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quiz2_sol - Massachusetts Institute of Technology Physics...

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Unformatted text preview: Massachusetts Institute of Technology - Physics Department Exam #2 SOLUTIONS Physics - 8.01 Fall 1999 Problem 1 42 points 6 pts a) max. extension: mg sin + smg cos = kxmax ) xmax = mkg (sin + s cos ) N=mg cos θ Fsp=kx F = µ smg cos θ fr mg max speed when a = 0 x = mg (sin + k cos ) k d) gravity: mg sin (xmax x) 2 spring: + 1 k(xmax x2 ) 2 fricion: k mg cos (xmax x) e) Substitute in the results of d) for x = 0 . Add up the total work done by the three forces. If this is zero or larger than zero, the spring will become as short as l. Thus, 12 2 kxmax mgxmax(sin + k cos ). ; ; ; ; ; 10 pts 10 pts ) 10 pts b) c) net force along slope once the block starts moving: kx ; mg sin ; k mg cos = ma 6 pts 1 Problem 2 32 points a) As the object goes up, the component of the resistive force in the vertical (y) direction 6 pts is in the same direction as the gravitational force mg (see gure). Fres,y Fres,y mg When the object is on the way down, the y component of the resistive force is in the opposite direction. Thus, the component of the force in the y direction is smaller on the way down than on the way up. Thus, it will take longer than 2 sec to come down. q q b) For zero acceleration, T = 2 gl . For an acceleration a downwards, T = 2 g l a . If a were 10 m/sec2 , we'd have free fall, and the pendulum would be weightless (no oscillations, T ! 1). Since a = 5 m=sec2 , T = T 2. C c) The viscous term will dominate when vterm � vcrit where vcrit = C21r . When the viscous m term dominates, vterm = C1gr . The mass of the oil drop is given by m = 4 r3 . Thus, 3 4 r3 g � C1 . 3C1 r C2 r 0 6 pts mg 0 6 pts )r 8 pts 6 pts p ; 2 3C1 4 gC2 !1 3 1 d) ! = 2, T = 2! , f = T = 2! = 1 Hz e) The speed is maximum when x = 0, i.e., when sin(2t + 4 ) = 0 This happens when 2t + 4 = n where n = 0 �1 �2 �3 : : : ) t = 2 (n ; 14 ) Alternatively, v = ;0:6 cos(2t + 4 ), and the max speed is when cos(2t + 4 ) = �1. This leads to the same times. 2 Problem 3 26 points v1 CM m2 v r2 r1 m1 2 acceleration must equal that of gravity. m star 1: !2 r1 = (r1 +2rG)2 2 m star 2: !2 r2 = (r1 +1rG)2 2 Add these two equations and substitute ! = 2T to nd 4T 22 (r1 + r2 ) = (mr11+m22))2G . ( +r Thus, 3 T = 21 (r1 + r2) 2 1 G 2 (m1 + m2) 2 3 10 pts 5 pts 5 pts a) m b) F = (m11+r22G2 r) m m c) a1 = (r1+2rG)2 and a2 = (r1+1rG)2 2 2 d) Let ! be the angular velocity, which must be the same for both stars. The centripetal 6 pts ...
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