* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Third Midterm Exam Solutions
Old quantum theory wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Fictitious force wikipedia , lookup
Tensor operator wikipedia , lookup
Equations of motion wikipedia , lookup
Jerk (physics) wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Length contraction wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Relativistic mechanics wikipedia , lookup
N-body problem wikipedia , lookup
Photon polarization wikipedia , lookup
Accretion disk wikipedia , lookup
Rotational spectroscopy wikipedia , lookup
Centripetal force wikipedia , lookup
Moment of inertia wikipedia , lookup
Hunting oscillation wikipedia , lookup
Angular momentum wikipedia , lookup
Work (physics) wikipedia , lookup
Angular momentum operator wikipedia , lookup
Rigid body dynamics wikipedia , lookup
PHYS 2013, General Physics I, Spring 2015 Name: ID: Third Exam Problem Max. Points 1 20 2 20 3 20 4 20 5 20 6 (bonus) 20 Score Total: Comments: * You must clearly show your work in order to receive full credit on any problem. * You are allowed to use a graphing calculator on this exam. You may NOT use a smartphone or any other device with internet capabilities. * Turn in your equation sheet with your exam. * If you need more space for calculations, a blank page is included at the end of the exam. * You will lose points for numerical answers with missing or incorrect units. * You will lose points for inappropriate use of sig-figs. 1 (This page is intentionally blank) 2 Problem 1 (20 points): A bicycle wheel has an initial angular velocity of 1.10 rad/s. a) If its angular acceleration is constant and equal to 0.275 rad/s2 , what is its angular velocity at time t = 2.10 s? b) Through what angle has the wheel turned between time t = 0 s and t = 2.10 s? For rotational motion with constant acceleration ω = ω0 + αt = 1.10 rad/s + 0.275 rad/s2 (2.10 s) = 1.68 rad/s and (angular) displacement 1 1 0.275 rad/s2 (2.10 s)2 = 2.92 rad. θ = ω0 t + αt2 = (1.10 rad/s) (2.10 s) + 2 2 3 Problem 2 (20 points): Small blocks, each mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. a) Compute the moment of inertial of the system about an axis perpendicular to the rod and passing through the center of the rod. b) Compute the moment of inertial of the system about an axis perpendicular to the rod and passing through a point one-forth of the length from one end. The moment of inertia is given by I = mr12 + mr22 + mr32 For part a: I=m and for part b: I=m L 4 L 2 2 2 L +0+ 2 L + 4 2 4 2 ! 3L + 4 1 = mL2 2 2 ! = 11 mL2 16 Problem 3 (20 points): The flywheel of an engine has a moment of inertia 2.70 kg·m2 about its rotation axis. a) What constant torque is required to bring it up to an angular speed of 430 rev/min in a time of 8.50 s starting from rest. b) What is its final kinetic energy? Torque required to bring the flywheel up to speed can be determined from (rotational analog of impulse- momentum theorem) τ ∆t = Iω and so 2.70 kg·m2 ((430 rev/min) · (2πrad/rev) · (1min/60s)) Iω = = 14.3 N·m τ= ∆t (8.50 s) And the final kinetic energy is 1 1 I = Iω 2 = 2.70 kg·m2 ((430 rev/min) · (2πrad/rev) · (1min/60s))2 = 2730 J 2 2 5 Problem 4 (20 points): A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of Ii = 18 kg·m2 . She then tucks into a small ball, decreasing this moment of inertia to If = 3.6 kg·m2 . While tucked, she makes two complete revolutions in t1 = 1.1 s. If she hadn’t tucked at all, how many revolutions would she have made in the t2 = 1.8 s from board to water? The energy is not always conserved, but momentum is and so we can use conservation of angular momentum Ii ωi = If ωf to express angular velocity before tucking ωi = If ωf . Ii With final (after tucking) angular velocity she made two complete revolutions and thus ωf t1 = 4π 4π ωf = t1 Therefore her initial (before tucking) angular velocity is If ωi = Ii 4π t1 with which in t2 seconds she would have rotated by angle θ = ωi t2 = 4π If t2 Ii t1 or would make θ If t2 3.6 · 1.8 =2 =2 = 0.65 rev 2π Ii t1 18 · 1.1 revolutions. 6 Problem 5 (20 points): A uniform ladder X = 5.0 m long rests against a frictionless, vertical wall with lower end L = 3.0 m from the wall. The ladder weights Wl = 160 N. The coefficient of static friction between the foot of the ladder and the ground is µs = 0.40. A man weighting Wm = 740 N climbs slowly up the ladder. a) What is the maximum frictional force that the ground can exert on the ladder at its lower end? b) What is the actual frictional force when the man has climbed 1.0 m along the ladder? c) How far along the ladder can the man climb before the ladder starts to slip? The angle of inclination is given by 5 = 0.93 rad 3 α = arccos This is a 2D equilibrium problem which gives us two translational conditions µs N1 − N2 = 0 (1) N1 − Wl − Wm = 0 (2) (where N1 and N2 are normal forces from the ground and from the wall respectively) and one more rotational conditions which we choose to be expressed for rotations around the foot of the ladder N1 · 0 + µs N1 · 0 − Wl L cos α − Wm z cos α + N2 L sin α = 0 2 or 1 z cos α N2 = Wl + Wm . 2 L sin α Using (2) the largest fictional force can be is µs N1 = µs (Wl + Wm ) = 0.40 · 900 N = 360 N and from (1) and (3) for z = 1.0 m the frictional force is µs N1 = N2 = 1 cos α 1 Wl + Wm = 170 N. 2 5 sin α And so the ladder would start slipping when 160 N z + 740 N 0.75 = 360 N 2 5.0 m or z = 2.7 m. 7 (3) (Bonus) Problem 6 (20 points): Two round rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subject to a tensile force with magnitude 4000 N. (Young’s modulus for steel is 20 × 1010 Pa and for copper is 11 × 1010 Pa) a) For steel rod, what is the strain? Strain for the steel rod is given by 4000 N F/A = 1.13 × 10−4 = 2 −2 10 Y (20 × 10 Pa) 3.14 · (1.50 × 10 /2) b) For steel rod, what is the elongation? And thus elongation is ∆L = L F = 8.49 × 10−5 m YA c) For copper rod, what is the strain? Strain for the copper rod is given by 4000 N F/A = 2.06 × 10−4 = 2 10 −2 Y (11 × 10 Pa) 3.14 · (1.50 × 10 /2) d) For copper rod, what is the elongation? And thus elongation is F ∆L = L = 1.54 × 10−4 m YA 8