E = K + U. p mv. p i = p f. F dt = p. J t 1. a r = v2. F c = m v2. s = rθ. a t = rα. r 2 dm i. m i r 2 i. I ring = MR 2.

Transcription

1 v = v i + at x = x i + v i t at2 E = K + U p mv p i = p f L r p = Iω τ r F = rf sin θ v 2 = v 2 i + 2a x F = ma = dp dt = U v dx dt a dv dt = d2 x dt 2 A circle = πr 2 A sphere = 4πr 2 V sphere = 4 3 πr3 ρ = m V J t2 t 1 F dt = p a r = v2 r = rω2 F c = m v2 r ω dθ dt α dω dt = d2 θ dt 2 s = rθ v = rω a t = rα τ = Iα = dl dt d 2 x dt 2 = ω2 x x(t) = A cos(ωt + φ) ω = 2πf = 2π T ω spring = k m E SHO = 1 2 ka2 T = 2π L g F s = k s I r 2 dm i m i r 2 i F g = G m 1m 2 r 2 W sf s i F ds P W = dw t dt K linear 1 2 mv2 U g = mgh U s = 1 2 k s2 I ring = MR 2 I disk = 1 2 MR2 I sphere = 2 5 MR2 I rod,cm = 1 12 ML2 I = I CM + MR 2 K rotational 1 2 Iω2 g = F g m = G m r 2 U g = G m 1m 2 r ( 4π T 2 2 = GM v esc = ) r 3 2GM r F b = ρv g 1

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3 Name: Score: / Physics 204A Exam 4 12/12/12 Short-answer problems: Do any seven problems 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 stick has variable linear density λ = m L 3 x 2, where L is the total length of the stick and x is the distance from the end. What is the rotational inertia I for this stick around the point x = 0? 2. Find the rotational inertia of a sphere around a point on the edge of the sphere. 3

4 3. A torsion spring follows the equation τ = κθ. If this spring is attached to a disk of mass m and radius r, what will be the period of oscillation of the disk and spring? 4. A satellite orbits Earth at an altitude of 300 km. Which is greater, the energy required to lift it to that altitude from the ground or the kinetic energy required to maintain a circular orbit at that altitude? Explain your answer. 4

5 5. Felix Baumgartner s balloon (used to lift him to the edge of space so he could make the highest parachute jump in history) had a volume V = m 3. The balloon skin had a mass of 1700 kg, and it was filled with helium which has density ρ He = 0.18 kg/m 3. What was the payload capacity of this balloon at ground level, where the density of air is ρ air = 1.3 kg/m 3? 6. Bubba ( m = 100 kg) runs tangentally to a merry-go-round at a playground at a speed v = 7.0 m/s. As he passes the merry-go-round, he leaps onto the very edge of it, at R = 1.5 m. The merry-go-round (with Bubba on it) then rotates at ω = 0.70 rad/s. What is the rotational inertia I of this merry-go-round? 5

6 7. Bubba (M = 100 kg) leans a ladder against a plate-glass window. The ladder (L = 3.0 m) has mass m = 10 kg, and it is at an angle of 70 to the ground. The window will break when the normal force on it is F N = 255 N. How far up this ladder will Bubba get before the window breaks? (Suggestion: use the bottom of the ladder as a pivot, and consider torques.) 8. Bubba s diving board consists of a stiff plank (L = 2.5 m) of negligible mass, hinged at the back end. There is a large coil spring (k = 20, 000 N/m) under the plank at point L/2. What is the frequency of Bubba s small-amplitude motion if he stands on the end of the board? Bubba s mass is still m = 100 kg. 6

7 Physics 204A Key Exam 4 12/12/12 Short-answer problems: Do any seven problems 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: 2. Answer: I = L 0 r 2 dm = L 0 dm = λdx = m L 3 x2 dx x 2 ( m L 3 x2) dx = m L 3 L 0 x 4 dx = 1 5 ml2 I = I cm + MR 2 = 2 5 MR2 + MR 2 = 7 5 MR2 3. Answer: τ = κθ = d2 θ dt 2 = κ I θ This is the equation for simple harmonic motion, so κ ω = I and 4. Answer: The change in potential is T = 2π ω = 2π I κ = 2π mr 2 ( 1 U = GMm r 1 ) R where r is the orbital radius and R is the radius of the Earth. The kinetic energy is From F c = F g we can obtain a value of v 2 : so K = 1 2 mv2 2κ m v2 r = GMm r 2 = v 2 = G M r ( 1 GMm r 1 ) 1 R 2 GM r 1 r 1 R 1 2r Since R > 1 2r, the potential energy (left side) is the smaller term. 1

8 5. Answer: The buoyant force on the balloon is F b = ρ air gv = N We must subtract the weight of the balloon and the helium, though: W balloon = m balloon g = (m skin + ρ He V ) g = N Which gives us a total lifting capacity of N, or 4, 000 kg. 6. Answer: Just before the collision, Bubba s momentum is P = mv and his angular momentum is L = r P = mvr. Angular momentum is conserved, so L i = L f : mvr = (I Bubba + I mgr )ω I mgr = mvr ω mr2 = 1280 kg m 2 7. Answer: The ladder remains in equilibrium as long as the net torque is zero. τ cl = mg L cos θ + Mgx cos θ 2 τ ccl = F N L sin θ where x is how far up the ladder Bubba has gone. Set τ cl = τ ccl and solve for x: mg L 2 cos θ + Mgx cos θ = F NL sin θ x = F NL sin θ mg L 2 cos θ Mg cos θ = F NL Mg tan θ ml 2M = 1.99 m 8. Answer: You can use either forces or torques for this one: here s an answer using torque. For small angles, cos θ 1 so L 2 τ = Iα ( k L2 ) θ cos θ = ml 2 d2 θ dt 2 d 2 θ dt 2 = kθ 4m cos θ d 2 ( ) θ k dt 2 θ 4m This is the equation for Simple Harmonic Oscillation, with ω = 1 k 2 m Frequency is f = ω 2π, so f = 1 k = 1.13 Hz 4π m 2