F s = k s I ring = MR 2 I disk = 1 2 MR2 I sphere = 2 5 MR2 d 2 x dt 2 = ω2 x x(t) = A cos(ωt + φ) ω = 2πf = 2π T U g = G m 1m 2 r ( 4π T 2 2 = GM ) r I rod,cm = 1 12 ML2 I rod,end = 1 ML2 I = I CM + MR 2 τ r F = rf sin θ τ = Iα = dl dt ω spring = k m E SHO = 1 2 ka2 T = 2π L g F g = G m 1m 2 r 2 g = F g m = G m r 2 v esc = 2GM r F b = ρv g P = ρgh A 1 v 1 = A 2 v 2 Constant Symbol Approximate Value Gravitational Constant G 6.67 10 11 Nm 2 /kg 2 Acceleration of Gravity (near Earth) g 9.80 m/s 2 Mass of Earth m e 5.98 10 24 kg Mass of Sun m s 1.99 10 0 kg Radius of Earth r e 6.7 10 6 m Mass of Mars M M 6.46 10 2 kg Radius of Mars orbit R mo 2.28 10 11 m Speed of light in vacuum c 2.998 10 8 m/s Standard atmospheric pressure P o 101. 10 Pa 1
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Name: Score: / Physics 204A Exam 4 12/7/16 Short-answer problems: Do any seven problems (not including the bonus problem) and clearly mark the one you wish to omit by drawing a diagonal line through the answer space. Show your work for complete credit. Six points each. 1. A long thin rod of mass m and length L is perfectly balanced on its end, on the ground. If nudged slightly, the rod will fall over. What is the speed of the upper end of the rod when it hits the horizontal ground? Assume that there is enough friction that the lower end of the rod does not slide. 2. Calculate by direct integration the rotational inertia of a thin rod of mass m and length L about an axis a distance D from one end. Confirm that your answer agrees with the rotational inertia of a thin rod rotating around one end, I = 1 ml2, when D = 0.
. A satellite in just the right orbit will travel around the Earth once every 24 hours. Since the Earth rotates once every 24 hours, this satellite will appear to remain in one fixed spot relative to the ground. (Very useful for satellite dishes: you don t have to re-align the dish halfway through The Walking Dead.) This is called geosynchronous orbit. What is the radius of geosynchronous orbit for Earth? 4. A neutron star has a mass equal to the mass of our Sun, but a radius of only 10 km. What would be your weight on the surface of this star? 4
5. A wooden board is 15 cm wide, cm thick, and 50 cm long. The density of the board is 650 kg/m. It s floating, flat, in a puddle of water. How far above the surface of the water will be the top of the board? 6. What force would it take to hold a regulation basketball under water? A regulation basketball is 75 cm in circumference, and has a mass of 624 grams. 5
7. Bubba is going bungee jumping with his girlfriend, Betty Sue. Bubba jumps first, and after the fall bounces up and down on the end of the bungee cord with period T 1 = 2.5 s. Betty Sue goes second, with the same bungee cord, and you time her oscillation period as T 2 =.1 s. What is Betty Sue s mass? (Bubba, of course, has mass m = 80 kg.) 8. An old-school mechanical pocket-watch keeps time using the oscillations of a rotary pendulum. Instead of a swinging pendulum, the watch uses the motion of a balance wheel with rotational inertia I that oscillates back and forth under the influence of a torsion spring. (For a torsion spring, the equivalent to Hooke s Law is τ = κθ.) Show that the motion of a balance wheel is simple harmonic, and find the period of the balance wheel in terms of I and κ. 9. Two-Point Bonus Question: The Earth orbits the sun once every 65.25 days. In that time, how many rotations does the Earth make about its axis? 6
Physics 204A Key Exam 4 12/7/16 Short-answer problems: Do any seven problems (not including the bonus problem) and clearly mark the one you wish to omit by drawing a diagonal line through the answer space. Show your work for complete credit. Six points each. 1. Answer: Use conservation of energy. E i = mg L 2 E f = 1 2 Iω2 I = 1 ml2 mg L 2 = 1 ( ) 1 2 ml2 ω 2 g = 1 g Lω2 = ω = L v = Lω = gl 2. Answer: I = L D D I = M L x 2 M L dx x L D D I = M [ L L 2 D + LD 2] L If D = 0, then I = 1 ML2 as expected. I = 1 M [ L 2 LD + D 2]. Answer: One way of doing this is to use Kepler s Third Law: ( ) 4π T 2 2 = r 2 GM ( T 2 GM r = 4π 2 ) 1 = 42, 164 km 4. Answer: This is a completely ridiculous question. For starters you would not exist on a neutron star, as you would be instantly flattened into a microscopic film of nuclear material. But anyway... F g = G Mm = 1 10 14 N r 2 Answers will vary, depending on the value of m used, but the answer should be on the order of 10 14 N. 1
5. Answer: The dimensions of the board aren t really necessary here; only the thickness t matters. The force of buoyancy on the board depends on the amount of the board under water, let s call that x. F b = ρ w gax where A is the area, which depends on length and width. The gravitational force on the board is F g = ρ b gat These two forces are equal, so ρ w x = ρ b t = x = t ρ b ρ w = 0.65t So if x = 0.65t is the amount of board under water, the amount of board above the water is 0.5t = 1.05 cm 6. Answer: The radius of the basketball is 7. Answer: and the volume of a sphere is r = C 2π V = 4 πr = 4 π ( C 2π ) 1 = C So the force necessary to hold the ball down can be calculated from the sum of forces on the ball: ( ) C F g + F = F b = F = F b mg = ρg 6π 2 mg = 6.7 N ω = k m = T = 2π m k Take the ratio of their periods: T 1 m1 = T 2 m 2 ( ) 2 T2 m 2 = m 1 = 12 kg T 1 That s about 270 pounds. This girl should probably not be bungee jumping... 6π 2 8. Answer: This is simple harmonic motion. τ = Iα = κθ α = d2 θ dt 2 = κ I θ ω = κ I 2 I = T = 2π κ
9. Answer: 66.25 As the Earth spins around the sun, whatever reference point you re using faces the sun exactly once per day. But as the year progresses, the direction of facing the sun changes! This adds one revolution to the rotation of the Earth.