# Quiz 5

## Transcription

Quiz 5

Name________________________________________ Physics 110 Quiz #5, November 1, 2013 Please show all work, thoughts and/or reasoning in order to receive partial credit. The quiz is worth 10 points total. I affirm that I have carried out my academic endeavors with full academic honesty. ____________________________ € € 1. Two objects of equal mass and radius are shown on the right. A disk (on the right with moment of inertia I = 12 MR 2 ) and a hoop (on the left with moment of inertial I = MR 2 ) have a string wrapped around their circumferences. Hanging from the string, halfway between the disk and the hoop, is a block of mass m . The disk and the hoop are free to rotate about their centers. When the block is released from rest and allowed to fall, the block a. b. c. d. € moves to the right of the centerline. moves to the left of the centerline. moves along the centerline. moves both left and right of the centerline or it oscillates about the centerline. 2. Suppose that we have the set-up shown below consisting of a shaft of radius rshaft = 0.5cm around which a massless string is wound and a disk of moment of inertial I is placed. The string is passed over a massless pulley where a mass mh = 200g is hung and allowed to fall from rest through a height h = 1m acquiring a speed of v = 1.84m/s. a. Using energy ideas derive an expression for the moment of inertia of the disk and then evaluate your expression using the information given in the problem above. ΔE = 0 = ΔKE T + ΔKE R + ΔU g = ( 12 mh v 2f − 0) + ( 12 Iω 2f − 0) + (0 − mh gy i ) v 0 = ( 12 mh v 2f − 0) + 12 I f − 0 + (0 − mh gh ) r sh m 2gh 2 × 9.8 s2 ×1m 2 I = 2 −1 mh rsh2 = −1 (0.2kg)(0.005m) = 2.4 ×10−5 kg ⋅ m 2 2 1.84 m vf s Disk 2 ( ) rshaft mh € b. Using ideas of torque and forces derive an expression for the moment of inertia of the disk and then evaluate your expression using the information given in the start of the problem. Disk rshaft mh ∑τ : rsh FT = rsh ( mh g − mh a) = Iα = I a rsh 2gh g g → I = −1mh rsh2 = 2 −1 mh rsh2 = 2 −1 mh rsh2 = 2.4 ×10−5 kg ⋅ m 2 v a vf f 2h ∑F y € Useful formulas: : FT − mh g = −mh a → FT = mh g − mh a Motion in the r = x, y or z-directions Uniform Circular Motion v2 ar = r v2 Fr = mar = m r 2πr v= T mm FG = G 1 2 2 r rf = r0 + v 0r t + 12 ar t 2 v fr = v 0r + ar t v fr 2 = v 0r 2 + 2arΔr Vectors Circles Triangles Spheres C = 2πr A = 12 bh A = 4 πr 2 A = πr 2 V = 43 πr 3 Quadratic equation : ax 2 + bx + c = 0, whose solutions are given by : x = −b ± b 2 − 4ac 2a Useful Constants € € 2 magnitude of avector = v x + v 2 y g = 9.8 m s2 Linear Momentum/Forces → → N A = 6.02×10 → p f = p i + F Δt → → F = ma → 1 2 K R = Iω Heat 2 TC = 59 [TF − 32] 2 TF = 95 TC + 32 Lnew = Lold (1+ αΔT ) U g = mgh → Fs = −k x U S = 12 kx 2 F f = µFN WT = FdCosθ = ΔE T Anew = Aold (1+ 2αΔT ) Vnew = Vold (1+ βΔT ) : β = 3α PV = Nk B T W R = τθ = ΔE R W net = W R + WT = ΔE R + ΔE T 3 2 ΔQ kA = ΔT Δt L ΔQ PR = = εσAΔT 4 ΔT ΔU = ΔQ − ΔW PC = ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss € Fluids kB T = 12 mv 2 ΔQ = mcΔT ΔE R + ΔE T + ΔU g + ΔU S = 0 Rotational Motion kg 2 vsound = 343 m s Work/Energy KT = 2 mv → 2 kB = 1.38×10−23 J K mole σ = 5.67×10−8 W m 2 K 4 €1 p = mv → G = 6.67×10−11 Nm 23 atoms v direction of avector → φ = tan−1 y vx € Geometry /Algebra Simple Harmonic Motion/Waves ω= € k 2π = 2πf = m T TS = 2π m k TP = 2π l g 1 v =± k x 2 2 A1− m A2 x ( t ) = x max sin(ωt ) or x max cos(ωt ) v ( t ) = v max cos(ωt ) or − v max sin(ωt ) a( t ) = −amax sin(ωt ) or − amax cos(ωt ) v max = ωx max ; amax = ω 2 x max Sound v = fλ = v = fλ = (331+ 0.6T) ms v 2L I = 2π 2 f 2 ρvA 2 I ; Io = 1×10−12 mW2 I0 v v f n = nf1 = n ; f n = nf1 = n 2L 4L f n = nf1 = n β = 10log € € FT µ