Physics 7A Lecture Fall 04 Final Solutions December, 04
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.
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
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)
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
is lost due to the inelastic collision with the conveyor belt, but is otherwise correct.
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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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
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)