Last revised 1999/11/02 |
#73.
tau = (2/5) m R2 (omega f / t f )tau 1 = 0.69 N m
tau 2 = 9.84 N m
(b)
F = (2/5) m R omega f / t fF1 = 3.05 newton
F2 = 11.5 newton
#76. [My answer agrees with an older edition of the text]:
(b) a = 2 R theta / t2 [same for each block]
(c)
T1 = M (g - 2 R theta / t2 )
T2 = M g - (2 theta / t2 ) (M R + I / R )
(b) 1.1 x 104 N m
(c) 1.3 x 106 Joule
#87 [My answers agree with text]
(b) (3/2) g sin (theta)
(c) theta = 41.8o = 0.73 radians
#15. Text's answers:
(b) (50/7) m g
#33 Proof.
#41 [The text gives absolute magnitudes of dL/dt and torque]
(b) dLz/dt = m g t v0 cos (theta0)
(c) torquez = m g t v0 cos (theta0)
#69
cos (theta) = 1 - 6 h m2/ [ lp (2 m + M) (3 m + M ) ]
I have used lp for the length of the pendulum to improve the clarity of the printing. This expression has the correct limits for m = 0, M = 0, and h = 0. Note that g has dropped out because there is no way to cancel out its factor of sec -2 .