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ENG102_Solution_LT

# ENG102_Solution_LT - DATE Solution to LateTerm exercise...

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Unformatted text preview: DATE: November 30, 2011 Solution to LateTerm exercise ENG102 Fall 2011 Closed Book / one single-page sheet of personal notes allowed. No cell phones or WiFi devices present. Do not leave the room without permission from the proctor. Address the proctor if you suspect the problem formulation is erroneous, but do not ask the proctor to interpret either the question or your answer. The exercise will be evaluated as a whole, but the numbered questions are intended to count equally. The exercise starts at 2:10pm and ends at 4:00pm. Write clearly your complete name, page number, and number of pages on each sheet of paper you turn in. M R y x O a m v 1. Look at the ﬁgure above. A system consisting of a thin ring with radius R and mass M is positioned on a smooth ﬂat surface. It is hit by a small particle with mass m and initial velocity v as illustrated on the ﬁgure. The particle becomes embedded in the ring after the collision. Answers should be provided using given parameters only. a. Identify the location of the center of mass RCM as a distance from the center of the ring at the time of the collision. R0 can now be considered a given parameter. The center of mass is located RCM = Mmm R from O in the direction of the + contact point. b. Determine the velocity of the center of mass of the system before and after the collision, and determine the momentum of the system before and after the collision. VCM = Mm m v . p = mv . + c. Determine the moment of inertia ICM of the rigid object around the center of mass after the collision, and determine the angular momentum HCM around the center of mass before and after the collision. The composite moment of inertia around CM is: 2 ICM = M (R2 + RCM ) + m(R − RCM )2 a HCM = mv (R − RCM ) R D EPARTMENT OF M ECHANICAL AND A ERONAUTICAL E NGINEERING | U NIVERSITY OF C ALIFORNIA | D AVIS , C ALIFORNIA 95616 T EL : 530.752.5335 | FAX : 530.752.4158 (1) (2) Solution to LateTerm exercise ENG102 Fall 2011 – November 30 Page 2 d. Determine the angular velocity ωCM of the rigid object around the center of mass after the collision. HCM = ICM ωCM a mv (R − RCM ) R ⇒ ωCM = 2 M (R2 + RCM ) + m(R − RCM )2 (3) (4) s=0 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 s 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 k 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 M 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 2. Look at the ﬁgure above. A system consisting of a mass M , a linear spring with spring constant k . We deﬁne the position of the right edge of the mass to be zero when the spring is relaxed. The mass can move on the surface with no friction. a. Find the equation of motion for the position s of the mass, and ﬁnd the frequency ω of oscillation. M s = −ks ¨ (5) k M (6) ω= b. Find the complete solution to the equation of motion, and determine the speciﬁc solution for the initial conditions s(0) = 0 and s(0) = v0 . ˙ s(t) = C1 cos ωt + C2 sin ωt v0 C1 = 0 , C2 = ω v0 sin ωt s(t) = ω 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 k 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 M 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 s 11111111 00000000 11111111 00000000 11111111 00000000 (7) (8) (9) We now consider a non-zero kinetic friction coefﬁcient µk describing the dissipation between the mass and the surface. The gravitational acceleration is g . c. Write the equations of motion for the coordinate s of M when the mass moves to the right and to the left. s=0 s > 0 : M s = −ks − µk Mg ˙ ¨ s < 0 : M s = −ks + µk Mg ˙ ¨ (10) (11) ∆s Now look at the ﬁgure to the left. We have rotated the system 90◦ clockwise such that the mass is now directly affected by gravity. There is no friction with the side wall. 2 Solution to LateTerm exercise ENG102 Fall 2011 – November 30 Page 3 d. Find the static extension ∆s of the new resting position. k ∆s = Mg → ∆s = M g . k e. Write the equation of motion for the coordinate s of the mass. Find the frequency of oscillation. M s = −ks + Mg = −k (s − ¨ d 2 (s − M g ) Mg k M = −k (s − ) 2 dt k k ω= M Mg ) k (12) (13) (14) f. Find the complete solution to the equation of motion, and determine the speciﬁc solution for the initial conditions s(0) = ∆s and s(0) = v0 . ˙ Mg + C1 cos ωt + C2 sin ωt k v0 C1 = 0 , C2 = ω Mg v0 s(t) = + sin ωt k ω s(t) = Niels Grønbech-Jensen 3 (15) (16) (17) ...
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