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77777 77777 Instructor(s): Profs. Acosta, Hamlin, Muttalib PHYSICS DEPARTMENT Exam 2 PHY 2048 Name (print, last first): March 24, 2017 Signature: On my honor, I have neither given nor received unauthorized aid on this examination. YOUR TEST NUMBER IS THE 5-DIGIT NUMBER AT THE TOP OF EACH PAGE. (1) Code your test number on your answer sheet (use lines 76–80 on the answer sheet for the 5-digit number). Code your name on your answer sheet. DARKEN CIRCLES COMPLETELY. Code your UFID number on your answer sheet. (2) Print your name on this sheet and sign it also. (3) Do all scratch work anywhere on this exam that you like. Circle your answers on the test form. At the end of the test, this exam printout is to be turned in. No credit will be given without both answer sheet and printout. (4) Blacken the circle of your intended answer completely, using a #2 pencil or blue or black ink. Do not make any stray marks or some answers may be counted as incorrect. (5) The answers are rounded off. Choose the closest to exact. There is no penalty for guessing. If you believe that no listed answer is correct, leave the form blank. (6) Hand in the answer sheet separately. Axis R Axis Axis Annular cylinder (or ring) about central axis Hoop about central axis R1 Solid cylinder (or disk) about central axis R2 L R 1 1 I = 2 M(R 12 + R 22 ) I = MR 2 Axis Solid cylinder (or disk) about central diameter Axis L L I = 2 MR2 Thin rod about axis through center perpendicular to length Axis Solid sphere about any diameter 2R R 1 1 Axis 2 1 I = 4 MR2 + 12 ML2 I = 12 ML2 Thin spherical shell about any diameter Axis R I = 5 MR2 Hoop about any diameter Axis Slab about perpendicular axis through center 2R b a 2 I = 3 MR2 1 I = 2 MR2 1 I = 12 M(a2 + b2) 77777 77777 Formula-sheet: Exam 1 • For constant acceleration ~a: ~v = ~v0 + ~at 2 vx2 = vx0 + 2ax (x − x0 ) 1 ~r = ~r0 + ~v0 t + ~at2 2 2 vy2 = vy0 + 2ay (y − y0 ) Acceleration due to gravity: g = 9.8 m/s2 = 32 ft/s2 vertically down. • For force acting on a body of mass m: F~ = m~a • Frictional forces: fs,max = µs N ; fk = µk N. N : normal force. • For uniform circular motion: centripetal acceleration is ac = • Kinetic energy: K = v2 r 1 mv 2 2 • Work done by a constant force: W = F~ · d~ = F d cos (angle between F~ and d~ ) • Work-kinetic energy theorem: W = Kf − Ki ~ = îAx + ĵAy ; • Vectors (2d): A A= q A2x + A2y ; ~ and î); Ax = A cos (angle between A ~ and î) Ay = A sin (angle between A Formula-sheet: Exam 2 • Potential Energy: ∆U = −W = − Z ~ r2 F~ · d~r Fx = − ~ r1 Gravitational (y=up): Fy = −mg dU dx U (y) = mgy Elastic (x from spring equilibrium): Fx = −kx U (x) = 1 2 kx 2 • Mechanical Energy: Emec = K + U Emec = constant for an isolated system with conservative forces • Work-Energy: W (external) = ∆K + ∆U + ∆E(thermal) • Center of Mass: ~rcom = N 1 X mi~ri Mtot i=1 • Linear Momentum: p~ = m~v d~ p F~ = dt P~tot Mtot = N X mi i=1 Impulse: J~ = ∆~ p= Z tf F~ (t)dt = F~av ∆t ti d~ p If F~ = = 0 then p~ = constant and p~f = p~i dt N X dP~tot = Mtot ~acom F~net = pi ~ = Mtot ~vcom = dt i=1 77777 77777 • Elastic Collisions of Two Bodies, 1D v1f = m1 − m2 2m2 v1i + v2i m1 + m2 m1 + m2 • Rockets: Thrust = M a = vrel 2m1 m2 − m1 v1i + v2i m1 + m2 m1 + m2 Mi ∆v = vrel ln Mf v2f = dM dt • Rotational Variables angular position: θ(t) angular velocity: ω(t) = dθ(t) dt d2 θ(t) dω(t) = dt dt2 • For constant angular acceleration α: angular acceleration: α(t) = ω 2 = ω02 + 2α(θ − θ0 ) 1 θ = θ0 + (ω + ω0 )t 2 ω = ω0 + αt 1 θ = θ0 + ω0 t + αt2 2 • Angular to linear relationships for circular motion arc length: s = rθ velocity: v = rω tangential acceleration: aT = rα centripetal acceleration: ac = rω 2 • Rotational Inertia: I = N X mi ri2 (discrete) i=1 R I = r2 dm (uniform) Parallel Axis: I = Icom + Mtot d2 (d is displacement from com) 1 2 1 1 2 Iω Kroll = M vcom + Icom ω 2 2 2 2 = rω acom = rα • Rotational, Rolling Kinetic Energy: Krot = Rolling without slipping: xcom = rθ vcom • Torque: ~τ = ~r × F~ (where k̂ = î × ĵ gives directions for cross product) τ = rF sin (angle between ~r and F~ ) = rF⊥ ~ dL dt L = rp sin (angle between ~r and ~ p) = rp⊥ ~ dL ~ = constant and L ~f = L ~i = 0 then L If ~τnet = dt ~ = ~r × ~ • Angular Momentum: L p ~τ = • Work done by a constant torque: W = τ ∆θ = ∆Krot = 1 2 1 2 Iω − Iω 2 2 2 1 • Power done by a constant torque: P = τ ω • For torque acting on a body with rotational inertia I: ~τ = I~ α • Stress and Strain (Y = Young’s modulus, B = bulk modulus) F ∆L Linear: =Y A L F ∆V Volume: P = = −B A V • For torque acting on a body with rotational inertia I: ~τ = I~ α