Bryant Grigsby (Physics BSc) Vice President of Operations and New Product Introduction Lumenetix Scotts Valley, CA
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1 PHYSICIST PROFILE Bryant Grigsby (Physics BSc) Vice President of Operations and New Product Introduction Lumenetix Scotts Valley, CA Bryant first considered a business major but found it lacking in technical challenges. He switched to engineering, but having taken a physics intro class he was drawn to physics because it provided a little bit of understanding of everything. I expected that getting a job as an engineer or scientist would be easy, but it wasn t. Perhaps having an advanced degree or minoring in EE or ME would have made it easier. Bryant and partner in an Olympic canoe team trial Physics gave Bryant the broad background necessary to lead a multidisciplinary science and engineering team dealing with electronic design, optics, thermal and stress analysis, statistics, and software. When you lead a large engineering team you are asked to make decisions that will greatly affect what you are designing, often in an area that you have never worked in. A BS in physics gives you a broad enough understanding in everything such that, if you have any decent problem solving skills, you can interpolate and come to a sound decision.
2 Class November 2017 Review for Exam 4 Gravitational Potential Energy Final Exam: Saturday, December 16, 2017, 6:30-9:30 pm Make-Up Exam: Monday, December 18, 8:00-11:00 am (Must request makeup by 4:00 pm, November 29.
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4 2015 Midterm Exam 4 MC 1 An aircraft performs a maneuver called an aileron roll. During this maneuver, the plane turns like a screw as it maintains a straight flight path, which sets the wings in circular motion. If it takes it 35.0 s to complete the circle and the wingspan of the plane is 11.0 m, what is the acceleration of the wing tip? [ Wingspan is the distance from the end of one wing to the end of the other wing.] (a) 0.18 m/s 2 (b) 5.6 m/s 2 (c) 1.0 m/s 2 (d) 0.57 m/s 2 (e) 2.3 m/s 2 4
5 2015 Midterm Exam 4 MC 1 An aircraft performs a maneuver called an aileron roll. During this maneuver, the plane turns like a screw as it maintains a straight flight path, which sets the wings in circular motion. If it takes it 35.0 s to complete the circle and the wingspan of the plane is 11.0 m, what is the acceleration of the wing tip? [ Wingspan is the distance from the end of one wing to the end of the other wing.] (a) 0.18 m/s 2 a =! 2 r! = 2 T (b) 5.6 m/s 2 (c) 1.0 m/s 2 (d) 0.57 m/s 2 (e) 2.3 m/s 2 a = 4 2 d T 2 2 =0.18 m/s2 5
6 2015 Midterm Exam 4 MC 2 Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 8.5 times that of particle B. The period of particle B is 2.0 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to (a) r A /r B = 2.1 (b) r A /r B = 0.24 (c) r A /r B = 17 (d) r A /r B = 21 (e) r A /r B = 4.3 6
7 2015 Midterm Exam 4 MC 2 Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 8.5 times that of particle B. The period of particle B is 2.0 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to (a) r A /r B = 2.1 (b) r A /r B = 0.24 (c) r A /r B = 17 (d) r A /r B = 21 (e) r A /r B = 4.3 a = r r = T 2 a T 4 2 r A = (1)2 (8.5) r B (2) 2 =2.1 (1) 7
8 2015 Midterm Exam 4 MC 3 A torque of 12 N-m is applied to a solid, uniform disk of radius 0.50 m, causing the disk to accelerate about an axis perpendicular to the face of the disk and through its center of mass at 5.7 rad/s 2. What is the mass of the disk? (a) 13 kg (b) 4.3 kg (c) 8.5 kg (d) 17 kg (e) 5.7 kg 8
9 2015 Midterm Exam 4 MC 3 A torque of 12 N-m is applied to a solid, uniform disk of radius 0.50 m, causing the disk to accelerate about an axis perpendicular to the face of the disk and through its center of mass at 5.7 rad/s 2. What is the mass of the disk? (a) 13 kg (b) 4.3 kg (c) 8.5 kg (d) 17 kg (e) 5.7 kg m = 2 r 2 = (2)(12) (0.5) 2 (5.7) = I = 1 2 mr2 = 16.8 kg 9
10 2015 Midterm Exam 4 MC 4 A 23 kg mass is connected to a nail on a frictionless table by a massless string 1.3 m long. There is no appreciable friction between the nail and the string. If the tension in the string is 51 N while the mass moves in a uniform circle on the table, how long does it take for the mass to make one complete revolution? (a) 5.2 s (b) 4.5 s (c) 4.8 s (d) 3.8 s (e) 2.1 s 10
11 2015 Midterm Exam 4 MC 4 A 23 kg mass is connected to a nail on a frictionless table by a massless string 1.3 m long. There is no appreciable friction between the nail and the string. If the tension in the string is 51 N while the mass moves in a uniform circle on the table, how long does it take for the mass to make one complete revolution? r (a) 5.2 s F = ma = 4 2 mr 4 2 mr (b) 4.5 s T 2 T = F r (c) 4.8 s 4 2 (23)(1.3) T = =4.8 s (d) 3.8 s 51 (e) 2.1 s 11
12 2015 Midterm Exam 4 MC 5 A metal bar is hanging from a hook in the ceiling when it is suddenly struck by a ball that is moving horizontally (see figure). The ball is covered with glue, so it sticks to the bar. During this collision (a) both the angular momentum of the system (ball and bar) and its kinetic energy are conserved. (b) the angular momentum of the system (ball and bar) is not conserved because the hook exerts a force on the bar. (c) the angular momentum of the system (ball and bar) is conserved about the hook because neither the hook nor gravity exerts any torque on this system about the hook. (d) both the linear momentum and the angular momentum of the system (ball and bar) are conserved. (e) the angular momentum of the system (ball and bar) is conserved about the hook because only gravity is acting on the system. 12
13 2015 Midterm Exam 4 MC 5 A metal bar is hanging from a hook in the ceiling when it is suddenly struck by a ball that is moving horizontally (see figure). The ball is covered with glue, so it sticks to the bar. During this collision (a) both the angular momentum of the system (ball and bar) and its kinetic energy are conserved. (b) the angular momentum of the system (ball and bar) is not conserved because the hook exerts a force on the bar. (c) the angular momentum of the system (ball and bar) is conserved about the hook because neither the hook nor gravity exerts any torque on this system about the hook. (d) both the linear momentum and the angular momentum of the system (ball and bar) are conserved. (e) the angular momentum of the system (ball and bar) is conserved about the hook because only gravity is acting on the system. 13
14 2015 Midterm Exam 4 Problem m long, 10.0 kg ladder, 45 above horizontal 80.0 kg you, 2.0 m from the top Smooth wall; µ with ground = N rope at base of ladder Force diagram, equations, friction force with no rope, does rope break and fall 14
15 2015 Midterm Exam 4 Problem m long, 10.0 kg ladder, 45 above horizontal 80.0 kg you, 2.0 m from the top Smooth wall; µ with ground = N rope at base of ladder Force diagram, equations, friction force with no rope, does rope break and fall F W F f F R =0 F G m 1 g m 2 g =0 F W L sin 135 m 1 g(l 2) sin 135 m 2 g(l/2) sin 135 =0 15
16 2015 Midterm Exam 4 Problem 1 F G = 883 N F W F f F R =0 F G m 1 g m 2 g =0 F W L sin 135 m 1 g(l 2) sin 135 m 2 g(l/2) sin 135 =0 F F = 441 N 7.07F W = 0 F W = 677 N NO Rope : F f = F W µ = F W (m 1 + m 2 )g =0.77 Rope : F R = F W F f = 677 µ(m 1 + m 2 )g = 236 N 0 Rope Doesn 0 t Break; Saved 16
17 2015 Midterm Exam 4 Problem 2 Wheel 3.0 kg, 66.0 cm diameter, mass in rim 25.0 N horizontal force on axle w i = 2.0 radians/s, w f = 5.0 radians/s Moment of inertia, torque, angular acceleration, travel 17
18 2015 Midterm Exam 4 Problem 2 Wheel 3.0 kg, 66.0 cm diameter, mass in rim 25.0 N horizontal force on axle w i = 2.0 radians/s, w f = 5.0 radians/s Moment of inertia, torque, angular acceleration, travel I = mr 2 = (3)(0.33) 2 =0.327 kg m 2 ma = F F f I = = F f r a = r Fr =2I =2 m r = F I r = Fr 2 =4.125 N m = I = 12.6 radians/s2 18
19 2015 Midterm Exam 4 Problem 2 Wheel 3.0 kg, 66.0 cm diameter, mass in rim 25.0 N force on axle w i = 2.0 radians/s, w f = 5.0 radians/s Moment of inertia, torque, angular acceleration, travel! 2 f! 2 i =2 =0.833 radians x = r =0.275 m 19
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