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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