Name Section THE PENNSYLVANIA STATE UNIVERSITY Department of Engineering Science and Mechanics Engineering Mechanics 12 Final Exam May 5, 2003 8:00 9:50 am (110 minutes) Problem 1 Problem 2 Problem 3 Problem 4 Total 1
Problem 1 Two identical homogeneous and uniform disks of mass m and radius R are connected at their centers by a spring of constant k and undeformed length L 0. The disks are released from rest when the distance between their centers is 3L 0. Find the speed of their centers when the disks first impact with one another. Assume that the disks roll without slipping. m, R 3L 0 k, L 0 The problem can be solved using the work-energy theorem. Let the subscript 1 denote the instant at which the disks are released. Also, let the subscript 2 denote the instant of first impact between the disks. Then the work-energy theorem can be given the form: T 1 + V 1 + (U 1 2 ) NC = T 2 + V 2, (1) where T denotes the kinetic energy of the system, V denotes the potential energy of the system, and (U 1 2 ) NC denotes the work of all the forces, whether internal or external, whose work is not accounted for via the potential energy V. As the system is released from rest, we have T 1 = 0. (2) Also, since the disks are homogeneous and the rolling occurs on a straight horizontal surface, gravity s contribution to the potential energy can be disregarded. Hence, the only contribution of the potential energy function is that of the spring s: V 1 = 1 2 k(3l 0 L 0 ) 2 = 2kL 2 0. (3) As far as the kinetic energy at 2 is concerned, recalling that when a disk is rolling without slip on a stationary surface, the point of contact is the (instantaneous) center of rotation for the disk, denoting by Q such a point, we have that ( ) 1 T 2 = 2 2 I Qω2 2, (4) where, by the parallel axis theorem I Q = I G + mr 2 = 1 2 mr2 + mr 2 = 3 2 mr2, (5) and where ω 2 denotes the angular speed of the disks at 2. Notice that, by symmetry, the kinetic energy of one disks is equal to that of the other disk. The total kinetic energy is 2
therefore equal to twice that of a single disk. This fact is reflected in the presence of the coefficient 2 right after the equal sign of Eq. (4). Now, recalling that when rolling without slipping on a stationary surface we have that v C = Rω, (6) where v C is the speed of one of the disks, we have that the total kinetic energy of the system at 2 can be given the form: T 2 = 3 2 m ( ) vc 2. (7) 2 As far as the potential energy at 2 is concerned, we have that Finally, as far as the term (U 1 2 ) NC is concerned, we have that V 2 = 1 2 k(2r L 0) 2. (8) (U 1 2 ) NC = 0, (9) because of the fact that, due to the rolling without slipping condition, the contact force between the disks and the ground does not perform any work. Now that every term in the work-energy equation has been properly characterized, we can rewrite Eq. (1) as 2kL 2 0 = 3 2 m ( ) vc 2 + 1 2 2 k(2r L 0) 2. (10) Solving the above equation with respect to (v C ) 2 we have k (v C ) 2 = 3L 2 0 + 4RL 0 4R 3m 2. (11) 3
Problem 2 A hinged bar with mass m = 10 kg and length L = 2 m is falling toward the ground. There is no friction. When θ = 30, the bar s angular velocity is ω = 2 rad/s. Find: θ (a) the angular acceleration, α, of the bar, and (b) the force exerted by the hinge on the bar. ω = 2 rad/s The moment of inertia for the bar about its center of mass is I G = ml 2 /12. The free body and mass acceleration diagrams are as shown at right, where A and G denote the locations of the hinge and center of mass, respectively. Applying F = ma G, using the usual x-y coordinates, we get: (x) A x = mr G/A α cos(θ) mr G/A ω 2 sin(θ), (1) (y) A y mg = mr G/A α sin(θ) mr G/A ω 2 cos(θ). (2) Further, applying M A = I A α (with positive rotations taken as clockwise), results in: ( ) ( ) L 1 2 sin(θ) mg = 12 ml2 + mrg/a 2 α, (3) where we ve employed the parallel axis theorem to get I A from I G. The above Eqs. (1-3) provide three equations in the three unknowns A x, A y and α. Noting that r G/A = L/2, substituting the given information into Eqs. (1) (3) gives: Equations (4) (6) can be readily solved to obtain: A x = 5 3 α 20, (4) A y 98.1 = 5 α 20 3, (5) 49.05 = (13.333) α. (6) A x = 11.859 N (7) A y = 46.065 N (8) α = 3.679 rad/s (clockwise). (9) 4
Problem 3 Car A weighs 3300 lb, and car B weighs 2500 lb. The initial velocity of car A is (v A ) 1 = 45 m.p.h. in the direction shown. The initial velocity of car B is (v B ) 2 = 35 m.p.h. in the direction shown. The cars impact plastically. They stick together and slide as shown. If the kinetic coefficient of friction between the tires and the road is µ k = 0.55, how far do the cars slide? and in what direction θ? A (v A ) 1 B A (v B ) 1 θ B This is 2D impact. The initial velocity vectors are ( ) 5280 ft/mi ( v A ) 1 = (v A ) 1 î = (45 m.p.h.) î = 66 î ft/s, 3600 s/h ( ) 5280 ft/mi ( v B ) 1 = (v B ) 1 ĵ = (35 m.p.h.) ĵ = 51.3 ĵ ft/s. 3600 s/h During impact, momentum is conserved. The initial and final momenta are Since the cars stick together after impact, so that Conserving momentum, ( p 1 = p 2 ), p 1 = m A ( v A ) 1 + m B ( v B ) 1, p 2 = m A ( v A ) 2 + m B ( v B ) 2. ( v B ) 2 = ( v A ) 2, p 2 = (m A + m B ) ( v A ) 2. ( v A ) 2 = m A ( v A ) 1 + m B ( v B ) 1 = W A ( v A ) 1 + W B ( v B ) 1 m A + m B W A + W B (3300)(66) î + (2500)(51.3) ĵ = 3300 + 2500 The direction of travel is thus ( ) 22.1 θ = tan 1 37.6 = 0.531 rad = 30.4. = 37.6 î + 22.1 ĵ ft/s. The distance the cars slide (d) is found from work energy. The initial speed is (v A ) 2 = ( v A ) 2 = (37.6) 2 + (22.1) 2 ft/s = 43.6 ft/s. 5
Setting the work done by friction to the change in kinetic energy U 2 3 = T 3 T 2, µ k (m A + m B )gd = 1 2 (m A + m B ) [ (v A ) 2 3 (v A) 2 ] 1 2 = 2 (m A + m B ) [ 0 (v A ) 2 ] 2. Solving for d, d = [ (va ) 2 2] 2µ k g = (43.6) 2 ft = 53.7 ft. (2)(0.55)(32.2) 6
Problem 4 An unpowered spacecraft of mass m performs a fly by of the planet Venus, following the trajectory AB as shown. The coordinates relative to Venus at point A are (x, y) = ( 5.0, 1.0) 10 8 ft. At point B, (x, y) = (3.960, 3.382) 10 8 ft. Find the speed of the vehicle at point B. y Venus A v A = 16, 000 ft/s B 38.41 x The gravitational force acting on the spacecraft is centrally-directed: that is, it always points toward the origin O of the coordinate system. Thus, the angular momentum about O is conserved as the spacecraft moves from point A to point B: (h O ) A = (h O ) B. (1) Using the given information, we have (taking counterclockwise as positive): (h O ) A = (1 10 8 )(m)(16, 000) (2) (h O ) B = (3.960 10 8 )(m)v B cos(38.41 ) (3.382 10 8 )(m)v B sin(38.41 ), (3) which, upon equating the expressions for (h O ) A and (h O ) A gives 16, 000 = v B ((3.960) cos(38.41 ) (3.382) sin(38.41 )). (4) This is easily solved to give v B = 15, 971 ft/s. 7