Classical Mechanics

Oscillations and Waves

Quantum and Thermal Physics

Lecture_31_23.11.2011.pdf-Lecture 32 23.11.2011 Review of selected

Suggested Problems # 3 Solutions QUESTION 1 - The triatomic molecule problem We look at an idealized version of a triatomic molecule, with a larger mass M at the center, and two smaller masses m on each side, connected by springs of stiffness k , which represent the binding force between the atoms. We consider simple harmonic oscillations between the atoms, of the form η j = η 0 j cos ( ωt ) . Part A: EOM and normal frequencies Using a free-body diagram of the situation, it is possible to note that the force felt by the mass on the left (mass 1) will be proportional to the displacement of mass 2, minus the displacement of mass 1. For mass 2, the force will depend on the extension of the springs on both sides, whereas for mass 3, the situation will be very similar to mass 1, except for the fact the spring is connected to the left of the mass. This is represented mathematically as: m ¨ η 1 =( η 2 − η 1 ) k (1) M ¨ η 2 = − ( η 2 − η 1 ) k +( η 3 − η 2 )=( η 1 + η 3 − 2 η 2 ) k (2) m ¨ η 3 = − ( η 3 − η 2 ) k (3) Since η j = η 0 j cos ( ωt ) , we ﬁnd that ¨ η j = − ω 2 η j . Replacing these in the above equations, and rearranging: − mω 2 η 0 1 cos ( ωt )+ � η 0 1 cos ( ωt ) − η 0 2 cos ( ωt ) � k =0 (4) − Mω 2 η 0 2 cos ( ωt )+ � 2 η 0 2 cos ( ωt ) − η 0 1 cos ( ωt ) − η 0 3 cos ( ωt ) � k =0 (5) − mω 2 η 0 3 cos ( ωt )+ � η 0 3 cos ( ωt ) − η 0 2 cos ( ωt ) � k =0 (6) Dividing by cos ( ωt ) and by the respective masses m and M , − ω 2 η 0 1 + � η 0 1 − η 0 2 � k m =0 (7) − ω 2 η 0 2 + � 2 η 0 2 − η 0 1 − η 0 3 � k M =0 (8) − ω 2 η 0 3 + � η 0 3 − η 0 2 � k m =0 (9) In matrix form, this gives:      k m − ω 2 − k m 0 − k M 2 k M − ω 2 − k M 0 − k m k m − ω 2           η 1 η 2 η 3      =0 (10) For a non-trivial solution, we look at situations where the determinant of the 3x3 matrix will be equal to 0. The determinant of an arbitrary square matrix is given by the Leibniz formula, which for a 3x3 matrix A ij takes the form: det ( A ij )= � � � � � � � � � a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 � � � � � � � � � = a 1 1( a 22 a 33 − a 32 a 23 )+ a 21 ( a 32 a 13 − a 12 a 33 )+ a 31 ( a 12 a 23 − a 22 a 13 ) (11) In the case of our 3x3 matrix above, we look at: � � � � � � � � � k m − ω 2 − k m 0 − k M 2 k M − ω 2 − k M 0 − k m k m − ω 2 � � � � � � � � � =0 (12) This is, based on equation (11): � k m − ω 2 � 2 � 2 k M − ω 2 � − 2 k 2 Mm � k m − ω 2 � =0 (13) 2 Rearranging this equation, we obtain: � ω 2 − k m � � � ω 2 � 2 − � k m − 2 k M � ω 2 � =0 (14) � ω 2 − k m � � ω 2 − k ( M +2 m ) mM � ω 2 =0 (15) It is thus easy to show, from equation (15) that the three solutions will be, as presented in the problem statement: ω 2 1 =0 (16) ω 2 2 = k m (17) ω 2 3 = k ( M +2 m ) mM (18) Part B: Normal modes and atom motion CASE 1: ω 2 1 =0 We plug ω 1 =0 into equation (40), which becomes      k m − k m 0 − k M 2 k M − k M 0 − k m k m           η 1 η 2 η 3      =0 (19) Solving this matrix, we obtain the following three equations: k m η 1 − k m η 2 =0 (20) − k M η 1 + 2 k M η 2 − k M η 3 =0 (21) − k m η 2 + k m η 3 =0 (22) 3

PHY | 293 | Oscillations and Waves | assessment | Suggested Problems # 3 Solutio

Lecture 29 18.11.2011 1. Friction on the rolling object and torque due to friction 2. Angular momentum about an origin: definition, units, calculation 3. Angular momentum of a particle 4 Angular momentum of a rotating rigid 4 . Angular momentum of a rotating rigid body about a fixed axis 5 Torque and change of angular momentum 5 . Torque and change of angular momentum in a non-isolated system 6 C ti f l t i 6 . C onserva ti on o f angu l ar momen t um i n an isolated system 1. Friction on the rolling object and torque due to friction ( disc ) • Rolling over a surface is possible due to F 0 (a) is possible due to existence of the force of friction F F sf 0 of friction • Without friction: in (a) a wheel is sliding down the incline down the incline without rotation; in (b) a wheel rotates (b) in (b) a wheel rotates without linear motion F sf 2. Angular momentum about an origin: definition, units, calculation definition, units, calculation • Angular momentum f ti l f L r o f a par ti c l eo f mass m, moving with velocity , b t th ii 0 i v r a b ou t th e or i g i n 0 i s v r m p r L r r r r r × = × = F f v r m p r L × = × = • F or a system o f particles: L L r r Σ = φ i L L Σ = 2. Angular momentum about an origin: definition, units, calculation definition, units, calculation • The SI unit is (kg . m 2 )/ s • The magnitude of L is L = r mv sin φ φ is the angle between φ is the angle between and p r r r • Direction of is perpendicular to the L r φ perpendicular to the plane formed by and r r p r 3. Angular momentum of a particle 1P ti l i li ti 1 : P ar ti c l e i n a li near mo ti on L = r s in φ mv = d mv 2 r sin φ = d = arm = distance from the origin to trajectory 2 from the origin to trajectory For v =const, L = const φ 2: Particle in a circular 1 motion: L = r sin φ mv = mvr = 1 = m ( r ω ) r = (mr 2 ) ω = I ω Direction of is L r 4. Angular momentum of a rotating rigid body about a fixed axis • Each element of the body can be considered a can be considered a particle, which rotates in the xy - plane about the z - axis xy plane about the z axis with an angular speed ω • The angular momentum of an individual p article L i = m i r i 2 ω = I i ω • L and ω are directed • L i and ω i are directed along the z -axis 5. Torque and change of angular momentum in a non-isolated s y stem ( fixed axis ) • The angular momentum of the entire object ( ) 2 L L I ∑ ∑ r ( ) 2 z i ii i i L L mr I ω ω = = = ∑ ∑ I z ω z I ω = L r [R ll ] r r where L , I, and ω are taken about same axis • New derivation of equation of rotational motion [R eca ll ] v m p = z z I d I dL α ω τ = = = Σ r • In a non - isolated system τ et ≠ 0 and the z z z z ex t I d t I d t α τ Σ In a non isolated system τ e x t ≠ 0 , and the angular momentum is not conserved

PHY | 180 | Classical Mechanics | lecture notes | Lecture 29 18.11.2011 1. Frict

2011_Test_1_solutions.pdf-Page 1 of 4 PHY180H1F Test

1 Question 1 5 points in total At one instant, a 17.5-kg sled is moving over a horizontal surface of snow at 3.50 m/s. After 8.75 s elapsed, the sled stops. (a) Find the average friction force acting on the sled while it was moving (b) Assuming the deceleration of the sled was uniform, find the coefficient of friction between the sled runners and snow Solution First, we are setting the x -axis for 1-D motion in positive x -direction (left or right) (a) Correct answer is F avg = -7.00 î N (b) | F x | = k n , where k is the coefficient of kinetic friction; n is the normal force y : n y – mg = 0 → n = mg = 4.08x10 -2 x = - 7.00 N mg F n F k x x = = 2 Question 2 8 points in total Galilean transformation equations give relationships between kinematic quantities measured in different inertial frames of reference. Write down any three out of four possible Galilean transformation equations for two inertial frames of reference A and B whose relative velocity is v AB . Define all variables you are using in these equations. Solution Relative location of two inertial frames of reference A and B with velocity vector v AB must be shown somehow like to the right. There are 4 Galilean transformation equations: v PA = v PB + v BA (all velocities are vectors) a PA = a PB (accelerations are vectors) Δ t A = Δ t B or t A = t B Any three of them may be written by a student. All vector quantities must be shown as vectors with correct notation All variables must be correctly defined in words. The correct definition of v PA is “velocity of a point [mass] P relative to the frame of reference A”. PA PB BA t = + r r v r r r 3 Question 3 12 points in total A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 o past the lowest point on its way up, its total acceleration is (-22.5 i ˆ + 20.2 j ˆ ) m/s 2 . For that instant of time (a) sketch a vector diagram for the ball showing the components of its acceleration; (b) determine the speed of the ball; (c) determine the velocity of the ball. Solution As the total acceleration is given with unit vectors î and ĵ , and for any given values we always consider them to be measured in the laboratory frame of reference, the Cartesian system of coordinates must be set first, for example as to the right. (a) A diagram for trajectory and acceleration at the given instant of time may be separated from the defined axes y and x or may be combined with them. Possible diagrams (all are correct, but not all are complete) are: L (1) (2) (3) (4) The other possible and correct presentation is with two diagrams combined above in one (3) [diagram (3) shows the total acceleration as a vector sum of a y - and a x - components and at the same time as a vector sum of centripetal and tangential components]. Note: it is wrong to direct the total acceleration to the pivot, because the centripetal acceleration is not the total acceleration. (b) For the instantaneous centripetal acceleration: a c = a x cos( π /2 – α ) + a y cos α = a x sin α + a y cos α 29.7 m/s 2 L a v R v a c c = → = 2 Speed v = (29.7· 1.50) 1/2 = 6.67 m/s y x α a y a x L 36.9 o a y a x a L α a y a x L

All courses in PHY

Popular Documents