Phys221Sp2000Exam3Sol

Phys221Sp2000Exam3Sol - P221/32000/Problem 3A. A particle...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P221/32000/Problem 3A. A particle of mass m = 0.5 kg oscillates in one dimension with simple harmonic motion so that its displacement a: from its equilibrium position varies with time t according to the figure below. Notice that a: is in meters and t is in seconds. 5.0 x (meter) ) C) (b) [1 point] Determine the amplitude of the oscillation. _ The displacement x varies from — 5.0 m to + 5.0 in, a range of 10.0 m, so the amplitude, which is half the range, is 5.0 m. (c) [2 points] Determine the period and frequency of the oscillation. The maximum displacement occurs at 0.5 s, 1.5 s, 2.5 5, etc, so the period is apparently T = l0 s. ' The frequency f of the oscillation is thenf = 1/T = 10 Hz. (d) [2 points] Determine the equation for x(t) that describes this simple harmonic motion. At time t = 0.0 s. x is at its minimum (most negative) displacement, suggesting a function of the form z(t) = — A cos wt. Since A = 5.0 m (from part (b) above, and w = 27Tf = 27r(1.0 Hz) = 27r rad/s, x(t) = - (5.0 m) cos [(6.285rad/s) t] (e) [3 points] Determine the total energy (in joules) of the particle at t = 0.00 s. The total energy of the particle is = lie/i2 = §mw2A2 = §(0.5 kg)(27r 5‘1)2(5.0 m)2 = 2571'2 J z 250 J. (f) [2 points] Determine the acceleration of the particle at t = 0.00 s. The acceleration is a(t) = — w2x(t) fOr sinusoidal motion, so att = 0.00 s, a(t) = -— (271')2( — 5.0 m) z +m rn/s2 2W P221/SZOOO/Problem 3B. A satellite of mass m is placed in a geosynchronous circular orbit about the earth. (Geosynchronous means the period T of the satellite in its orbit is 24 hours.) Assume that the gravitational constant G, the earth's mass ME, and the earth's radius RE are known quantities. Express all your answers as algebraic expressions involving some or all of the Quantities m, T, G, ME, and R; (b) [4 points] Determine the radius R of this orbit and the satellite's speed 1). Newton's second law of motion, together with the gravitational force law and the centripetal acceleration a = 112/7 gives GMEm/R2 = mv2/R so 122 = GME/R 2 __, circumference __ 27TH 277R _ GM; However,v— ——pefi0d —— T so ( T ) — R and Rs : GMET2/47r2 which yields R = Thenv = i/GME/ = GME/S/ GMECZ‘Q/47r2 (c) [4 points] What is the escape speed of an object lomted at the distance R calculated in part (b)? In other words, what speed must it have to be able to escape from that distance to infinity? The mechanical energy of an object at a distance R from the earth is E=K+U=§mv2—G—"g‘4h. To be able to escape, E Z 0 so my? — ElgflL 2 0 or v2 > 2GME/R ‘= zoME/e/GMET2/47r2 so the escape speed is vesc = i /2GME/«3/ GMETQ /47r2 (d) [2 points] What is the total energy of the satellite in its geosynchronous orbit? The total energy of the satellite, which has the orbital radius R and the orbital speed 1) determined in part (b), is lmv2_ GmME ~ lm GME _ Grail/IE __ _ GmIWE _ _ GmM 2 R — 2 R R ’ 2R _ Z37GMET2/47r2‘ (e) [2 bonus points] Prove that for any circular satellite orbit, the kinetic energy K is equal to half the magnitude of the gravitational potential energy U, if we take U = 0 at r = 00. Taking U = O at infinity, the gravitational potential energy of the earth-satellite system is U = — Egg-ML, which has magnitude —CL"}§!E-. The satellite's kinetic energy is §mv2, which, since 112 = GME/R for a circular orbit, is éc—"fibflg or half the magnitude of the potential energy. P221/SZOOO/Problem 3C. A solid nonconducting sphere of radius R has a total positive charge Q uniformly distributed throughout its volume. Use Gauss' law in determining the electric field for this arrangement. (b) [2 points] What is the appropriate Gaussian surface to choose for this physical situation? Why? The appropriate Gaussian surface is a sphere of radius 7*, where r can be larger or smaller than R. This is dUe to the fact that the physical situation described has spherical symmetry, so the electric field can only be of the form EX?) = Er(r) ’r‘. (c) [2 points] Determine the magnitude and direction of the electric field for r > R, where 'r is the distance from the center of the sphere. Applying Gauss' law to the case r > R, the surface integral $1113} (121 becomes 47rr2ET and since the charge inside is + Q, 47rr2Er(r) = 62/60 so 47rr2ET(r) = Q/47reor2. (d) [4 points] Determine the magnitude and direction of the electric field for 'r < R. For T < R the surface integral has the same expression, 47rr2ET(r), as in part (c), but the charge is just the charge inside the sphere of radius 7'. This charge is most easily expressed as a fraction of the total charge Q, the fraction being volume of sphere of radius r __ 3; TTT3 _ r3 volume of sphere of radius R _ grim — R3 ‘ 2 _ Q’s .. L _ i. Thus 47m” E,(r) — 605,3 and E,(r) ~ AWTQEORs — 47mm (e) [2 points] Graph the electric field magnitude E('r) as a function of 1- for the range 0 _<_ r S 2R. MULTIPLE-CHOICE QUESTIONS Notice that these questions are numbered 21 -3 0. These questions are worth 2 points each. 21. A uniform stick of length L is pivoted about a horizontal axis through a small hole near its 0 cm mark. (The moment of inertia ofa uniform rod of mass M and length L about an axis through one end, perpendicular to the rod, is I = ML2/3.) What is the approximate period of the stick when it swings back and forth? (A) 0.5s (B) 1.05 *(C) 2.05 (D) 4.05 (E) 8.0s Solution: The distance from the pivot to the center of mass is L/2, so u; = ; /—Mgd/ = ,/Mg-;-L/(ML2/3) .—. fry/2L and T = 27T/w = 27n/2L/3g % 277 2(l.50 m)/3(10 m/sg) % 27n/152/1 % 2 s. 22. A single pendulum swings from a rope of length L with its angle 9 with the vertical direction given by 6(t) = (0.05 rad) cos [(2.00 rad/s) t — 0.8 rad]. What is its length L? (A) 5.0 m (B) 4.5 m (C) 3.5 m *(D) 2.5 m v (E) 1. O m Solution: From the formula for 6(t), we see that w = 2.00 rad/s. Since w = x/g/L for a simple pendulum, L = g/Lo2 = (10 111/52)/(2.OO/s)2 = 10/4 m = 2,5 m. 23. An electron in a hydrogen atom makes a transition from the n, = 4 to the n = 3 energy state. The energy of the emitted photon is: (A) 0.25 eV *(B) 0.66 eV (C) 1.05 eV (D) 2.51 eV (E) 10.2 eV Solution: The energy ofthe n = 4 state is — 13'166‘V = -— 0.85 eV and that ofthe n = 3 state is — 132 W = — 1.51 eV. The difierence between these is 0.66 eV, which is the energy of the emitted photon. 24. For an oscillation subjected to a damping force proportional to its velocity, which of the following statements is true: ‘ (A) The displacement is a sinusoidal function of time (B) The velocity is a sinusoidal function of time. (C) The frequency is a decreasing function of time. (D) The mechanical energy is constant. *(E) None of (A) through (D) are true. 25. Two uncharged metal spheres, L and V g R, are in contact, as shown below. They rest on insulating stands. A negatively I ‘ charged rod is brought close to L, but not a touching it, as shown. The two spheres are slightly separated, after which the rod - " is then withdrawn. As a result, (A) both spheres are neutral. (B) both spheres are positively charged. (C) both spheres are negatively charged. (D) L is negative and R is positive *(E) L is positive and R is negative Solution: The negative rod will induce a positive charge on L and a negative charge on K which become permanent after the two spheres are separated. 26. The diagram shows the electric field lines in a region of space containing two. small charged spheres, Y and Z. Which of the following statements is true? (A) Y is negatively charged and Z is positively charged. (B) The magnitude of the electric field is the same everywhere. (C) The electric field is strongest midway between Y and Z. *(D) A small negatively—charged particle placed at X would be pushed to the right. (E) The charges on Y and Z must have the same sign. Solution: Since electric field lines leave Y and are directed towards Z, Y is a positive charge and Z is a negative charge. The magnitude of the electric field is not constant, but is greatest near the charges Y and Z. 27. A-point charge is placed at the center of a spherical Gaussian surface. The total electric flux <PE through this Gaussian surface will change (A) if the sphere is replaced by a cube of the same volume. (B) if the radius of the sphere is increased. (C) if the point charge is moved slightly off center (but remains inside the surface). *(D) if the point charge is moved just outside the sphere (E) if a second point charge is placed just outside the sphere. _________________—_.____—..—————-——-——-—-—-————--—- 28. In planetary motion, the line from the star to the planet sweeps out equal areas in equal times. This is a direct consequence of: (A) the conservation of electric charge (B) the conservation of mass *(C) the conservation of angular momentum (D) the conservation of linear momentum (E) the conservatiOn of mechanical energy _________________—_____..__————-_—————-——-——- 29. The gravitational force on an object at the surface of the earth has a magnitude of 90 N. What is the magnitude of the gravitational force on the same object if it is placed at a distance of 3B from the center of the earth, where R is the radius of the earth? *(A) 10 N (B) 30 N (C) 90 N (D) 270 N (E) 810 N Solution: Moving an object three times as far away reduces the gravitational force to (1/3)2 = 1/9 of its initial value. 1/9 of 90 N is 10 N. 30. A planet is in a circular orbit around the sun. Its distance from the sun is four times the average distance of the earth from the sun. The period of this planet, in earth years, is: (A) 2 (B) 4 *(C) 8 . (D) 16 (E) 64 Solution: By Kepler‘s third law, the period squared is proportional to the radius cubed, so the period is proportional to the 3/2 power ofthe radius. The 3/2 power of4 is the cube of 2, or 8. Solutions mflflflfllfl fl MB III-.- ...
View Full Document

Page1 / 7

Phys221Sp2000Exam3Sol - P221/32000/Problem 3A. A particle...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online