equation of motion for a simple harmonic oscillator can always be written in the standard form \large{\frac{d^2x}{dt^2}=\omega^2 x}. To solve this problem.com/myct/itemView?assignmentProblemID=54120380&view=print&offset=next 3/4 . Hint 2. Compute the torque What is the magnitude of the torque \texttip{\tau \left(\theta \right)}{tau(theta)} that the electric field exerts about the center of mass of the dipole when the dipole is oriented at an angle \texttip{\theta }{theta} with respect to the electric field? Express the magnitude of the torque in terms of quantities given in the problem introduction and \texttip{\theta }{theta}.) Start with the angular analogue of Newton's second law: \tau=I\alpha. it denotes the frequency of the dipole's oscillation. An easier way to see this is to use the trigonometric identity 1 − cos θ = 2 sin 2 θ 2 to write ωmax as \large{2 \sin{\frac{\theta_0}{2}}\sqrt{\frac{qED}{I}}}.9/16/2015 MasteringPhysics: Ch 21 HW Correct Thus ωmax increases with increasing θ 0 . ANSWER: https://session. What is the period \texttip{T}{T} of the dipole's oscillations in this case? Express your answer in terms of \texttip{\pi }{pi} and quantities given in the problem introduction. is equal to the second derivative of \texttip{\theta }{theta}. you need to write the equation of motion for the dipole in the standard form with \texttip{x}{x} replaced by the angular variable \texttip{\theta }{theta}. The dipole moment When it makes an angle \texttip{\theta }{theta} with the positive y axis. the angular acceleration. which has a simple relationship to the period of oscillation. Hint 1. Hint 2. the dipole moment of the electric dipole \texttip{\vec{P}}{P_vec} can be written as \vec{P} = qD \left(\sin{\theta}\; \hat{i}+ \cos{\theta}\; \hat{j}\right).masteringphysics. the variable \texttip{\omega }{omega} does not represent the angular velocity of the dipole; rather. the torque can be related to the potential energy \texttip{U\left(\theta \right)}{U(theta)} by \large{\tau = \frac{dU}{d\theta}}. Alternatively. Formula for torque on a dipole The torque on a dipole with dipole moment \texttip{\vec{P}}{P_vec} in an electric field \texttip{\vec{E}} {E_vec} is given by \tau = \vec{P}\times\vec{E}. Hint 1. just as linear acceleration is equal to the second derivative of position. torque is a vector quantity. The smallangle approximation Because \texttip{\theta _{\rm 0}}{theta_0} is small.9/16/2015 MasteringPhysics: Ch 21 HW \texttip{\tau \left(\theta \right)}{tau(theta)} = q E D {\sin}\left({\theta}\right) Hint 3. (Recall that small angular displacements can be treated as vectors. the torque \texttip{\vec{\tau }}{tau_vec} and the (small) angular displacement \texttip{\vec{\theta }}{theta_vec} must be in opposite directions. since they obey vector addition. After all. Express your answer in terms of quantities given in the problem introduction. Compare this to the standard form \large{\frac{d^2x}{dt^2}=\omega^2 x} for a simple harmonic oscillator to obtain the oscillation frequency \texttip{\omega }{omega} for the motion of the dipole. you can apply the smallangle approximation to the expression \tau(\theta)=q E D\sin(\theta) for torque. The relationship between (angular) oscillation frequency and period The relationship between \texttip{\omega }{omega}. ANSWER: \texttip{T}{T} = \large{2{\pi}\sqrt{\frac{I}{qED}}} Correct https://session.) If you did the vector algebra carefully. For a system to oscillate. Find the oscillation frequency Putting together what you have so far yields \large{I\alpha=I\frac{d^2\theta}{dt^2}=q E D \theta}. while large angles do not. the angular oscillation frequency of the dipole. Hint 4. Now let's think about the direction. and take the torque to be \tau(\theta)=q E D \theta. keeping in mind that \texttip{\tau }{tau} now represents the component of \texttip{\vec{\tau }}{tau_vec} in the \texttip{\hat{\theta }}{theta_unit} direction.masteringphysics. For future purposes we will write this as \tau(\theta)= q E D \theta.com/myct/itemView?assignmentProblemID=54120380&view=print&offset=next 4/4 . and the period of oscillation \texttip{T}{T} is given by \large{\omega = \frac{2 \pi}{T}}. rather than the magnitude of \texttip{\vec{\tau }}{tau_vec}. the torque must be a restoring torque; that is. you would find that the correct vector equation is \vec{\tau}\left( \theta \right) = qED \vec{\theta}. ANSWER: \texttip{\omega }{omega} = \large{\sqrt{\frac{q E D}{I}}} Hint 5. Up to this point we have been interested only in the magnitude of the torque.