* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Mid Term Test 2012 Answers File
Tensor operator wikipedia , lookup
Routhian mechanics wikipedia , lookup
N-body problem wikipedia , lookup
Fictitious force wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Specific impulse wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Old quantum theory wikipedia , lookup
Center of mass wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Classical mechanics wikipedia , lookup
Moment of inertia wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Seismometer wikipedia , lookup
Accretion disk wikipedia , lookup
Angular momentum operator wikipedia , lookup
Angular momentum wikipedia , lookup
Hunting oscillation wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Equations of motion wikipedia , lookup
Photon polarization wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Centripetal force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Newton to Einstein Mid-Term Test. 9am., Thursday 15th Nov. 2012 Duration: 50 minutes. Use g = 10 m s-2 for numerical calculations. You may use the following expressions for the dot and cross products: u.v u x v x u y v y u z v z u v u yvz u z v y , u z vx u xvz , u xv y u y vx Section A: Answer ALL Questions A1. Given the vectors u = (1, 0, 0) and v = (1, 1, 0), find a) u + v (1+1, 1+0, 0+0) = (2, 1, 0) b) u.v 11 + 01 + 00 = 1 c) u × v (00 – 01, 01 – 10, 11 – 01) = (0, 0, 1) [3] A2. State Newton’s Second Law of Motion, and explain what it says about momentum. F = ma (OR F = dp/dt) – it defines momentum as that quantity whose rate of change is force. [3] A3. A ramp is at 30º to the horizontal, and a mass m sits on it without sliding. a) Give the force exerted by the ramp on the mass, and resolve it into vertical and horizontal components. F = mg upwards, Fvert = mg, Fhoriz = 0 b) Resolve the force exerted by the ramp on the mass into components parallel and perpendicular to the ramp. Fpar = mg sin 30° = ½mg. Fperp = mg cos 30° = Sqrt[3]/2 mg. c) What is the minimum coefficient of friction? min = Fpar / Fperp = 1/Sqrt[3] = 0.58. [6] A4. A steel ball weighs 2kg. a) The ball is lifted through 20 metres vertically. Calculate its gravitational potential energy. mgh = 21020 = 400J b) The ball is then dropped. Calculate its kinetic energy and velocity when it has fallen 20 metres. mgh = ½mv2 so v2 = 2gh = 400; v = 20ms–1 [3] A5. A dumbbell consists of two 1kg point masses, connected by a light rod one metre long. Calculate its moment of inertia about a) an axis perpendicular to the rod and passing through the centre, I = Sum[mr2] = 1½2 + 1½2 = ½ kgm2 b) an axis perpendicular to the rod and passing through one of the masses. I = Sum[mr2] = 102 + 112 = 1 kgm2 [4] A6. State Kepler’s Third Law. Derive it for a planet in a circular orbit in an inverse-square law gravitational field. T2 ∂ r3. Derive by equating centripetal and gravitational forces: mr2 = mr(2/T) 2 = GMm/r2, so r/T2 ∂ 1/r2, so T2 ∂ r3. [3] A7. A comet is travelling at 12 km s-1 as it crosses Jupiter’s orbit on its way to the Sun. What is its speed later in its orbit when it crosses Jupiter’s orbit again? Explain your reasoning. [3] 12 kms–1, because KE + PE = const; PE is the same both times, so KE and hence speed must be the same. Section B. Answer ONE Question B1. a) An empty cylindrical can (with no ends) weighing 100g rolls along the floor at 4 ms-1. Calculate its total kinetic energy (translational plus rotational). [5] Translational KE = ½mv2. Rotational KE = ½I2 = ½ (mr2) (v/r)2 = ½mv2. So total KE = mv2 = 0.1 16 = 1.6J. b) A pendulum consists of a 10 kg point bob hanging on a light string of length 10m. It swings through an angle 2, i.e. each side of vertical. (i) Find the tension T in the string at the end-points of the motion, i.e. when the pendulum is at the angle . T mg The diagram shows the forces acting on the bob when it is stationary at the angle . The acceleration and hence the resultant force has to be at right angles to T. Using vectors, we thus have (T + mg).T = 0 or T2 + mgTsin(–) = 0, i.e. T = mgsin. Give the value of the tension in the string at the end-points if the angle is 90. T = mgsin0 = 0. (ii) (iii) Give the value of the acceleration of the bob at the end-points if the angle is 90. T = 0, so F = mg, so acceleration is 10ms–2 vertically downwards. [6] c) The pendulum of part (b) is swinging with the angle = 90. By considering potential and kinetic energy, find (i) the velocity of the bob at the lowest point of the motion PE at = 90° is mgh = 1000J. PE =0 at the bottom, so KE = ½mv2 = 1000J, so v = Sqrt[2gh] = Sqrt[200] = 14.14ms–1. (ii) the tension in the string at this point. T = mg + ma = mg + mv2/r = 100 + 200 = 300N. [6] d) Given that G = 6.67 10-11 N m2 kg-2, the mass of the Earth M = 5.97 × 1024 kg and the radius of the earth is R = 6380 km, (i) Find the escape velocity for a rocket at 200km altitude. PE per unit mass is –GM/r, and PE + KE = 0 for escape velocity. So v = Sqrt[2 GM/r] = 11.2 kms–1. (ii) Find the orbital speed of a satellite in a circular orbit at 200 km altitude. So total KE = mv2 = 0.1 16 = 1.6J. B2. [7] a) State Kepler’s Second Law and explain how it relates to angular momentum. A planet sweeps out equal areas in equal times. The area is ½v.r per unit time. This is proportional to the angular momentum mv.r, so the constancy of the area is equivalent to the angular momentum being conserved. [5] b) A flywheel with a moment of inertia I = 5 kg m2 is mounted on a light axle. It is initially stationary. A torque of 100 N m is applied to the axle for 100 s. (i) Calculate the final angular velocity of the flywheel. v = u + at = u + F/m t, so f = I + t = 0 + /I t = 2000 rads–1. (ii) Calculate the final angular momentum of the flywheel. L = If = 5 2000 = 10000 kgm2s–1. (iii) Calculate the final rotational energy of the flywheel. E = ½I2 = ½ 5 4 106 = 107J. (iv) If a torque of 10 N m is now applied about an axis perpendicular to the axle of the flywheel, what will the motion be? Explain your answer. [9] This is a gyroscope, so it will precess around the third axis. The rate of addition of angular momentum around this axis is the torque, so L moves to L + dt in time dt, i.e. with precessional angular velocity /L = 10/1000 = 10–3 rads–1. c) The potential energy of a body varies with position as U 2x 2 . (i) Give an expression for the force F on the body and sketch F(x). F = –dU/dx = –4x. F 1 x –4 (ii) At what position will the body be in stable equilibrium? Explain [5] your answer. At x = 0, since here the force is zero and U is minimum. . F d) Given two vectors u and v, with magnitudes u = 1 and v = 2, and given that the magnitude of the cross product u v = 1, find (i) The angle between u and v. u v = uv sin , so 1 = 2 sin , so sin = ½ and = 30° (ii) F The value of the dot product u.v. [6] u v = uv cos = 2 cos30° = Sqrt[3]. End of paper Prof. D.J. Dunstan