UNIVERSITY OF TORONTO Faculty of Arts and Science

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ω = k/m x = A cos (ωt + ϕ 0 ) L = I ω a x = ω 2 x P = F v P = de sys J = F dt = p w = m g F G = Gm 1m 2 D = 1 2 CρAv2 a r = v2

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UNIVERSITY OF TORONTO Faculty of Arts and Science DECEMBER 2013 EXAMINATIONS PHY 151H1F Duration - 3 hours Attempt all questions. Each question is worth 10 points. Points for each part-question are shown in square brackets in the left margin. Write your name and student number on every examination book you use. Allowed Aids: calculator, ruler, paper dictionary and attached formula sheets. 1. Provide short answers to TWO of the following questions. Each answer is worth 5 points. If you answer more than two, indicate which two you would like graded. (a) State Kepler s three laws of planetary motion. (b) Two particles both of mass 1.0 kg, travelling in perpendicular directions at speeds 4.0 m/s undergo a perfectly inelastic collision. How much energy is dissipated in the collision? (c) The scalar product of two vectors is defined to be A B = A B cos θ, where θ is the angle between the two vectors. Using this definition only (i.e. the second definition given in the formula sheet may not be assumed), show that A ( B + C) = A B + A C. (d) Explain the terms conservative force and non-conservative force. Give two examples of each. (e) Use dimensional analysis to deduce an expression for the speed of transverse waves in a string in terms of the Tension in the the string, T and the mass per unit length of the string, ρ. (NOTE: there is no FIG. 1 associated with Q.1) 2. A disk of mass M and radius R starts from rest at height h above the bottom of the bowl shown in Fig. 2. It rolls without slipping down the left side of the bowl. Express your answers to the following questions in terms of the variables M, R, g, and h only. [4] (a) Staring from the definition of moment of inertia, I = r 2 dm, derive the moment of inertia of the disk abouts center. Show all steps in your derivation. (b) Whas the speed of the disk s center of mass at the bottom of the bowl? (c) The right side of the bowl, starting at the bottom, is frictionless, so the disk slips completely. This means that on the right side of the bowl, there is no torque to change the angular velocity of the disk. To what maximum height does the disk rise on the right side of the bowl? FIG. 2: Disk and bowl for Q. 2. Page 1 of 6

3. Recently, my Aunt sent me a lovely ornament. It had a mass M O ; to display it, I built a shelf of mass M S and length L and positioned the ornament ats centre, as shown in Fig.3. Unfortunately, I am a rotten carpenter, and the right hand support for the shelf soon disintegrated, so that the shelf fell down, pivoting about the left hand support as it did so. You may assume that the dimensions of the ornament are small, so that may be treated as a point-like object. Express your answers in terms of M O, M S, L, g and µ S, the coefficient of static friction between the ornament and the shelf. (a) If moment of inertia of the shelf abouts centre is M S L 2 /12 and the ornamens not slipping along the shelf, find the total moment of inertia of the shelf plus the ornament about the pivot. (b) Neglecting air resistance and assuming that static friction is holding the ornament fixed with respect to the shelf, find the angular velocity of the shelf about the pivot after it has fallen through an angle θ. (c) Draw the free-body diagram for the ornament before it starts to slip off the shelf. Whas the acceleration of the ornament? (d) Show that the angle at which the ornament starts to slide down the shelf is ( )] M S θ crit = ArcTan [µ S 10M S + 9M O FIG. 3: Ornament on a shelf, for Q.3 FIG. 4: Pully and weights for Q.4. 4. A light string, whose ends are attached to two weights m 1 and m 2, is wound over a pulley, of mass M, radius R and moment of inertia MR 2. The pulley is supported from the ceiling by a rope (see Fig.4). Assume that the string does not slip on the pulley. (a) Whas the acceleration of m 1? (b) Find the tension in the rope which supports the pulley. Page 2 of 6

5. Suppose a perfectly straight railway tunnel, a chord to the earth, was constructed to connect London and Paris. Gravity would accelerate the train at one end, so it picked up speed, without the need of a propulsion system, reaching the maximum velocity at the centre of the tunnel, where gravity would begin to slow it down until, in the absence of any friction, the train came to a stop exactly at the other end of the tunnel. (a) Newton s Shell Theorem states that a hollow, spherical shell exerts zero gravitational force at any point inside the shell. Use this to find an expression for the gravitational force on a body of mass m at some distance r < R 0 from the centre of earth (assume that the Earth has uniform density). Express your answer in terms of g, r and R 0. (b) Show that, when kinetic friction is present, the acceleration of the train is given by: a x = g R 0 x µ k g cos θ 0, [4] where x is the distance of the train from the centre of the tunnel (point O in the figure). (c) Suppose a train starts from resn London and begins its journey towards Paris. If the coefficient of kinetic friction is µ k = 0.002, the radius of the earth, R 0, is 6378km and the θ 0 (as shown in FIG.5) is 1.53 o, how far from Paris does the train come to a halt? FIG. 6: Wall and beam for Q.6. FIG. 5: London-to-Paris Railway for Q.5. 6. One end of a uniform beam of length L and mass M rests against a smooth, frictionless vertical wall. The other end is held by a string which is attached to the wall as shown in the Fig.6. The beam is neither rotating nor translating. (a) Draw a free body diagram for the beam, labelling all the forces acting on the beam at the points where they are acting. (b) Whas the magnitude of the normal force acting on the beam by the wall? Express your answer in terms of M, β and g only. (c) Find the angle α in terms of β. Page 3 of 6

FIG. 7: Potential energy function U(x) for Q.7. 7. Figure 7 shows the total potential energy U (in Joules) as a function of position x (in meters) for a particle of mass m = 3.0 kg. At time t = 0 seconds the particle is located at x = 0 and is at rest. [4] (a) Estimate the acceleration (in m s 2 ) experienced by the particle at x=0 m. (b) Estimate the time taken for the particle to reach x=2 for the first time. (c) Estimate the maximum and minimum values of x(t) during the entire motion (starting from t=0). (d) Now suppose there is a kinetic frictional force of constant magnitude F fric = 0.1N acting everywhere. Where will the particle ultimately come to rest, and whas the total distance it will travel. This is the end of the exam. HAPPY HOLIDAYS EVERYONE! Page 4 of 6

Mathematical Formulas Formulas You may use these formulas without proof, unless you are specifically requested to derive the result as part of a question. You may detach the formula sheef you wish. Vectors Scalar Product: F G = F G cos θ = Fx G x + F y G y + F z G z, Vector Product: F G = F G sin θ A B = B A, ( i j = k, j k = i, d k i = j [ F G] df = ) G + F ( dg ) Differentiation Polynomials: f(t) = c(t B) n df = cn(t B)n 1 c(t B) n = c(t B)n+1 n + 1 t, (n 1), Trig functions: f(t) = A cos(bt) df df = AB sin(bt), f(t) = A sin(bt) = AB cos(bt) d df Sum of two functions: (f + g) = + dg, Product Rule: d df (fg) = g + f dg, d df dg t Chain Rule: [f(g(t))] = Integral of a sum: [f(t ) + g(t )] = f(t ) + g(t ), dg Trigonometric Functions Definitions: sin θ = opposite hypotenuse tan θ = cos θ = opposite adjacent adjacent hypotenuse Useful property (from Pythagoras): sin 2 θ + cos 2 θ = 1 Sum rules: cos(θ + φ) = cos θ cos φ sin θ sin φ, sin(θ + φ) = sin θ cos φ + sin φ cos θ Solutions of Algebraic Equations Quadratic: ax 2 + bx + c = 0 x = b ± b 2 4ac 2a Page 5 of 6

Physics Formulas and constants: v = d r, d v a =, v(t) = v() + a(t ), r(t) = r( ) + v(t ), F net = m a = d p, p = m v, F k = µ k n, 0 F s µ s n, w app = n r = r + vt, u = u + v, a = a, s = Rθ, v = rω = r dθ, 2π ω = T, r(t) = x i + y j = R cos[θ(t)] i + R sin[θ(t)] j = R e r, v(t) = Rω sin[θ(t)] i + Rω cos[θ(t)] j = Rω e θ, a = a r e r + Rα e θ = Rω 2 e r + R dω e θ, ω(t) = ω( ) + J = f α(t ), θ(t) = θ( ) + ω(t ), F (t ) = p, K = 1 2 mv2, U g = mgy, F spring = k s, U spring = 1 2 k( s)2, E mech = K + U, K = W net, W = sf s i F s ds, W = F 0 r, U = W c, F s = du ds, E mech = W nc, E sys = E mech + E th, E th = W diss, W nc = W diss + W other, P = de sys = F v, x cm = 1 mi x i = 1 x dm, M M τ = r F, τ net = Iα, τ g = Mgx cm, I = m i ri 2 = r 2 dm, I = I cm + Md 2, K rot = 1 2 Iω2, L dl = r p, = τ net, ω = ω k, L = I ω (sometimes), F g = Gm 1m 2 r 2 e r, U g = Gm ( ) 1m 2 4π, T 2 2 = r 3, g = 9.8 m/s 2, G = 6.67 10 11 Nm 2 /kg 2 r GM Total Marks = 70 Total Pages = 6 Page 6 of 6