Physics 7A Lecture 2 Fall 2014 Final Solutions. December 22, 2014

Transcription

1 Physics 7A Lecture Fall 04 Final Solutions December, 04

2 PROBLEM The string is oscillating in a transverse manner. The wave velocity of the string is thus T s v = µ, where T s is tension and µ is the linear mass density. We can relate the wavelength λ of the string to the angular frequency ω of the string like so: ω = π v λ. The vertical acceleration of the string given amplitude A min is a = ω A min. The bug will just lose contact when the string is accelerating more than gravity, which is the condition that a = g. We therefore get that g = ω A min = A min = g ω = gλ 4π v = gλ µ 4π T s.

3 Problem Part (a) Since the rod is hinged in the center, the torque due to gravity is 0 Let the extension of the spring be x For very small angles θ Thus, the torque is x = l sinθ () x l θ () (3) τ = kx l cosθ (4) τ kl θ 4 (5) (6) Thus, writing torque-acceleration equation gives Comparing with α = ω θ ml α = τ (7) ml α = kl θ 4 (8) α = 3k m θ (9) (0) ω = 3k m () Part (b) After the rod is cut, mass m m/ and length l l/. Also, the gravity acts as the center of mass and provides torque.the position of the center of mass is l/4 from the hinge. For very small angles θ, the torque becomes

4 τ = m g l sinθ 4 () τ = mgl 8 θ (3) (4) Total torque thus becomes The moment of inertia is τ = mgl 8 θ + kl 4 θ (5) (6) I = 3 m l (7) (8) Comparing with α = ω θ 4 ml α = τ (9) 4 ml α = mgl 8 θ + kl 4 θ (0) α = ( mgl 8 + kl 4 ) θ () 4 ml () ω = 6k m + 3g l (3)

5 Problem 3 Solution Physics 7A Section Final (Hallatschek) December 04 The power required to run conveyor is given by F net v, where F net is the net force the conveyor belt must exert upward along the ramp (parallel to surface of ramp) and v is the constant velocity of the belt itself. The terms that make up F net come from three sources: Force from inelastic collision on ramp: The falling mass is continuously being accelerated up to the speed v as it hits the ramp. The external force required to sustain this acceleration of mass flow is given by P t = ρ tv = ρv. t Force to hold up mass already on ramp against gravity: This force is the usual mg sin θ, and since the mass on the ramp at time t is given by ρ t, we have for this force ρ tg sin θ. Force to counteract friction force F F r : This force is simply F F r. Summing the above three forces and multiplying by v to get the power, we get P ower = (ρv + ρ tg sin θ + F F r ) v () Note on solutions: Some students attempted to calculate the power by calculating the change in energy as a function of time, where energy includes both potential and kinetic energy of the mass on conveyor. This method ignores the energy that

6 is lost due to the inelastic collision with the conveyor belt, but is otherwise correct.

7 Physics 7A Section Prof. Hallatschek Problem 4 Trevor Grand Pre December 0, 04 (a)to solve this problem we set up te force equation of this system. F = GM Em r = M E a E Solving for the acceleration, we get a E = mg r. (b) From this, we can see that the the acceleraton for the closest and farthest point on earth is a far = mg (r+r E and a ) near = mg (r r E. ) When we subtract the two and then simplify, we get a near a far = = mg (r r E ) mg (r + r E ) 4mGr E r (r r E )

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9 Physics 7A, Lecture (Hallatschek) Final Exam, Fall 04 Problem 6 Part C Part A Sum vectors about y-axis II tt ww ii = (II tt + II rr )ww ff ww ff = ww iiii tt II tt + II rr Part B ANS Sum vectors about y-axis LL + JJ = LL II tt ww ii + ττ(δtt) = II tt ww ff ττ = II tt ww ff ww ii Δtt Part D () ANS Initial Kinetic Energy Final Kinetic Energy Sub from Part A KK ii = II ttww ii ww ii = KK ii II tt () KK ff = (II tt + II rr )ww ff KK ff = (II tt + II rr ) ww iiii tt II tt + II rr KK ff = (II tt + II rr ) ww iiii tt II tt + II rr KK ff = (II tt + II rr ) ww iiii tt II tt + II rr KK ff = (II tt + II rr )ww ii II tt (II tt + II rr ) Sub from Equation () KK ff = KK iiii tt (II tt + II rr ) ANS Friction acting on ring dddd = μμ(dddd) = μμμμ mm rr ππrr ππππππππ = μμμμmm rr RR rrrrrr Torque due to friction ring ττ = rrrrrr ττ = μμμμmm rr rr dddd 0 RR RR = 3 μμμμmm rrrr = 4μμμμII rr 3RR Sub into () 4μμμμII rr 3RR = II tt ww ff ww ii Δtt Δtt = 3RRRR tt ww ff ww ii 4μμμμII rr () ANS

10 Lecture Final Problem 7 Solution (a): Consider a differential volume element of water. Newton s second law tells us: PouterdA - PinnerdA = ρ da dr ω r The da from each term cancel. Pouter - Pinner becomes dp, since it is a difference in pressure across a differentially small volume element. dp = ρ dr ω r Integrating both sides gives us: P P0 = ½ ρ ω (r r0 ) P = P0 + ½ ρ ω (r r0 ) (b): Use Bernoulli s principle. Equate a point just inside the test tube to a point just outside. Realize that a point just inside will have the pressure found in part (a) with r = L. A point just outside will have a pressure P0 (since it is exposed to the atmosphere). Therefore: P0 + ½ (ρ v ) = P0 + ½ ρ ω (L r0 ) v = ω Sqrt[L r0 ] (c): Av = Av, so (a v / A) = dx/dt. Solve for dt, then integrate. The bounds on the dt side are 0 to t. The bounds on the dx side are r0 to L. dt = (A/a) dx ( / ω Sqrt[L x ]) t = (A/ ω a) Cos - (r0 / L)