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Physics 3210Week 5 clicker questionsThe central-force Lagrangian isWhat is the Lagrangian equation of motion in r?A.B. C.D.   2 2 21r r U r2L       2r r F r     2r r U r     r r F r     2r r U r   A particle is observed to move in a spiral orbit r=k. What is the force law that produces this orbit?A.B. C. 22322kF r 1rr   22222kF r 1rr  D. E. 2232F r 1rr 2223kF r 1rr   232kF r 1rrA particle moves in an orbit where its radius alternately increases and decreases with time. What determines the turning points of the motion?A.B. C.r0D. E.dr0dt22dr0dtdr0d22dr0dIn one transit from rminback to rmin, an orbit moves by an angle D. What condition on Dmust hold if the orbit closes on itself?A.B. C.0DD. E.2D 2 a, a integerD 2, a integeraD 2a, a and b integersbD Physics 3210 Week 5Wednesday clicker questionsExam grade distributionMedian = 66Standard deviation = 22Current course grade=55% exam, 45% homeworkIn one transit from rminback to rmin, an orbit moves by an angle D. What condition on Dmust hold if the orbit closes on itself?A.B. C.0D D. E.2D 2 a, a integerD 2, a integeraD 2a, a and b integersbD Which of these plots show a physically possible V(r) for the gravitational potential?A.B.C.D.For motion in gravitational potential characterized by V(r) as plotted, what type of motion corresponds to total energy E2?A. Bounded circular motion with a fixed r.B. Bounded motion between a minimum and maximum r.C. Unbounded motion with a minimum r but no maximum r.D. Unbounded motion with neither a minimum nor a maximum r.For motion in gravitational potential characterized by V(r) as plotted, what type of motion corresponds to total energy E1?A. Bounded circular motion with a fixed r.B. Bounded motion between a minimum and maximum r.C. Unbounded motion with a minimum r but no maximum r.D. Unbounded motion with neither a minimum nor a maximum r.Given the equationand the constantswhat is the correct way to rewrite the orbit equation?A.B. C.D. E.22211krcos2E1k2222E, 1kk    =1+ cosrr=1+ cos=1 cosr r=1 cos r= 1+ cos  We showed thatWhat is the correct value of  when E=Vmin?A.B. C.D. E.2min2kV2011  22  Physics 3210 Week 5Friday clicker questionsWe showed thatWhat is the correct range of  when Vmin<E<0?A.B. C.D. E.2min2kV20101  12  10  Which orbit corresponds to motion with larger total energy?C. The orbits correspond to motion with equal energy.D. It cannot be determined from the information given.What is the correct value of  when E=0?A.B. C.D. E.011  22  Which parabolic orbit corresponds to motion with larger angular velocity (at a given angle)?C. The orbits correspond to motion with equal angular velocity.D. It cannot be determined from the information given.What is the correct range of  when E>0?A.B. C.D. E.001  12  10Which hyperbolic orbit corresponds to motion with larger total energy?C. The orbits correspond to motion with equal energy.D. It cannot be determined from the information


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CU-Boulder PHYS 3210 - Week 5

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