PHYSICS 218 Exam 3 Fall, 2013

Size: px

Start display at page:

Download "PHYSICS 218 Exam 3 Fall, 2013"

Michael Doyle
5 years ago
Views:

1 PHYSICS 218 Exam 3 Fall, 2013 Wednesday, November 20, 2013 Please read the information on the cover page BUT DO NOT OPEN the exam until instructed to do so! Name: Signature: Student ID: Section Number: Rules of the exam: You have 75 minutes to complete the exam. Formulae are provided on the last page of the exam packet. You may NOT use any other formula sheet. You might use any type of handheld calculator. Be sure to put a box around your final answers and clearly indicate your work to your grader. Partial credit can be given only if your work is clearly explained and labeled. No credit will be given if we can t figure out which answer you are choosing, or which answer you want us to consider. If the answer marked does not obviously follow from the shown work, even if the answer is correct, you will not get credit for the answer. Multiple choice questions do not need to show any work to get credit. Have your TAMU ID ready when submitting your exam to the proctor. Please check that no pages are missing in your copy of the exam: pages are numbered and there should be 10 pages in total. Put your initials here after reading the above instructions: 1

2 Short Problems (20) Problem 2 (20) Problem 3 (20) Problem 4 (20) Problem 5 (20) Total (100) 2

3 Problem 1 (20 points) Problem 1.1.: (5 points) A star of mass M and radius R is spinning with an angular velocity ω. Assume the mass is uniformly distributed. The radius suddenly shrinks to R/4 and no mass is lost. Assuming that the star can be treated as a solid body before and after the compression, the new angular velocity is: A. 16ω B. 4ω C. 9ω D. 2ω E. none of the above Problem 1.2.: (5 points) 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. Assume that the collision happens instantaneously. During this collision (i.e. up until the instance when the bar has started moving) A. The angular momentum of the system (ball and bar) about the hook is conserved because only gravity is acting on the system. B. The angular momentum of the system (ball and bar) about the hook 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 angular momentum of the system (ball and bar) and its kinetic energy are conserved. E. Both the linear momentum and the angular momentum of the system (ball and bar) are conserved. 3

4 Problem 1.3.: (5 points) A light triangular plate OAB is in a horizontal plane and is pivoted about an axis perpendicular to the page through point O. Three forces F1= 7 N, F2 = 3 N and F3 = 8 N act on the plate as shown in the figure. Consider counterclockwise torques as positive. The sum of the torques about this axis is closest to: A N*m B N*m C. zero D N*m E. 0.3 N*m Problem 1.4.: (5 points) Three identical balls A, B, and C are each attached to a very light string, and each string is wrapped around the rim of one of three frictionless pulleys, each of mass M and radius R attached to the ceiling. The only difference is that the pulley for ball A is a thin solid disk, the pulley for ball B is a ring (a hollow disk), and the pulley for ball C is a solid cylinder. If all balls are released from rest and fall the same distance, how will the kinetic energy of the balls compare to each other? A. KE(A) > KE(B) >KE(C) B. KE(A) = KE(C) >KE(B) C. KE(B) > KE(A)=KE(C) D. KE(B) > KE(A) >KE(C) E. KE(A) = KE(B) >KE(C) 4

5 Problem 2 (20 points): A target in a shooting gallery consists of a vertical square wooden board, m on a side and with mass kg, that pivots on a horizontal axis along its top edge. The board is struck faceon at its center by a bullet with mass 1.50g that is traveling at 300 m/s and that remains embedded in the board. A) (8 points) What is the angular speed of the board just after the bullet's impact? B) (7 points) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? C) (5 points) What minimum bullet speed would be required for the board to swing all the way over after impact? 5

Problem 3 (20 points): A large roll of paper of mass m and radius R rests on the rough surface and is held by a massless cable attached to a rod through the center of the roll.

6 Problem 3 (20 points): A large roll of paper of mass m and radius R rests on the rough surface and is held by a massless cable attached to a rod through the center of the roll. The horizontal rod (oriented into the page) turns without friction in a bracket at the end of the cable, and the moment of inertia of the paper and rod about the axis is I. The other end of the cable is attached by a frictionless hinge to the floor such that the cable makes an angle with the floor. The weight of the cable/bracket system is negligible. The coefficient of kinetic friction between the paper and the surface is. A constant horizontal force F is applied to the paper as shown in the Figure, and the paper unrolls. A) (5 points) Draw a free body diagram for the roll of paper. B) (8 points) In terms of the quantities given, what is the magnitude of the force that the cable exerts on the paper as it unrolls? C) (7 points) Again, in terms of the quantities given, what is the magnitude of the angular acceleration of the roll? 6

7 Problem 4 (20 points): A uniform horizontal beam of length L is attached by a frictionless pivot to a wall. A cable making an angle of with the horizontal is attached to the beam a distance D from the pivot point. The beam has a mass M. The breaking point of the cable is Tmax. A man of mass m walks out along the beam. A) (5 points) Draw a free-body diagram for the beam with the man walking on it as described. B) (15 points) In terms of the quantities given, find the maximum distance that the man can walk out on the beam before the cable will break. 7

8 Problem 5 (20 points): Your starship lands on a mysterious Planet X. As chief engineer, you make the following measurements: A 2.50 kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 6.00 s; the radius of Planet X is 3.2 X 10 4 km; and there is no appreciable atmosphere on the planet. From this information find the following: A) (10 points) What is the mass of Planet X? B) (10 points) If your starship goes into a circular orbit 30,000 km from the center of the planet, how many hours will it take the ship to make one complete orbit? 8

9 Phys 218 Exam III Formulae General math: h φ θ h a h o A = A x î + A y ĵ + A zˆk h a = h cos θ = h sin φ h o = h sin θ = h cos φ h 2 = h 2 a + h 2 o tan θ = h o h a ax 2 + bx + c = 0 x = b ± b 2 4ac 2a A B = A x B x + A y B y + A z B z = AB cos θ = A B = AB A B = (A y B z A z B y )î + (A z B x A x B z )ĵ + (A x B y A y B x )ˆk = AB sinθ = A B = AB df/dt = nat n 1 If f(t) = at n t2, then t 1 f(t)dt = a n+1 (tn+1 2 t n+1 1 ) (for n 1) f(t)dt = a n+1 tn+1 + C (for n 1) translational Forces, Energy and Momenta: translational K trans = 1 2 Mv2 cm W = F const d r F r force P = dw dt = F v p cm = m 1 v 1 + m 2 v = M v cm J = Fdt = p Fext = M a cm = d p cm dt Fint = 0 rotational τ = r F and τ = F r K rot = 1 2 I totω 2 W = τdθ const torque τ θ P = dw dt = τ ω L = I 1 ω 1 + I 2 ω = I tot ω = r p τext = I tot α = d L dt τint = 0 Both translational and rotational W = K = K trans,f + K rot,f K trans,i K rot,i E tot,f = E tot,i + W other Equations of motion: rotational constant (linear/angular) acceleration only r(t) = r + v t at2 v(t) = v + at v 2 x = v 2 x,0 + 2a x (x x 0 ) (and similarly for y and z) r(t) = r ( v i + v f )t v = r2 r1 t 2 t 1 a = v2 v1 t 2 t 1 θ(t) = θ + ω t αt2 ω(t) = ω + αt ω 2 f = ω 2 + 2α(θ θ ) θ(t) = θ (ω i + ω f )t always true v = d r dt a= d v dt = d2 r dt 2 r(t) = r + t 0 v(t )dt v(t) = v + t 0 a(t )dt ω = θ2 θ1 t 2 t 1 α = ω2 ω1 t 2 t 1 ω = dθ dt α= dω dt = d2 θ dt 2 θ(t) = θ + t 0 ω(t)dt ω(t) = ω + t 0 α(t)dt K f + U f = K i + U i + W other U = F d r ; Ugrav = Mgy cm ; U elas = 1 2 k(r r equil) 2 [ F x (x) = du(x)/dx F = U = U x î + U y ĵ + U ˆk ] z Circular motion: Relative velocity: Constants/Conversions: g = 9.80 m/s 2 = ft/s 2 (on Earth s surface) G = N m 2 /kg 2 R = m M = kg R = m M = kg 1 km = mi 1 mi = km 1 ft = m 1 m = ft 1 hr = 3600 s 1 s = hr 1 kg m s = 1 N = lb 2 1 lb = N 1 J = 1 N m 1 W = 1 J/s 1 rev = 360 = 2π radians 1 hp = W Forces: Newton s: F = m a, FB on A = F A on B Hooke s: Felas = k(r r equil )ˆr Centre-of-mass: r cm = m 1 r 1 + m 2 r m n r n m 1 + m m n Gravity: friction: a rad = v2 R v A/C = v A/B + v B/C v A/B = v B/A f s µ s n, f k = µ k n (and similarly for v and a) T = 2πR v s = Rθ v tan = Rω a tan = Rα F grav = G M 1M 2 R12 2 ˆr U grav = G M 1M 2 R 12 T = 2πa3/2 GM

10 For a point-like particle of mass M a distance R from the axis of rotation: I = MR 2 Parallel axis theorem: I p = I cm + Md 2

PHYSICS 218 Exam 3 Spring, 2014

PHYSICS 218 Exam 3 Spring, 2014 Wednesday, April 16, 2014 Please read the information on the cover page BUT DO NOT OPEN the exam until instructed to do so! Name: Signature: Student ID: E-mail: Section