Vector mechanics PHY1021: Jan 2011 exam- Hints and tips. Lecturer: Dr. Stavroula Foteinopoulou

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1 Vector mechanics PHY1021: Jan 2011 exam- Hints and tips Lecturer: Dr. Stavroula Foteinopoulou 1(i) W<0ifforcecausesa KE < 0contribution. Friction is not possible to do negative work in translationalmotion. W = KE W = B F A xdx = B ma A xdx = B m dv x dx = B v A dt A xdv x = K B K A 1(ii) minimum velocity at the bottom of the circle v A is: mv2 A = mg(2l) v A =5.6 ms 1 Force at the top of the circle is: F Rod = mg =4.9 N First find velocity at this point v B from work energy theorem, when mass has completed a quarted of a circle. 1 2 mv2 B 1 2 mv2 A = mgl v B = 2gL Force of rod, when the mass has compelted a quarter of a circle is: F Rod = mv2 B L F Rod =9.8 N 2(i) 1

2 (a)v =(4i +4j) ms 1 (b)a = dv =(2i +4j) ms 2 dt (c) â = a = 1 a 20 (2i +4j) (d) F = ma =(2i +4j) N (e) v 1 = v(2s) =4 2 ms 1 v 2 = v(4s) = 320 ms 1 From work-energy theorem: W = 1 2 m(v2 2 v2 1 )=144J (f) v 1 = v(2s) =(4i +4j) ms 1 v 2 = v(4s) =(8i +16j) ms 1 Magnitude of impulse is: J = mv 2 mv 1 =12.65 Kgms 1 Direction- find angle θ between J and x-axis. cos(θ) = J i θ = J 2(ii) The location of the support should be the center of gravity, which is the same as the center of mass (constant g), and has coordinate x CM : 1 L x CM = M 1 +M 2 +M 3 (M 1 + m L + M L) 4 x CM =1.714 m 2

3 3 [3] zero [2] Position A: cylinder on top of incline Positon B: cylinder at the bottom of incline Conservation of mechanical energy: K A + U A = K B + U B U A = K B Mgh = 1 2 Mv2 CM Iω2 since v CM = ωr and I = 1 2 MR2 we have: 4 v CM = 3 v CM =4.43ms 1 2v CM =8.85ms 1 We have: 3

4 τ = f s R = I a /R So Since: I = 1 2 MR2 we get: f s = 1 2 M a But: M a = Mgsinθ -f s which together gives: f s = 1 Mgsinθ =6.54 N 3 For rolling without slipping to occur friction must be static. So: f s µ s N = µ s Mgcosθ So we get for the coefficient of friction µ s, µ s (i) (a) This is the only way to conserve momentum along the y-direction (perpendicular to x-direction). (The momentum along the y-direction is initially zero). (b) Conservation of momentum in the y-direction: mv f1 sin45 0 = mv f2 sin45 0 v f1 = v f2 Accordingly, conservation of momentum in the x-direction becomes: mv =(mv f1 + mv f2 )cos45 0 The two above equations yield: v f1 = v f2 = v f = v 2 (c) 4

5 K i K f = 1 2 mv mv2 f = 1 2 mv2 m( v 2 ) 2 =0 Which means initial and final kinetic energy are equal i.e. collision is ellastic. 4(ii) (a) τ = RF =1N m (b) W = τ θ =4π J (c) Work energy theorem for rotational motion: 1 2 Iω2 0=W which gives ω =7.1rads 1 (d) ω = ω 0 + αt with ω 0 =0,andα = τ I which gives t =1s 5(i) (a) The work of a force must be indepedent of the path for the force to be conservative. (b) K + U = constant K: Kinetic energy U: Potential energy (c) No, only when only conservative forces are present. In the case that non-conservative forces are present mechanical energy is still conserved if the latter contribute zero work (e.g. friction in rolling without slipping motion, normal force) (d) 5

6 Work energy theorem: W A B = K B K A Conservative force: W A B = U A U B which indicates that work depends only on start and end point but not on the particular path taken. So: K B K A = U A U B which gives: U A + K A = U B + K B which expresses conservation of mechanical energy 5(ii) F r = du dr = = 12U 0 R 0 [( R 0 r ) 13 ( R 0 r ) 7 ] F r =0whichoccursar: r = R 0 6

7 6(i) proper time: Time interval between two events measured in a reference frame where the events took place at the same location. proper length: The length of an object in a reference frame where its end points are not moving, or the distance between two points in a reference frame where they are stationary. 7

8 6(ii) Orepresentstheevent(atx =0,t=0). The black lines represent x = ct and x = ct s 2 = c 2 t 2 x 2 6(iii) (a) Lorentz scalar: A quantity that is invariant under the Lorentz transformation. (The norm (or norm squared) of a Minkowskian four-vector, is a Lorentz scalar, as all scalar products between four-vectors). (b) P =(E/c,p) So: P P = E2 p 2 c 2 But: P P is a Lorentz scalar so: E 2 c 2 p 2 is a Lorentz scalar. Using the enery-momentum relationship we get that m 0 c 2 is a Lorentz scalar, and since the speed of light in vacuum is invariant 8

9 (Einstein s second postulate), the rest mass m 0,mustalsobeaninvariant. 6(iii) m = γm 0 γ =2 γ = q 1 1 v2 c So 2 v =0.866 c 9

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19 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHY3129 Name of module Device Physics Date of examination January (i) Luminescence definitions, operation of electroluminescent device - see lecture notes. ac operation avoids formation of a space charge that cancels the accelerating electric field within the particle The non-linear (threshold type) dependence of optical emission upon drive voltage favors matrix addressing: there is minimal emission from half-selected cells. 1(ii) Rate of injection of angular momentum "!J e ˆ M 1 ( ) The absorbed transverse component is " = #!J M ˆ 1 # ( ˆ e M 1 $ M ˆ ) ˆ 2 M 2 Using a " ( b " c) = b( a # c) $ c( a # b), #!J the torque is rewritten as " = M 2 2 $ M 2 $ M ˆ 1 e M 2 d2 ( ) per unit volume of layer 2, which has the same vector form as the Landau Lifshitz damping if ˆ M 1 is parallel to B. Critical current density is J c = " e µ 2 0M 2d2! perimeter is 24 mt. so I c = 1.2 ma, and Oersted field at 2.(i) Director definition, elastic deformations, HAN cell see lecture notes t decay = 65 ms, t rise = 0.28 ms. These values are well-suited to a video rate application. A pixel can be switched on quickly and then holds its state until refreshed at 50 Hz rate. 2(ii) Operation of bipolar transistor see lecture notes.

20 The common base circuit has low input and high output impedances and is commonly used for impedance matching. Base transport factor, emitter efficiency see lecture notes. I c /I b = (i) Hard disk drive operation see lecture notes. Relaxation time obeys and Arrhenius relation " = " 0 exp ( µb k B T) where the magnetic moment µ is the product of the magnetization and the particle volume, and! 0 = 1/f FMR. Setting! = 10 years, critical diameter = 9.6 nm. Reduced grain sizes require larger anisotropy for thermal stability, which then requires larger write fields. 3(ii) Operation of MESFET see lecture notes. I D = 0 when L d = 0.15 µm, so N d = 2.8 " m #3 Begin from Q = AeN d L d and differentiate. 4(i) Derivation of guiding equation see lecture notes. ($ 1 ) min = 80.8, 12 allowed modes (counting both possible polarization states) Intermodal dispersion (see lecture notes) yields a pulse duration of 675 fs. Shorter pulse duration may be achieved by reducing the core thickness so that the waveguide supports only a single mode. 4(ii) MOSFET and CMOS see lecture notes In 3 years the rate of heat dissipation will increase by a factor of 4 under constant voltage scaling.

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24 PHY3148 The nature of the material and hence the examination means Solutions and Hints are not applicable to this module. The level of detail expected in each answer is that given in the lectures. The answer is expected to provide a focussed and logical response to the question asked, rather than a random selection of more-or-less relevant facts regurgitated from the notes. For more guidance contact the lecturer.

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28 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code Name of module PHYM432 Relativity and Cosmology Date of examination January First law of thermodynamics, connect a change in volume to a change in total energy for an ideal cosmological fluid. Substitute and take the time derivative to get the fluid equation Use conservation of energy for a test particl of unit mass on the surface of a unifomr ideal sphere enclosing mass M at density!. 1

29 3. Constant comes from GR and is "kc 2. The energy equation: In the LI, the metric tensor in the Minkowski metric, transform to the accelerated frame: Of the 16 terms, only the diagonal terms are non-zero Eqn of motion in accelerated frame is non-zero terms are for the equation with " =1. Using the formula gives with a diagonal metric g 11 = [ g 11 ] "1 so 4. Singularity at r = R S = 2GM c 2, which is a coordinate singularity and at r = 0, which is a gravitational singularity. R S = 2GM c 2 = 2969 m 2